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Applications of Integration

2.3 Centre of Mass and Torque

32 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1

Using symmetry, find the centroid of the finite region between the curves y=(x−1)2y=(x-1)^2 and y=−x2+2x+1y=-x^2+2x+1.

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

Hint

It might help to know that −x2+2x+1=2−(x−1)2-x^2+2x+1 = 2-(x-1)^2.

Answer

(1,1)(1,1)

Full solution

Note −x2+2x+1=2−(x−1)2-x^2+2x+1 = 2-(x-1)^2. So, both parabolas are symmetric about the line x=1x=1, and the xx-coordinate of the centroid is x=1x=1.

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

The parabolas meet when:

(x−1)2=2−(x−1)22(x−1)2=2∣x−1∣=1x=0,x=2\begin{align*} (x-1)^2&=2-(x-1)^2\\ 2(x-1)^2&=2\\ |x-1|&=1\\ x=0,\quad x&=2 \end{align*}

At both these points, y=1y=1, so we see the figure is symmetric about the line y=1y=1. Then the yy-coordinate of the centroid is 11.

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

Therefore, the centroid is at (1,1)(1,1).

Q2Stage 1

Using symmetry, find the centroid of the region inside the unit circle, centred on the origin, and outside a rectangle, also centred on the origin, with width 1 and height 0.5.

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

Hint

The centroid of a region doesn't have to be a point in the region.

Answer

(0,0)(0,0)

Full solution

The circle and the cut-out rectangle are symmetric about the xx-axis, and about the yy-axis, so the centroid is the origin.

Remark: the centroid of a region doesn't have to be a point in the region!

Q3Stage 1

A long, straight, thin rod has a number of weights attached along it. True or false: if it balances at position xx, then the mass to the right of xx is the same as the mass to the left of xx.

Hint

Read over the very beginning of Section 2.3 in the CLP-2 text, specifically Equation 2.3.1.

Answer

In general, false.

Full solution

In general, this is false: weights farther out from the centre “count more" when we calculate the centre of mass. For instance, a rod with a 1-kg weight at x=−10x=-10 and a 10-kg weight at x=1x=1 will balance at x=0x=0. There's far more mass to one side of x=0x=0 than the other.

Q4Stage 1

A straight rod with negligible mass has the following weights attached to it:

  • A weight of mass 1 kg, 1m from the left end,

  • a weight of mass 2 kg, 3m from the left end,

  • a weight of mass 2 kg, 4m from the left end, and

  • a weight of mass 1 kg, 6m from the left end.

Where is the centre of mass of the weighted rod?

Hint

Use Equation 2.3.1 in the CLP-2 text.

Answer

3.53.5 metres from the left end

Full solution

Following Equation 2.3.1 in the CLP-2 text, the centre of mass of the rod is at:

xˉ=∑(mass)×(position)∑(mass)=1×1+2×3+2×4+1×61+2+2+1=216=72\begin{align*} \bar x &=\frac{\sum \text{(mass)}\times\text{(position)}}{\sum \text {(mass)}} = \frac{1\times 1 + 2\times 3 + 2\times 4 + 1 \times 6}{1+2+2+1}=\frac{21}{6}=\frac{7}{2} \end{align*}

That is, the centre of mass is 3.53.5 metres from the left end.

Q5Stage 1

For each picture below, determine whether the centre of mass is to the left of, to the right of, or along the line x=ax=a, or whether there is not enough information to tell. The shading of a region indicates density: darker shading corresponds to a denser area. In part (d), the right hand side of the right hand B×AB\times A rectangle has x=2ax=2a.

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 10

Figure from prob_s2.3, line 10

Figure from prob_s2.3, line 17

Figure from prob_s2.3, line 17

Figure from prob_s2.3, line 23

Figure from prob_s2.3, line 23

Figure from prob_s2.3, line 38

Figure from prob_s2.3, line 38

Hint

Imagine cutting out the shape and setting it on top of a pencil, so that the pencil lines up with the vertical line x=ax=a. Will the figure balance, or fall to one side? Which side?

Answer

(a) to the left (b) to the left (c) not enough information
(d) along the line x=ax=a (e) to the right

Full solution

(a) If we were to set this figure on a pencil lined up along the vertical line x=ax=a, it seems pretty clear that it would fall to the left. So, the centre of mass is to the left of the line x=ax=a. The same is true in (b): the added density on the left makes it only more lopsided. However, in (c), the right side is denser than the left, which could counterbalance the left. Without knowing more about the dimensions and the density, we can't say where the centre of mass is in relation to the line x=ax=a.

(d) Consider a section of the figure, consisting of all points (x,y)(x,y) in the figure with b≤x≤cb\le x\le c, and its “mirror" section on the other side of the line x=ax=a. These two sections, which are drawn in red in the sketches below, will have the same area, at the same distance from x=ax=a. Since we only care about the xx-coordinate of the centre of mass, it doesn't matter that the two halves are at different yy-coordinates. The centre of mass falls along the line x=ax=a.

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 26

Figure from prob_s2.3, line 26

(e) There is the same amount of area to the left and right of the line x=ax=a, as in part (d). However, the area to the right is “stretched out" more, so that it occupies space farther away from the line x=ax=a. So, the centre of mass will be to the right of the line x=ax=a.

Q6Stage 1

Tank AA is spherical, of radius 1 metre, and filled completely with water. The bottom of tank AA is three metres above the ground, where Tank BB sits. Tank BB is tall and rectangular, with base dimensions 2 metres by 1 metre, and empty. Calculate the work done by gravity to drain all the water from Tank AA to Tank BB by modelling the situation as a point mass, of the same mass as the water, being moved from the height of the centre of mass of AA to the height of the centre of mass of the water after it has been moved to BB.

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

You may use 10001000 kg/m3^3 for the density of water, and g=9.8g=9.8 m/sec2^2 for the acceleration due to gravity.

Hint

You can find the heights of the centres of mass using symmetry.

Answer

39200π9(12−π)≈121,212 J\displaystyle\frac{39200\pi}{9}(12-\pi)\approx 121,212~\text{J}

Full solution

  • The volume of water in Tank A is 43π(1)3=43π\frac{4}{3}\pi(1)^3 = \frac{4}{3}\pi cubic metres.

  • The mass of water is 40003π\frac{4000}{3}\pi kg.

  • By symmetry, the centre of mass of the water when it fills Tank AA is exactly in the centre of the sphere, at height yˉ1=4\bar y_1=4 metres above the ground (one metre above the bottom of Tank AA, which is three metres above the ground).

  • When the water is entirely in Tank BB, its height is 23π\frac{2}{3}\pi metres. (The base of Tank BB has area 2 m2^2, and the volume of water is 43π\frac{4}{3}\pi m3^3.) By symmetry, the centre of mass is exactly halfway up, at height yˉ2=13π\bar y_2=\frac{1}{3}\pi metres.

  • So, the point mass in our model is moved from yˉ1=4\bar y_1=4 to yˉ2=13π\bar y_2=\frac{1}{3}\pi, a distance of 4−13π4-\frac{1}{3}\pi metres, by gravity.

  • The work involved is:

    (40003π kg)×(4−13π m)×(9.8 msec2)=39200π9(12−π)≈121,212 J\left(\frac{4000}{3}\pi~\text{kg}\right)\times\left(4-\frac{1}{3}\pi~\text{m}\right)\times\left(9.8~\frac{\text{m}}{\text{sec}^2}\right) = \frac{39200\pi}{9}(12-\pi)\approx 121,212~\text{J}
Q7Stage 1

Let SS be the region bounded above by y=1xy=\frac{1}{x} and and below by the xx-axis, 1≤x≤31 \le x \le 3. Let RR be a rod with density ρ(x)=1x\rho(x)=\frac{1}{x} at position xx, 1≤x≤31 \le x \le 3.

  1. What is the area of a thin slice of SS at position xx with width dx\dee{x}?

  2. What is the mass of a small piece of RR at position xx with length dx\dee{x}?

  3. What is the total area of SS?

  4. What is the total mass of RR?

  5. What is the xx-coordinate of the centroid of SS?

  6. What is the centre of mass of RR?

Hint

Think about whether your answers should have repetition.

Answer

(a), (b) 1x dx\dfrac{1}{x}\,\dee{x} (c), (d) log⁡3\log 3 (e), (f) 2log⁡3\dfrac{2}{\log 3}

Full solution

  1. A thin slice of SS at position xx has height 1x\frac{1}{x}, so if its width is dx\dee{x}, its area is 1x dx\frac{1}{x}\,\dee{x}.

  2. A small piece of RR at position xx has density 1x\frac{1}{x}, so if its length is dx\dee{x}, its mass is 1x dx\frac{1}{x}\,\dee{x}.

  3. Adding up all our tiny slices from (a) gives us the total area of SS:

    ∫131x dx=log⁡3\int_1^3 \frac{1}{x}\,\dee{x} = \log 3

  4. Adding up all our tiny pieces from (b) gives us the total mass of RR:

    ∫131x dx=log⁡3\int_1^3 \frac{1}{x}\,\dee{x}=\log 3

  5. Using Equation 2.3.3 in the CLP-2 text, the xx-coordinate of the centroid of SS is

    ∫13x⋅1x dx∫131x dx=∫131 dxlog⁡3=2log⁡3\frac{\int_1^3 x\cdot\frac{1}{x}\,\dee{x}}{\int_1^3\frac{1}{x}\,\dee{x}} = \frac{\int_1^3 1\,\dee{x}}{\log 3} = \frac{2}{\log 3}

  6. Using Equation 2.3.2 in the CLP-2 text, the centre of mass of RR is

    ∫13x⋅1x dx∫131x dx=∫131 dxlog⁡3=2log⁡3\frac{\int_1^3 x\cdot\frac{1}{x}\,\dee{x}}{\int_1^3\frac{1}{x}\,\dee{x}} = \frac{\int_1^3 1\,\dee{x}}{\log 3} = \frac{2}{\log 3}

Remark: following the derivation of Equation 2.3.3 in the CLP-2 text, if we wanted to find the xx-coordinate of the centroid of SS, we would set up a rod that had exactly the characteristics of RR. That's why all the answers were repeated.

In Questions 8 through 10, you will derive the formulas for the centre of mass of a rod of variable density, and the centroid of a two-dimensional region using vertical slices (Equations 2.3.2 and 2.3.3 in the CLP-2 text). Knowing the equations by heart will allow you to answer many questions in this section; understanding where they came from will you allow to generalize their ideas to answer even more questions.

Q8Stage 1

Suppose RR is a straight, thin rod with density ρ(x)\rho(x) at a position xx. Let the left endpoint of RR lie at x=ax=a, and the right endpoint lie at x=bx=b.

  1. To approximate the centre of mass of RR, imagine chopping it into nn pieces of equal length, and approximating the mass of each piece using the density at its midpoint. Give your approximation for the centre of mass in sigma notation.

    Figure from prob_s2.3, line 2

    Figure from prob_s2.3, line 2

  2. Take the limit as nn goes to infinity of your approximation in part (a), and express the result using a definite integral.

Hint

The definition of a definite integral (Definition 1.1.9 in the CLP-2 text) will tell you how to convert your limits of sums into integrals.

Answer

(a) ∑i=1n[b−anρ(a+(i−12)(b−an))×(a+(i−12)(b−an))]∑i=1nb−anρ(a+(i−12)(b−an))\displaystyle\frac{\sum\limits_{i=1}^n\left[\frac{b-a}{n}\rho\left(\textcolor{red}{a+\left(i-\tfrac12\right)(\tfrac{b-a}{n})}\right)\times\left( \textcolor{red}{ a+(i-\tfrac12)\left(\tfrac{b-a}{n}\right) } \right)\right]}{\sum\limits_{i=1}^n\frac{b-a}{n}\rho\left(\textcolor{red}{ a+(i-\tfrac12)\left(\tfrac{b-a}{n}\right) } \right)} (b) xˉ=∫abxρ(x) dx∫abρ(x) dx\displaystyle\bar x=\frac{\int_a^b x\rho(x)\,\dee{x}}{\int_a^b\rho(x)\,\dee{x}}

Full solution

  1. If we chop RR into nn pieces, each piece has length b−a12n\frac{b-a}{\vphantom{\frac12}n}. Then our iith cut is at position a+i(b−an)a+i\left(\frac{b-a}{n}\right), so our iith piece runs from a+(i−1)(b−an)a+(i-1)\left(\frac{b-a}{n}\right) to a+i(b−an)a+i\left(\frac{b-a}{n}\right). The approximation of the mass of this piece comes from the density at its midpoint,

    mi=[a+(i−1)(b−an)]+[a+i(b−an)]2=a+(i−12)(b−an)m_i=\frac{\left[a+(i-1)\left(\frac{b-a}{n}\right)\right]+\left[a+i\left(\frac{b-a}{n}\right)\right]}{2} = a+(i-\tfrac12)\left(\frac{b-a}{n}\right)

    Figure from prob_s2.3, line 2

    Figure from prob_s2.3, line 2

    So, the iith piece has length b−a12n\frac{b-a}{\vphantom{\frac12}n}, with approximate density ρ(mi)=ρ(a+(i−12)(b−an))\rho\left(m_i\right)=\rho\left(a+(i-\tfrac12)\left(\frac{b-a}{n}\right)\right). We approximate that the iith piece has mass (b−an)⋅ρ(mi)\left(\frac{b-a}{n}\right)\cdot\rho\left(m_i\right) and position mim_i. Using Equation 2.3.1 in the CLP-2 text, the centre of mass of RR is approximately at position:

    xˉn=∑i=1n(mass of ith piece)×(position of ith piece)∑i=1n(mass of ith piece)=∑i=1n[b−anρ(mi)×mi]∑i=1nb−anρ(mi)=∑i=1n[b−anρ(a+(i−12)(b−an))×(a+(i−12)(b−an))]∑i=1nb−anρ(a+(i−12)(b−an))\begin{align*} \bar x_n &= \frac{\sum\limits_{i=1}^n\text{(mass of }i\text{th piece)}\times\text{(position of }i\text{th piece)}}{\sum\limits_{i=1}^n\text{(mass of }i\text{th piece)}}\\ &=\frac{\sum\limits_{i=1}^n\left[\frac{b-a}{n}\rho(\textcolor{red}{m_i})\times\textcolor{red}{m_i}\right]}{\sum\limits_{i=1}^n\frac{b-a}{n}\rho(\textcolor{red}{m_i})}\\ &=\frac{\sum\limits_{i=1}^n\left[\frac{b-a}{n}\rho\left(\textcolor{red}{a+\left(i-\tfrac12\right)(\tfrac{b-a}{n})}\right)\times\left( \textcolor{red}{ a+(i-\tfrac12)\left(\tfrac{b-a}{n}\right) } \right)\right]}{\sum\limits_{i=1}^n\frac{b-a}{n}\rho\left(\textcolor{red}{ a+(i-\tfrac12)\left(\tfrac{b-a}{n}\right) } \right)} \end{align*}

  2. Remember the definition of a midpoint Riemann sum:

    ∫abf(x) dx≈∑i=1nb−an⋅f(a+(i−12)(b−an))\int_a^b f(x)\,\dee{x} \approx \sum_{i=1}^n \frac{b-a}{n}\cdot f\left(a+(i-\tfrac12)\left(\frac{b-a}{n}\right)\right)

    The numerator of our approximation in part (a) is, therefore, a midpoint Riemann sum of ∫abρ(x)×x dx\int_a^b \rho(x)\times x\,\dee{x}, and the denominator is a midpoint Riemann sum of ∫abρ(x) dx\int_a^b \rho(x)\,\dee{x}.

    Using the definition of a definite integral (Definition 1.1.9 in the CLP-2 text), we see the limit of the approximation in (a) as xx goes to infinity is

    xˉ=∫abxρ(x) dx∫abρ(x) dx\bar x=\frac{\int_a^b x\rho(x)\,\dee{x}}{\int_a^b\rho(x)\,\dee{x}}

    This gives us the exact centre of mass of our rod.

    Remark: this is Equation 2.3.2 in the CLP-2 text.

Q9Stage 1

Suppose SS is a two-dimensional object and at (horizontal) position xx its height is T(x)−B(x)T(x)-B(x). Its leftmost point is at position x=ax=a, and its rightmost point is at position x=bx=b.

To approximate the xx-coordinate of the centroid of SS, we imagine it as a straight, thin rod RR, where the mass of RR from a≤x≤ba \le x \le b is equal to the area of SS from a≤x≤ba \leq x \leq b.

  1. If SS is the sheet shown below, sketch RR as a rod with the same horizontal length, shaded darker when RR is denser, and lighter when RR is less dense.

    Figure from prob_s2.3, line 2

    Figure from prob_s2.3, line 2

  2. If we cut SS into strips of very small width dx\dee{x}, what is the area of the strip at position xx?

  3. Using your answer from (b), what is the density ρ(x)\rho(x) of RR at position xx?

  4. Using your result from Question 8(b), give the xx-coordinate of the centroid of SS. Your answer will be in terms of aa, bb, T(X)T(X), and B(x)B(x).

Hint

In (a), the slices all have the same width, so the area of the slices is larger (and hence the density of RR is higher) where T(x)−B(x)T(x)-B(x) is larger.

Answer

(a)

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

(b) (T(x)−B(x)) dx(T(x)-B(x))\,\dee{x} (c) T(x)−B(x)T(x)-B(x) (d) xˉ=∫abx(T(x)−B(x)) dx∫ab(T(x)−B(x)) dx\bar x=\dfrac{\int_a^b x({T(x)-B(x)})\,\dee{x}}{\int_a^b({T(x)-B(x)})\,\dee{x}}

Full solution

  1. On the left-most corner of SS, T(x)=B(x)T(x)=B(x), so the height of SS is zero; that is, the area of a very small vertical strip is very close to zero, so the density of RR is close to 0. As we move closer to the position labeled a′a', the height of the strips increases, so the areas of the strips increases, so the density of RR increases. Then, between the points labeled a′a' and b′b', the height of SS remains constant, since T(x)T(x) and B(x)B(x) are parallel here, so the areas of the strips of SS remain constant, and the density of RR remains constant. Then, between b′b' and bb, the height of SS decreases, so the area of the strips decrease, so the density of RR decreases.

    Figure from prob_s2.3, line 2

    Figure from prob_s2.3, line 2

  2. At position xx, the height of SS is T(x)−B(x)T(x)-B(x), so a rectangle with width dx\dee{x} and this height would have area (T(x)−B(x)) dx(T(x)-B(x))\,\dee{x}.

  3. According to our model, the tiny section of RR at position xx with width dx\dee{x} has mass (T(x)−B(x)) dx(T(x)-B(x))\,\dee{x} (that is, the area of SS over this same tiny interval), so its density is ρ(x)=masslength=(T(x)−B(x)) dxdx=T(x)−B(x)\rho(x) = \frac{\text{mass}}{\text{length}} = \frac{(T(x)-B(x))\,\dee{x}}{\dee{x}} = T(x)-B(x).

  4. Imagine SS were a solid, of constant density. The mass of a portion of SS is proportional to the area of that portion. To find the xx-coordinate where the solid would balance, we imagine compressing together the vertical dimension of SS until it's a rod. That is, we would take a very thin vertical strip of SS, and turn it into a small segment of a rod, with the same mass. Then the centre of mass of that rod would be exactly the xx-coordinate of the centre of mass of the solid–that is, the xx-coordinate of the centroid of SS.

    The compressed rod we form in this way is exactly RR (perhaps multiplied by a constant, to account for the density of SS, but this doesn't affect where RR balances). So, the xx-coordinate of the centroid has the same position as the centre of mass of RR.

    Our result from Question 8(b) tells us the centre of mass of RR is

     ∫abxρ(x) dx∫abρ(x) dx\begin{align*}&~\frac{\int_a^b x\textcolor{red}{\rho(x)}\,\dee{x}}{\int_a^b\textcolor{red}{\rho(x)}\,\dee{x}}\end{align*}

    In (c), we found ρ(x)=T(x)−B(x)\rho(x)=T(x)-B(x). So, for the solid SS bounded by T(x)T(x) and B(x)B(x) on the interval [a,b][a,b],

    xˉ=∫abx(T(x)−B(x)) dx∫ab(T(x)−B(x)) dx\begin{align*}\bar x&=\frac{\int_a^b x(\textcolor{red}{T(x)-B(x)})\,\dee{x}}{\int_a^b(\textcolor{red}{T(x)-B(x)})\,\dee{x}}\end{align*}

    Remark: the denominator is the area of SS. This formula is the same as the formula found in Equation 2.3.3 of the CLP-2 text.

Q10Stage 1

Suppose SS is flat sheet with uniform density, and at (horizontal) position xx its height is T(x)−B(x)T(x)-B(x). Its leftmost point is at position x=ax=a, and its rightmost point is at position x=bx=b.

To approximate the yy-coordinate of the centroid of SS, we imagine it as a straight, thin, vertical rod RR. We slice SS into thin, vertical strips, and model these as weights on RR with:

  • position yy on RR, where yy is the centre of mass of the strip, and

  • mass in RR equal to the area of the strip in SS.

  1. If SS is the sheet shown below, slice it into a number of vertical pieces of equal length, approximated by rectangles. For each rectangle, mark its centre of mass. Sketch RR as a rod with the same vertical height, with weights corresponding to the slices you made of SS.

    Figure from prob_s2.3, line 2

    Figure from prob_s2.3, line 2

  2. Imagine a thin strip of SS at position xx, with thickness dx\dee{x}. What is the area of the strip? What is the yy-value of its centre of mass?

  3. Recall the centre of mass of a rod with nn weights of mass MiM_i at position yiy_i is given by

    ∑i=1n(Mi×yi)∑i=1nMi\frac{\sum\limits_{i=1}^n (M_i\times y_i) }{\sum\limits_{i=1}^n M_i}

    Considering the limit of this formula as nn goes to infinity, give the yy-coordinate of the centre of mass of SS.

Hint

Part (a) is a significantly different model from the last question.

Answer

(a) The strips between x=ax=a and x=a′x=a' at the left end of the figure all have the same centre of mass, which is the yy-value where T(x)=B(x)T(x)=B(x), x<0x<0. So, there should be multiple weights of different mass piled up at that yy-value.

Similarly, the strips between x=b′x=b' and x=bx=b at the right end of the figure all have the same centre of mass, which is the yy-value where T(x)=B(x)T(x)=B(x), x>0x>0. So, there should be a second pile of weights of different mass, at that (higher) yy-value.

Between these two piles, there are a collection of weights with identical mass distributed fairly evenly. The top and bottom ends of RR (above the uppermost pile, and below the lowermost pile) have no weights.

One possible answer (using twelve slices):

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

(b) The area of the strip is (T(x)−B(x)) dx(T(x)-B(x))\,\dee{x}, and its centre of mass is at height T(x)+B(x)2\dfrac{T(x)+B(x)}{2}.
(c) yˉ=∫ab(T(x)2−B(x)2) dx2∫ab(T(x)−B(x)) dx\displaystyle\bar y=\frac{\int_a^b \big(T(x)^2-B(x)^2\big)\,\dee{x}}{2\int_a^b\big( T(x)-B(x)\big)\,\dee{x}}

Full solution

  1. To begin with, we'll sketch some strips, and put a dot at the centre of mass of each one (its vertical centre).

    Figure from prob_s2.3, line 2

    Figure from prob_s2.3, line 2

    In our model, each of these strips corresponds to a weight on RR, positioned at its centre of mass (the height of the dot), and with a mass equal to the strip's area. For the portion of SS with a′≤x≤b′a'\le x \le b', each centre of mass is at a slightly different height, but the areas of the slices are the same. So, the corresponding weights along RR are at different heights, but all have the same mass, as shown below. (Note the rod RR below only contains the weights from the middle of SS–we'll add the rest later.)

    For clarity, the diagrams below are zoomed in.

    Figure from prob_s2.3, line 2

    Figure from prob_s2.3, line 2

    By contrast to the slices in the interval [a′,b′][a',b'], the slices of SS along [a,a′][a,a'] all have the same centre of mass, but different areas. So, there is one position along RR that has a number of weights all stacked on top of one another, of varying masses.

    The same situation applies to the slices of SS along [b,b′][b,b']. So, all together, our rod looks something like this:

    Figure from prob_s2.3, line 2

    Figure from prob_s2.3, line 2

    Remark: if we had sketched the density of RR, it would have looked something like this:

    Figure from prob_s2.3, line 2

    Figure from prob_s2.3, line 2

    because from our sketch, we see that the density of RR:

    • is 0 at either end,

    • is suddenly very high where the blue weights are, and

    • is constant and lower between the blue weights.

  2. At position xx, the height of SS is T(x)−B(x)T(x)-B(x), and the width of the strip is dx\dee{x}, so the area of the strip is (T(x)−B(x)) dx(T(x)-B(x))\,\dee{x}.

    Since the density of SS is uniform, the centre of mass of the strip is halfway up: at T(x)+B(x)2\dfrac{T(x)+B(x)}{2}.

  3. If we cut SS into nn strips, then the strip at position xix_i has area (T(xi)−B(xi))Δx(T(x_i)-B(x_i))\De x, where Δx=b−an\De x = \frac{b-a}{n}, and its centre of mass is at height T(xi)+B(xi)2\dfrac{T(x_i)+B(x_i)}{2}. So, our approximation of the centre of mass of the rod is:

    yˉn=∑i=1n(Mi×yi)∑i=1nMi=∑i=1n((T(xi)−B(xi))Δx)×(T(xi)+B(xi)2)∑i=1n(T(xi)−B(xi))Δx=∑i=1n(T(xi)2−B(xi)2)Δx2∑i=1n(T(xi)−B(xi))Δx\begin{align*}\bar y_n &=\frac{\sum\limits_{i=1}^n (M_i\times y_i) }{\sum\limits_{i=1}^n M_i}\\ &=\frac{\sum\limits_{i=1}^n \left((T(x_i)-B(x_i))\De x\right)\times\left(\dfrac{T(x_i)+B(x_i)}{2}\right) }{\sum\limits_{i=1}^n (T(x_i)-B(x_i))\De x} \\&=\frac{\sum\limits_{i=1}^n (T(x_i)^2-B(x_i)^2)\De x }{2\sum\limits_{i=1}^n (T(x_i)-B(x_i))\De x}\end{align*}

    We use the definition of a definite integral (Definition 1.1.9 in the CLP-2 text) to re-write the limit of the above function.

    yˉ=lim⁡n→∞∑i=1n(T(xi)2−B(xi)2)Δx2∑i=1n(T(xi)−B(xi))Δx=∫ab(T(x)2−B(x)2) dx2∫ab(T(x)−B(x)) dx\begin{align*}\bar y &=\lim_{n \to \infty}\frac{\sum\limits_{i=1}^n (T(x_i)^2-B(x_i)^2)\De x }{2\sum\limits_{i=1}^n (T(x_i)-B(x_i))\De x}\\ &=\frac{\int_a^b \big(T(x)^2-B(x)^2\big)\,\dee{x}}{2\int_a^b\big( T(x)-B(x)\big)\,\dee{x}}\end{align*}

    Remark: the denominator is twice the area of SS. This equation for the yy-coordinate of the centroid is the same as the one given in Equation 2.3.3 in the CLP-2 text.

Q11Stage 1Past exam · 2016Q4

Express the xx–coordinate of the centroid of the triangle with vertices (−1,−3)(-1,-{3}), (−1,3)(-1,{3}), and (0,0)(0,0) in terms of a definite integral. Do not evaluate the integral.

Hint

Which method involves more work: horizontal strips or vertical strips?

Answer

xˉ=−13∫−106x2 dx\displaystyle\bar x = -\frac{1}{3} \int_{-1}^0 6x^2\ \dee{x}

Full solution

We use vertical strips, as in the sketch below. (To use horizontal strips we would have to split the domain of integration in two: −3≤y≤0-3\le y\le 0 and 0≤y≤30\le y\le 3.)

Figure from prob_s2.3, line 886

Figure from prob_s2.3, line 886

The equations of the top and bottom of the triangle are

y=T(x)=−3xandy=B(x)=3x.\begin{equation*} y = T(x) = -3x \qquad\text{and}\qquad y = B(x) = 3x. \end{equation*}

The area of the triangle is A=12(6)(1)=3A=\frac12(6)(1)=3. Now, we can apply the vertical-slice versions of Equation 2.3.3 in the CLP-2 text.

xˉ=1A∫−10x[T(x)−B(x)] dx=13∫−10x[(−3x)−(3x)] dx=−13∫−106x2 dx\begin{equation*} \bar x = \frac{1}{A}\int_{-1}^0 x \big[T(x)- B(x)\big]\ \dee{x} = \frac{1}{3} \int_{-1}^0 x \big[({-}3x)-(3x)\big]\ \dee{x} = -\frac{1}{3} \int_{-1}^0 6x^2\ \dee{x} \end{equation*}

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Use Equations 2.3.2 and 2.3.3 in the CLP-2 text to find centroids and centres of mass in Questions 12 through 23.

Q12Stage 2

A long, thin rod extends from x=0x=0 to x=7x=7 metres, and its density at position xx is given by ρ(x)=x\rho(x) = x kg/m. Where is the centre of mass of the rod?

Hint

This is a straightforward application of Equation 2.3.2 in the CLP-2 text.

Answer

xˉ=143\bar x=\dfrac{14}{3}

Full solution

Applying Equation 2.3.2 in the CLP-2 text,

xˉ=∫07x⋅x dx∫07x dx=[13x3]07[12x2]07=13(73)12(72)=143\bar x = \frac{\int_0^7 x\cdot x\,\dee{x}}{\int_0^7 x\,\dee{x}} = \frac{ \left[\frac{1}{3}x^3\right]_0^7}{\left[\frac{1}{2}x^2\right]_0^7} = \frac{\frac{1}{3}(7^3)}{\frac{1}{2}(7^2)} =\frac{14}{3}
Q13Stage 2

A long, thin rod extends from x=−3x=-3 to x=10x=10 metres, and its density at position xx is given by ρ(x)=11+x2\rho(x) = \frac{1}{1+x^2} kg/m. Where is the centre of mass of the rod?

Hint

Remember the derivative of arctangent is 11+x2\frac{1}{1+x^2}

Answer

xˉ=log⁡10.12(arctan⁡10+arctan⁡(3))≈0.43\displaystyle \bar x=\frac{\log 10.1}{2(\arctan 10 + \arctan(3))}\approx 0.43

Full solution

Applying Equation 2.3.2 in the CLP-2 text,

xˉ=∫−310x⋅11+x2 dx∫−31011+x2 dx\begin{align*}\bar x &= \frac{\int_{-3}^{10} x\cdot \frac{1}{1+x^2}\,\dee{x}}{\int_{-3}^{10} \frac{1}{1+x^2}\,\dee{x}}\end{align*}

For the numerator, we use the substitution u=1+x2u=1+x^2, du=2x dx\dee{u}=2x\,\dee{x}.

=12∫101011u du[arctan⁡x]−310=12[log⁡u]10101arctan⁡10−arctan⁡(−3)=[log⁡101−log⁡10]2(arctan⁡10+arctan⁡(3))=log⁡10.12(arctan⁡10+arctan⁡(3))≈0.43\begin{align*}&= \frac{\frac{1}{2}\int_{10}^{101} \frac{1}{u}\,\dee{u}}{\Big[\arctan x \Big]_{-3}^{10} } =\frac{\frac{1}{2}\Big[\log u\Big]_{10}^{101}}{\arctan 10 - \arctan(-3)}\\ &=\frac{\Big[\log 101-\log 10\Big]}{2(\arctan 10 + \arctan(3))}=\frac{\log 10.1}{2(\arctan 10 + \arctan(3))}\approx 0.43\end{align*}

Since arctangent is an odd function, arctan⁡(−3)=−arctan⁡(3)\arctan(-3)=-\arctan(3); using logarithm rules, log⁡101−log⁡10=log⁡10110=log⁡10.1\log 101-\log 10 = \log \frac{101}{10}=\log 10.1.

Q14Stage 2Past exam · 2012A

Find the yy-coordinate of the centroid of the region bounded by the curves y=1y=1, y=−exy=-e^x, x=0x=0 and x=1x=1. You may use the fact that the area of this region equals ee.

Hint

This is a straightforward application of Equations 2.3.3 and 2.3.4 in the

CLP-2 text. Note that you're only asked for the yy-coordinate of the centroid.

Answer

yˉ=34e−e4\displaystyle\bar y = \frac{3}{4e}-\frac{e}{4}

Full solution

If we use horizontal strips, then we need to break the region into two pieces: y≥−1=−e0y \geq -1=-e^0, and y≤−1y \leq -1. However, if we use vertical strips, the equation of the top of the region is y=T(x)=1y=T(x)=1, and the equation of the bottom of the region is y=B(x)=−exy=B(x)=-e^x, for all xx from a=0a=0 to b=1b=1. So, we use vertical strips.

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

Using Equation 2.3.3 in the CLP-2 text, the yy-coordinate of the centre of mass is

yˉ=12A∫01[T(x)2−B(x)2] dx=12e∫01(1−e2x) dx=12e[x−12e2x]01=12e[1−e22−0+12]=34e−e4\begin{align*} \bar y & = \frac{1}{2A}\int_0^1 \big[T(x)^2-B(x)^2\big]\,\dee{x} = \frac{1}{2e}\int_0^1 \Big(1-e^{2x} \Big) \,\dee{x} = \frac{1}{2e}\bigg[x-\frac{1}{2}e^{2x}\bigg]_0^1 \\ &= \frac{1}{2e}\Big[1-\frac{e^2}{2} -0 + \frac{1}{2}\Big] =\frac{3}{4e}-\frac{e}{4} \end{align*}
Q15Stage 2Past exam · 1996A

Consider the region bounded by y=116−x2y=\frac{1}{\sqrt{16-x^2}}, y=0y=0, x=0x=0 and x=2x=2.

  1. Sketch this region.

  2. Find the yy–coordinate of the centroid of this region.

Hint

You can use a trigonometric substitution to find the area, then a partial fraction decomposition to find the yy-coordinate of the centroid. Remember sin⁡(1/2)=π/6\sin(1/2)=\pi/6.

Answer

(a)

Figure from prob_s2.3, line 1020

Figure from prob_s2.3, line 1020

(b) 3log⁡38π\dfrac{3\log 3}{8\pi}

Full solution

(a) The lines y=0y=0, x=0x=0, and x=2x=2 are easy enough to sketch. Let's get some basic information about y=T(x)=116−x2y=T(x)=\frac{1}{\sqrt{16-x^2}} on the interval [0,2][0,2].

  • For all xx in its domain, T(x)≥0T(x) \geq 0. In particular, it's always the top of our region (so T(x)T(x) is a reasonable name for it), while the bottom is B(x)=0B(x)=0.

  • T(0)=14T(0)=\frac{1}{4}, and T(2)=123T(2)=\frac{1}{2\sqrt{3}}

  • T′(x)=x(16−x2)3/2T'(x) = \frac{x}{(16-x^2)^{3/2}}, which is positive on [0,2][0,2], so T(x)T(x) is increasing.

    Remark: to see that T(x)T(x) is increasing, we can also just break it into pieces:

    • When x≥0x \ge 0, x2x^2 is increasing, so

    • 16−x216-x^2 is decreasing, so

    • 16−x2\sqrt{16-x^2} is decreasing, so

    • 116−x2=T(x)\frac{1}{\sqrt{16-x^2}}=T(x) is increasing.

  • T′′(x)=2x2+16(16−x2)5/2T''(x)=\frac{2x^2+16}{(16-x^2)^{5/2}}, which is positive, so T(x)T(x) is concave up.

Figure from prob_s2.3, line 1020

Figure from prob_s2.3, line 1020

Remark: If we only wanted to solve (b), it would still be nice to have a sketch of the region, but it wouldn't need to be so detailed. Knowing that T(x)T(x) is always greater than 0 would be enough to tell us we could use vertical slices with T(x)T(x) as the top and y=0y=0 as the bottom.

If we wanted to use horizontal slices (we don't... but we could!) we would additionally want to know that T(x)T(x) is increasing over [0,2][0,2], T(0)=14T(0)=\frac{1}{4}, and T(2)=123T(2)=\frac{1}{2\sqrt{3}}. This would tell us that:

  • the right endpoint of a horizontal strip is always x=2x=2,

  • the left endpoint is determined by T(x)T(x) from y=14y=\frac{1}{4} to y=123y=\frac{1}{2\sqrt{3}}, and

  • the left endpoint is x=0x=0 for 0≤y≤140 \le y \le \frac{1}{4}.

(b)

Figure from prob_s2.3, line 1028

Figure from prob_s2.3, line 1028

The part of the region with xx coordinate between xx and x+dxx+\dee{x} is a strip of width dx\dee{x} running from y=0y=0 to y=116−x2y=\frac{1}{\sqrt{16-x^2}}. It is illustrated in red in the figure above. So, the area of the region is

A=∫02116−x2 dx=∫0arcsin⁡(1/2) 14cos⁡t4cos⁡t dt=arcsin⁡12=π6\begin{equation*} A=\int_0^2 \frac{1}{\sqrt{16-x^2}}\,\dee{x} =\int_0^{\arcsin(1/2)}\,\frac{1}{4\cos t}4\cos t\,\dee{t} =\arcsin\frac{1}{2}=\frac{\pi}{6} \end{equation*}

where we made the substitution x=4sin⁡t, dx=4cos⁡tdt,16−x2=4cos⁡tx=4\sin t,\ \dee{x}=4\cos t \dee{t}, \sqrt{16-x^2}=4\cos t.

Using Equation 2.3.3 in the CLP-2 text,

yˉ=∫02[T(x)2−B(x)2] dx2A=∫02[(116−x2)2−02] dx2A=12A∫02116−x2 dx=12A∫021(4−x)(4+x) dx\begin{align*}\bar y &= \frac{\int_0^2\big[T(x)^2-B(x)^2\big]\,\dee{x}}{2A} =\frac{\displaystyle\int_0^2\left[\left(\frac{1}{\sqrt{16-x^2}}\right)^2-0^2\right]\,\dee{x}}{2A}\\ &= \frac{1}{2A}\int_0^2 \frac{1}{16-x^2}\,\dee{x} = \frac{1}{2A}\int_0^2 \frac{1}{(4-x)(4+x)}\,\dee{x}\end{align*}

Using the method of partial fractions, we see 116−x2=1/84+x+1/84−x\displaystyle\frac{1}{16-x^2} = \frac{1/8}{4+x}+\frac{1/8}{4-x}.

=12A∫02[1/84+x+1/84−x] dx=116A∫02[1x+4−1x−4] dx=116A[log⁡∣x+4∣−log⁡∣x−4∣]02=616π[log⁡6−log⁡2−log⁡4+log⁡4]=3log⁡38π\begin{align*}&= \frac{1}{2A}\int_0^2 \Big[\frac{1/8}{4+x}+\frac{1/8}{4-x}\Big]\,\dee{x} = \frac{1}{16A}\int_0^2 \Big[\frac{1}{x+4}-\frac{1}{x-4}\Big]\,\dee{x} \\ &= \frac{1}{16A} \Big[\log|x+4|-\log|x-4|\Big]_0^2 = \frac{6}{16\pi} \big[\log6-\log2-\log4+\log 4\big] \\ &=\frac{3\log 3}{8\pi}\end{align*}
Q16Stage 2Past exam · 2014A

Find the centroid of the finite region bounded by y=sin⁡(x)y = \sin(x), y=cos⁡(x)y = \cos(x), x=0x = 0, and x=π/4x = \pi/4.

Hint

Vertical slices will be easier than horizontal. An integration by parts might be helpful to find xˉ\bar x, while trigonometric identities are important to finding yˉ\bar y.

Answer

xˉ=π42−12−1\displaystyle\bar x=\frac{\frac{\pi}{4}\sqrt{2}-1}{\sqrt{2}-1} and yˉ=14(2−1)\displaystyle\bar y=\frac{1}{4(\sqrt{2}-1)}

Full solution

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

The top of the region is y=T(x)=cos⁡(x)y=T(x)=\cos(x) and the bottom of the region is y=B(x)=sin⁡(x)y=B(x)=\sin(x). So, the area of the region is

A=∫0π/4(T(x)−B(x)) dx=∫0π/4(cos⁡(x)−sin⁡(x)) dx=[sin⁡(x)+cos⁡(x)]0π/4=[12+12]−[0+1]=2−1\begin{align*} {}\hskip0.5inA&=\int_0^{\pi/4}\big(T(x)-B(x)\big)\,\dee{x} =\int_0^{\pi/4}\big(\cos(x)-\sin(x)\big)\,\dee{x} =\Big[\sin(x)+\cos(x)\Big]_0^{\pi/4}\\ &=\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right]-\left[0+1\right]=\sqrt{2}-1 \end{align*}

If we use horizontal slices, we'll need to break up the object into two regions, so let's use vertical slices. Using Equation 2.3.3 in the CLP-2 text, the region has centroid (xˉ,yˉ)(\bar x,\bar y) with:

xˉ=1A∫0π/4x(T(x)−B(x)) dx=1A∫0π/4x(cos⁡(x)−sin⁡(x)) dx\begin{align*}\bar x&= \frac{1}{A}\int_0^{\pi/4} x\big(T(x)-B(x)\big)\,\dee{x} =\frac{1}{A}\int_0^{\pi/4} x\big(\cos(x)-\sin(x)\big)\,\dee{x}\end{align*}

We use integration by parts with u=xu=x, dv=(cos⁡x−sin⁡x)dx\dee{v}=(\cos x - \sin x)\dee{x}; du=dx\dee{u}=\dee{x}, v=sin⁡x+cos⁡xv=\sin x + \cos x.

=1A([x(sin⁡x+cos⁡x)]0π/4−∫0π/4(sin⁡x+cos⁡x) dx)=1A[xsin⁡(x)+xcos⁡(x)+cos⁡x−sin⁡x]0π/4=1A[(π4⋅12+π4⋅12+12−12)−1]=π42−1A=π42−12−1\begin{align*}&=\frac{1}{A}\left(\Big[x(\sin x + \cos x )\Big]_0^{\pi/4} - \int_0^{\pi/4} (\sin x + \cos x)\,\dee{x}\right)\\ &=\frac{1}{A}\Big[x\sin(x)+x\cos(x)+\cos x -\sin x\Big]_0^{\pi/4}\\ &=\frac{1}{A}\left[\left(\frac{\pi}{4}\cdot\frac{1}{\sqrt2}+\frac{\pi}{4}\cdot\frac{1}{\sqrt{2}}+ \frac{1}{\sqrt2}-\frac{1}{\sqrt2}\right)- 1\right]\\ &=\frac{\frac{\pi}{4}\sqrt{2}-1}{A}=\frac{\frac{\pi}{4}\sqrt{2}-1}{\sqrt{2}-1}\end{align*}

Again using Equation 2.3.3 in the CLP-2 text,

yˉ=12A∫0π/4(T(x)2−B(x)2) dx=12A∫0π/4(cos⁡2(x)−sin⁡2(x)) dx=12A∫0π/4cos⁡(2x) dx=12A[12sin⁡(2x)]0π/4=14(2−1)\begin{align*}\bar y&= \frac{1}{2A}\int_0^{\pi/4} \big(T(x)^2-B(x)^2\big)\,\dee{x} =\frac{1}{2A}\int_0^{\pi/4} \big(\cos^2(x)-\sin^2(x)\big)\,\dee{x} \\ &=\frac{1}{2A}\int_0^{\pi/4} \cos(2x)\,\dee{x} =\frac{1}{2A}\Big[\frac{1}{2}\sin(2x)\Big]_0^{\pi/4} =\frac{1}{4(\sqrt{2}-1)}\end{align*}
Q17Stage 2Past exam · 1996D

Let AA denote the area of the plane region bounded by x=0x=0, x=1x=1, y=0y=0 and y=k1+x2y=\dfrac{k}{\sqrt{1+x^2}}, where kk is a positive constant.

  1. Find the coordinates of the centroid of this region in terms of kk and AA.

  2. For what value of kk is the centroid on the line y=xy=x?

Hint

No trigonometric substitution is necessary if you're clever with your uu-substitutions, and remember the derivative of arctangent.

Answer

(a) xˉ=kA[2−1]\displaystyle\bar x = \frac{k}{A}\big[\sqrt{2}-1\big], yˉ=k2π8A\displaystyle\bar y = \frac{k^2\pi}{8A} (b) k=8π[2−1]\displaystyle k=\frac{8}{\pi}\big[\sqrt{2}-1\big]

Full solution

(a) Since kk is positive, k1+x2>0\frac{k}{\sqrt{1+x^2}}>0 for every xx. Then the top of our region is defined by T(x)=k1+x2T(x) = \frac{k}{\sqrt{1+x^2}}, and the bottom is defined by B(x)=0B(x)=0.

If we make vertical slices, we don't have to turn our region into two parts, so let's use vertical slices. The question asks for our final answer in terms of the area AA of the region, so we don't need to find AA explicitly.

Using Equation 2.3.3 in the CLP-2 text, the xx–coordinate of the centroid is

xˉ=1A∫01x(T(x)−B(x)) dx=1A∫01xk1+x2 dx\begin{align*}\bar x &=\frac{1}{A}\int_0^1 x(T(x)-B(x))\,\dee{x}= \frac{1}{A}\int_0^1x\frac{k}{\sqrt{1+x^2}}\,\dee{x}\end{align*}

Although we have a quadratic function underneath a square root, we find an easier method than a trig substitution: the substitution u=1+x2, du=2x dxu=1+x^2,\ \dee{u}=2x\,\dee{x}. This changes the limits of integration to 1+02=11+0^2=1 and 1+12=21+1^2=2, respectively.

=1A∫12ku du2=k2A[u1/2]12=kA[2−1]\begin{align*}&= \frac{1}{A}\int_1^2\frac{k}{\sqrt{u}}\,\frac{\dee{u}}{2} =\frac{k}{2A}\left[\frac{\sqrt{u}}{1/2}\right]_1^2 =\frac{k}{A}\big[\sqrt{2}-1\big]\end{align*}

Again using Equation 2.3.3 in the CLP-2 text, the yy–coordinate of the centroid is

yˉ=12A∫01(T(x)2−B(x)2) dx=12A∫01k21+x2 dx=k22A∫0111+x2 dx=k22A[arctan⁡1−arctan⁡0]=k22A⋅π4=k2π8A\begin{align*} \bar y &= \frac{1}{2A}\int_0^1(T(x)^2-B(x)^2)\,\dee{x} = \frac{1}{2A}\int_0^1\frac{k^2}{1+x^2}\,\dee{x}\\ &= \frac{k^2}{2A}\int_0^1\frac{1}{1+x^2}\,\dee{x} =\frac{k^2}{2A}\Big[\arctan 1-\arctan 0\Big] =\frac{k^2}{2A}\cdot\frac{\pi}{4} =\frac{k^2\pi}{8A} \end{align*}

(b) We have xˉ=yˉ\bar x=\bar y if and only if

kA[2−1]=k2π8A\begin{align*}\frac{k}{A}\big[\sqrt{2}-1\big]&=\frac{k^2\pi}{8A}\end{align*}

Since kk and A are a positive constants (hence neither is equal to 0), we can divide both sides by kk and multiply both sides by AA:

2−1=kπ8k=8π[2−1]\begin{align*}\sqrt{2}-1&=\frac{k\pi}{8} \\k&=\frac{8}{\pi}\big[\sqrt{2}-1\big]\end{align*}
Q18Stage 2Past exam · 1997D

The region RR is the portion of the plane which is above the curve y=x2−3xy=x^2-3x and below the curve y=x−x2y=x-x^2.

  1. Sketch the region RR

  2. Find the area of RR.

  3. Find the xx coordinate of the centroid of RR.

Hint

In RR, the top function is x−x2x-x^2, and the bottom function is x2−3xx^2-3x.

Answer

(a)

Figure from prob_s2.3, line 1239

Figure from prob_s2.3, line 1239

(b) 83\dfrac{8}{3} (c) 11

Full solution

(a)

The curve y=x2−3xy=x^2-3x is a parabola, pointing up, with xx-intercepts at x=0x=0 and x=3x=3.

The curve y=x−x2y=x-x^2 is a parabola, pointing down, with xx-intercepts at x=0x=0 and x=1x=1.

To find where the two curves meet, we set them equal to each other:

x2−3x=x−x22x2−4x=02x(x−2)=0x=0andx=2\begin{align*} x^2-3x&=x-x^2\\ 2x^2-4x&=0\\ 2x(x-2)&=0\\ x&=0\quad\text{and}\quad x=2 \end{align*}

This is enough information to sketch the figure, on the left below.

Figure from prob_s2.3, line 1239

Figure from prob_s2.3, line 1239

Figure from prob_s2.3, line 1250

Figure from prob_s2.3, line 1250

(b) As we found in (a), the curves cross when x=0, x=2x=0,\ x=2. The corresponding values of yy are y=0y=0 and y=2−22=−2y=2-2^2=-2. Note the top curve is T(x)=x−x2T(x)=x-x^2, and the bottom curve is B(x)=x2−3xB(x)=x^2-3x. Using vertical strips, as in the figure on the right above, the area of RR is

∫02[(x−x2)−(x2−3x)] dx=∫02[4x−2x2] dx=[2x2−23x3]02=8−163=83\begin{align*} \int_0^2\big[(x-x^2)-(x^2-3x)\big]\,\dee{x} =\int_0^2\big[4x-2x^2\big]\,\dee{x} =\left[2x^2-\frac{2}{3}x^3\right]_0^2 =8-\frac{16}{3}=\frac{8}{3} \end{align*}

(c) Using Equation 2.3.3 in the CLP-2 text, the xx–coordinate of the centroid of RR (i.e. the weighted average of xx over RR) is

xˉ=38∫02x[(x ⁣− ⁣x2)−(x2 ⁣− ⁣3x)] dx=38∫02[4x2−2x3] dx=38[43x3−12x4]02=38[323−8]=1\begin{align*} \bar x &= \frac{3}{8}\int_0^2x\big[(x\!-\!x^2)-(x^2\!-\!3x)\big]\,\dee{x} = \frac{3}{8}\int_0^2\big[4x^2-2x^3\big]\,\dee{x} = \frac{3}{8}\Big[\frac{4}{3}x^3-\half x^4\Big]_0^2 = \frac{3}{8}\Big[\frac{32}{3}-8\Big] \\ &=1 \end{align*}
Q19Stage 2Past exam · 1998A

Let RR be the region where 0≤x≤10\le x\le 1 and 0≤y≤11+x20\le y\le\frac{1}{1+x^2}. Find the xx–coordinate of the centroid of RR.

Hint

Remember ddx{arctan⁡x}=11+x2\diff{}{x}\{\arctan x\} = \frac{1}{1+x^2}.

Answer

2πlog⁡2≈0.44127\dfrac{2}{\pi}\log 2\approx 0.44127

Full solution

Using Equation 2.3.3 in the CLP-2 text, the xx–coordinate of the centroid is

xˉ=∫01x11+x2 dx∫0111+x2 dx\begin{align*}\bar x&=\frac{\int_0^1 x\frac{1}{1+x^2}\,\dee{x}}{\int_0^1\frac{1}{1+x^2}\,\dee{x}}\end{align*}

We can guess the antiderivative in the numerator, or use the substitution u=1+x2u=1+x^2, du=2x dx\dee{u}=2x\,\dee{x}.

=12log⁡(1+x2)∣01arctan⁡x∣01=12log⁡2π/4=2πlog⁡2≈0.44127\begin{align*}&=\frac{\half\log(1+x^2)\big|_0^1}{\arctan x\big|_0^1} =\frac{\half\log 2}{\pi/4} =\frac{2}{\pi}\log 2 \approx 0.44127\end{align*}
Q20Stage 2Past exam · 2013A

Find the centroid of the region below, which consists of a semicircle of radius 33 on top of a rectangle of width 66 and height 22.

Figure from prob_s2.3, line 1325

Figure from prob_s2.3, line 1325

Hint

You can save quite a bit of work by, firstly, exploiting symmetry and, secondly, thinking about whether it is more efficient to use vertical strips or horizontal strips.

Answer

xˉ=0\bar x=0 and yˉ=1224+9π\bar y= \dfrac{12}{24+9\pi}

Full solution

By symmetry, the centroid lies on the yy–axis, so xˉ=0\bar x=0.

The area of the figure is the area of a half-circle of radius 3, and a rectangle of width 6 and height 2. So, A=12π(9)+6×2=92π+12A = \frac{1}{2}\pi(9)+6\times 2 = \frac{9}{2}\pi+12.

We'll use vertical strips as in the sketch below.

Figure from prob_s2.3, line 1345

Figure from prob_s2.3, line 1345

The top function of our figure is T(x)=9−x2T(x)=\sqrt{9-x^2}, and the bottom function of our figure is B(x)=−2B(x)=-2. Using Equation 2.3.3 in the CLP-2 text, the yy–coordinate of the centroid is:

yˉ=12A∫ab(T(x)2−B(x)2) dx=12A∫−33(9−x22−(−2)2) dx=12A∫−33(5−x2) dx=12A[5x−13x3]−33=12A[15−9+15−9]=6A=692π+12=129π+24\begin{align*} \bar y &=\frac{1}{2A}\int_a^b\big(T(x)^2-B(x)^2\big)\,\dee{x}\\ &=\frac{1}{2A}\int_{-3}^3 \left( \sqrt{9-x^2}^2-(-2)^2\right)\,\dee{x}\\ &=\frac{1}{2A}\int_{-3}^3 \left(5-x^2\right)\,\dee{x}\\ &=\frac{1}{2A}\left[5x-\frac{1}{3}x^3\right]_{-3}^3\\ &=\frac{1}{2A}\left[15-9+15-9\right]\\ &=\frac{6}{A} = \frac{6}{ \frac{9}{2}\pi+12} = \frac{12}{9\pi+24} \end{align*}
Q21Stage 2Past exam · 2015A

Let DD be the region below the graph of the curve y=9−4x2y=\sqrt{9-4x^2} and above the xx-axis.

  1. Using an appropriate integral, find the area of the region DD; simplify your answer completely.

  2. Find the centre of mass of the region DD; simplify your answer completely. (Assume it has constant density ρ\rho.)

Hint

Sketch the region, being careful the domain of 9−4x2\sqrt{9-4x^2}. You can save quite a bit of work by exploiting symmetry.

Answer

(a) 94π\dfrac{9}{4}\pi (b) xˉ=0\bar x = 0 and yˉ=4π\bar y = \dfrac{4}{\pi}

Full solution

(a) Notice that when x=0x=0, y=3y=3 and as x2x^2 increases, yy decreases until yy hits zero at x2=94x^2=\frac{9}{4}, i.e. at x=±32x=\pm\frac{3}{2}. For x2>94x^2>\frac{9}{4}, yy is not even defined. So, on DD, xx runs from −32-\frac{3}{2} to +32+\frac{3}{2} and, for each xx, yy runs from 00 to 9−4x2\sqrt{9-4x^2}. Here is a sketch of DD.

Figure from prob_s2.3, line 1393

Figure from prob_s2.3, line 1393

As an aside, we can rewrite y=9−4x2y=\sqrt{9-4x^2} as 4x2+y2=94x^2+y^2=9, y≥0y\ge 0, which is the top half of the ellipse which passes through (±a,0)(\pm a,0) and (0,±b)(0,\pm b) with a=32a=\frac{3}{2} and b=3b=3. The area of the full ellipse is πab=92π\pi ab=\frac{9}{2}\pi. The area of DD is half of that, which is 94π\frac{9}{4}\pi. But we are told to use an integral, so we will do so.

The area is

Area=∫−3/23/29−4x2 dx\begin{align*} \text{Area} &= \int_{-3/2}^{3/2} \sqrt{9-4x^2} \ \dee{x} \end{align*}

We can evaluate this integral by substituting x=32sin⁡θx=\frac{3}{2}\sin\theta, dx=32cos⁡θ dθ\dee{x} = \frac{3}{2}\cos\theta\,\dee{\theta} and using

x=±32  ⟺  sin⁡θ=±1\begin{equation*} x=\pm\frac{3}{2} \iff \sin\theta = \pm 1 \end{equation*}

So −π2≤θ≤π2-\frac{\pi}{2}\le\theta\le\frac{\pi}{2} and

Area=∫−π/2π/29−4(32sin⁡θ)2  32cos⁡θ dθ=∫−π/2π/29−9sin⁡2θ  32cos⁡θ dθ=92∫−π/2π/2cos⁡2θ dθ=92∫−π/2π/2cos⁡(2θ)+12dθ=94[sin⁡(2θ)2+θ]−π/2π/2=94π\begin{alignat*}{3} \text{Area} &= \int_{-\pi/2}^{\pi/2} \sqrt{9-4\big({\textstyle\frac{3}{2}}\sin\theta\big)^2} \ \ \frac{3}{2}\cos\theta\,\dee{\theta} = \int_{-\pi/2}^{\pi/2} \sqrt{9-9\sin^2\theta} \ \ \frac{3}{2}\cos\theta\,\dee{\theta} \\ &=\frac{9}{2} \int_{-\pi/2}^{\pi/2}\cos^2\theta\,\dee{\theta} = \frac{9}{2} \int_{-\pi/2}^{\pi/2}\frac{\cos(2\theta)+1}{2}\dee{\theta} =\frac{9}{4}\Big[\frac{\sin(2\theta)}{2}+\theta\Big]_{-\pi/2}^{\pi/2} =\frac{9}{4}\pi \end{alignat*}

(b) The region DD is symmetric about the yy axis. So the centre of mass lies on the yy axis. That is, xˉ=0\bar x=0. Since DD has area A=94πA=\frac{9}{4}\pi, top equation y=T(x)=9−4x2y=T(x)=\sqrt{9-4x^2} and bottom equation y=B(x)=0y=B(x)=0, with xx running from a=−32a=-\frac{3}{2} to b=32b=\frac{3}{2}, Equation 2.3.3 in the CLP-2 text gives us yˉ\bar y:

yˉ=12A∫ab[T(x)2−B(x)2] dx=29π∫−3/23/2[9−4x2] dx=49π∫03/2[9−4x2] dx=49π[9x−43x3]03/2=49π[9⋅32−43⋅3323]=49π[9⋅32−9⋅12]=4π\begin{alignat*}{3} \bar y &=\frac{1}{2A}\int_a^b\big[T(x)^2-B(x)^2\big]\ \dee{x} =\frac{2}{9\pi}\int_{-3/2}^{3/2}\big[9-4x^2\big]\ \dee{x} =\frac{4}{9\pi}\int_0^{3/2}\big[9-4x^2\big]\ \dee{x} \\ &= \frac{4}{9\pi}\Big[9x-\frac{4}{3}x^3\Big]_0^{3/2} = \frac{4}{9\pi}\Big[9\cdot\frac{3}{2}-\frac{4}{3}\cdot\frac{3^3}{2^3}\Big] = \frac{4}{9\pi}\Big[9\cdot\frac{3}{2}-9\cdot\frac{1}{2}\Big] = \frac{4}{\pi} \end{alignat*}
Q22Stage 2

The finite region SS is bounded by the lines y=arcsin⁡xy=\arcsin x, y=arcsin⁡(2−x)y=\arcsin(2-x), and y=−π2y=-\frac{\pi}{2}. Find the centroid of SS.

Hint

Horizontal slices will be easier than vertical.

Answer

(xˉ,yˉ)=(1,−2π)(\bar x, \bar y) = \left( 1,-\dfrac{2}{\pi}\right)

Full solution

Let's start by sketching the region at hand. We know the general shape of arcsine (it's like half a period of sine, if you swapped the xx and yy axes); we can sketch the curve y=arcsin⁡(2−x)y=\arcsin(2-x) by mirroring y=arcsin⁡xy=\arcsin x about the line x=1x=1.

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

If we use vertical strips, then we need two separate regions, because T(x)=arcsin⁡xT(x) = \arcsin x when x≤1x \le 1, and T(x)=arcsin⁡(2−x)T(x) = \arcsin(2-x) when x>1x >1. Also, we'd have to antidifferentiate functions that have arcsine in them. Let's think about horizontal strips. If y=arcsin⁡x\textcolor{red}{y=\arcsin x}, then x=sin⁡y\textcolor{red}{x=\sin y}, and if y=arcsin⁡(2−x)\textcolor{blue}{y=\arcsin(2-x)} then x=2−sin⁡y\textcolor{blue}{x=2-\sin y}. For all yy from −π2-\frac{\pi}{2} to π2\frac{\pi}{2}, the left endpoint of a strip is given by L(y)=sin⁡y\textcolor{red}{L(y) = \sin y}, and the right endpoint is given by R(y)=2−sin⁡y\textcolor{blue}{R(y) = 2-\sin y}.

First, let's use our horizontal slices (There's also a sneaky way to find the area of AA: look for a way to snip and rearrange bits of the figure to turn it into a rectangle!) to find the area of our region, AA.

A=∫−π/2π/2((2−sin⁡y)−(sin⁡y)) dy=∫−π/2π/2(2−2sin⁡y) dy=[2y+2cos⁡y]−π/2π/2=(π+0)−(−π+0)=2π\begin{align*} A&=\int_{-\pi/2}^{\pi/2} \big( (\textcolor{blue}{2-\sin y}) - (\textcolor{red}{\sin y})\big)\,\dee{y}=\int_{-\pi/2}^{\pi/2}\big(2-2\sin y\big)\,\dee{y}\\ &=\Big[2y+2\cos y\Big]_{-\pi/2}^{\pi/2} = \left(\pi+0\right) - \left(-\pi+0\right)=2\pi \end{align*}

From symmetry, it is clear that xˉ=1\bar x = 1. We find yˉ\bar y using Equation 2.3.3 in the CLP-2 text.

yˉ=∫−π/2π/2y[R(y)−L(y)] dyA=∫−π/2π/2y[(2−sin⁡y)−(sin⁡y)] dy2π=12π∫−π/2π/2y(2−2sin⁡y) dy=1π∫−π/2π/2y dy−1π∫−π/2π/2(ysin⁡y) dy\begin{align*}\bar y &= \frac{\int_{-\pi/2}^{\pi/2} y\left[\textcolor{blue}{R(y)} - \textcolor{red}{L(y)}\right]\,\dee{y}}{A}\\ &= \frac{\int_{-\pi/2}^{\pi/2}y \left[\textcolor{blue}{(2-\sin y)} - \textcolor{red}{(\sin y)}\right]\,\dee{y}}{2\pi}\\ &= \frac{1}{2\pi}\int_{-\pi/2}^{\pi/2} y(2-2\sin y) \,\dee{y} \\&= \frac{1}{\pi}\int_{-\pi/2}^{\pi/2} y \,\dee{y}-\frac{1}{\pi}\int_{-\pi/2}^{\pi/2} (y\sin y) \,\dee{y}\end{align*}

Since yy is an odd function, and the domain of integration is symmetric, the first integral evaluates to 0. Since ysin⁡yy\sin y is an even function (recall the product of two odd functions is an even function), we can simplify our limits of integration.

=−2π∫0π/2ysin⁡y dy\begin{align*}&=-\frac{2}{\pi}\int_0^{\pi/2} y\sin y\,\dee{y}\end{align*}

We use integration by parts with u=yu=y, dv=sin⁡y dy\dee{v}=\sin y\,\dee{y}; du=dy\dee{u}=\dee{y}, v=−cos⁡yv=-\cos y.

=−2π([−ycos⁡y]0π/2+∫0π/2cos⁡y dy)=−2π[−ycos⁡y+sin⁡y]0π/2=−2π[(0+1)−0]=−2π\begin{align*}&=-\frac{2}{\pi}\left( \big[-y\cos y\big]_0^{\pi/2}+ \int_0^{\pi/2} \cos y\,\dee{y} \right) \\&=-\frac{2}{\pi} \big[-y\cos y+\sin y\big]_0^{\pi/2}\\ &=-\frac{2}{\pi}\left[(0+1)- 0\right] = -\frac{2}{\pi}\end{align*}
Q23Stage 2

Calculate the centroid of the figure bounded by the curves y=exy=e^x, y=3(x−1)y=3(x-1), y=0y=0, x=0x=0, and x=2x=2.

Hint

Start with a picture: whether you use vertical slices or horizontal, you'll need to break your integral into multiple pieces.

Answer

(e2−3/2e2−5/2,e4−74e2−10)≈(1.2,2.4)\displaystyle\left(\frac{e^2-3/2}{e^2-5/2},\frac{e^4-7}{4e^2-10}\right)\approx (1.2,2.4)

Full solution

We'll start by sketching the region.

Figure from prob_s2.3, line 1

Figure from prob_s2.3, line 1

If we use horizontal slices, we need to divide our figure into three regions, as in the figure below, because the left and right functions change at the dashed lines.

Figure from prob_s2.3, line 1

Figure from prob_s2.3, line 1

If we use vertical slices, we only need two regions (shown below) to account for the different top and bottom functions. This seems easier than three regions, so we use vertical slices.

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

When 0≤x≤20 \leq x \leq 2, T(x)=exT(x) = e^x. When 0≤x≤10 \le x \le 1, B(x)=0B(x)=0, and when 1≤x≤21\le x \le 2, B(x)=3(x−1)B(x)=3(x-1).

The area of the figure is:

A=∫02(T(x)−B(x)) dx=∫01(ex−0) dx+∫12(ex−3(x−1)) dx=∫02ex dx−∫123(x−1) dx=[ex]02−[32(x−1)2]12=e2−1−32=e2−52\begin{align*}A&=\int_0^2 (T(x)-B(x))\,\dee{x} = \int_0^1 (e^x-0)\,\dee{x} + \int_1^2(e^x-3(x-1))\,\dee{x}\\ &=\int_0^2 e^x\,\dee{x} - \int_1^2 3(x-1)\,\dee{x}\\ &=\Big[e^x\Big]_0^2 - \left[\frac{3}{2}(x-1)^2\right]_1^2\\ &=e^2-1 -\frac{3}{2} = e^2 - \frac52\end{align*}

Using Equation 2.3.3 in the CLP-2 text:

xˉ=∫02x(T(x)−B(x)) dxA=1e2−5/2[∫01x(ex−0) dx+∫12x(ex−3(x−1)) dx]=1e2−5/2[∫02xex dx−∫123x(x−1) dx]\begin{align*}\bar x &=\frac{\int_0^2 x(T(x)-B(x))\,\dee{x}}{A} \\&= \frac{1}{e^2-5/2}\left[\int_0^1x\left(e^x-0\right)\,\dee{x} + \int_1^2 x\left(e^x-3(x-1)\right)\,\dee{x}\right]\\ &= \frac{1}{e^2-5/2}\left[\int_0^2 xe^x\,\dee{x} - \int_1^2 3x(x-1)\,\dee{x}\right]\end{align*}

For the left integral, we use integration by parts with u=xu=x, dv=ex dx\dee{v}=e^x\,\dee{x}; du=dx\dee{u}=\dee{x}, v=exv=e^x.

=1e2−5/2[[xex]02−∫02ex dx−3∫12(x2−x) dx]=1e2−5/2([xex−ex]02−3[13x3−12x2]12)=1e2−5/2((2e2−e2)−(−1)−3(83−2−13+12))=e2−3/2e2−5/2≈1.2\begin{align*}&= \frac{1}{e^2-5/2}\left[\left[xe^x\right]_0^2-\int_0^2 e^x\,\dee{x} - 3\int_1^2 (x^2-x)\,\dee{x}\right]\\ &= \frac{1}{e^2-5/2}\left(\left[xe^x-e^x\right]_0^2 - 3\left[\frac{1}{3}x^3-\frac{1}{2}x^2\right]_1^2\right)\\ &= \frac{1}{e^2-5/2}\left((2e^2-e^2)-(-1) - 3\left(\frac{8}{3}-2-\frac{1}{3}+\frac{1}{2}\right)\right)\\ &= \frac{e^2-3/2}{e^2-5/2}\approx 1.2\end{align*}

Using Equation 2.3.3 in the CLP-2 text again:

yˉ=∫02(T(x)2−B(x)2) dx2A=12(e2−5/2)[∫01(e2x−0) dx+∫12(e2x−9(x−1)2) dx]=12(e2−5/2)[∫02e2x dx−∫129(x−1)2 dx]=12(e2−5/2)([12e2x]02−[3(x−1)3]12)=12e2−5(12e4−12−3)=e4−74e2−10≈2.4\begin{align*}\bar y &=\frac{\int_0^2\left(T(x)^2-B(x)^2\right)\,\dee{x}}{2A}\\ &=\frac{1}{2(e^2-5/2)}\left[\int_0^1\left(e^{2x} - 0\right)\,\dee{x} + \int_1^2\left(e^{2x} - 9(x-1)^2\right)\,\dee{x}\right] \\&=\frac{1}{2(e^2-5/2)}\left[\int_0^2 e^{2x}\,\dee{x} - \int_1^2 9(x-1)^2\,\dee{x}\right] \\&=\frac{1}{2(e^2-5/2)}\left(\left[\frac{1}{2}e^{2x}\right]_0^2 - \Big[3(x-1)^3\Big]_1^2\right) \\&=\frac{1}{2e^2-5}\left(\frac{1}{2}e^{4}-\frac{1}{2} -3\right)\\ &=\frac{e^4-7}{4e^2-10}\approx 2.4\end{align*}

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q24Stage 3Past exam · 2016Q4

Find the yy-coordinate of the centre of mass of the (infinite) region lying to the right of the line x=1x=1, above the xx–axis, and below the graph of y=8/x3y=8/x^3.

Hint

For practice, do the computation twice — once with horizontal strips and once with vertical strips. Watch for improper integrals.

Answer

yˉ=85\bar y = \dfrac{8}{5}

Full solution

The area of the region is

A=∫1∞8x3 dx=lim⁡t→∞(∫1t8x3 dx)=lim⁡t→∞[−4x2]1t=lim⁡t→∞[−4t2+412]=0+4\begin{align*} A&=\int_1^\infty \frac{8}{x^3}\,\dee{x} = \lim_{t\to\infty} \bigg( \int_1^t \frac{8}{x^3}\,\dee{x} \bigg) = \lim_{t\to\infty} \bigg[ {-}\frac{4}{x^2}\bigg]_1^t = \lim_{t\to\infty} \bigg[ {-}\frac{4}{t^2} + \frac{4}{1^2} \bigg] = 0 + 4 \end{align*}

We'll now compute yˉ\bar y twice, once with vertical strips, as in the figure in the left below, and once with horizontal strips as in the figure on the right below.

Figure from prob_s2.3, line 1643

Figure from prob_s2.3, line 1643

Figure from prob_s2.3, line 1643

Figure from prob_s2.3, line 1643

Vertical strips:
The equation of the top of the region is y=T(x)=8x3y=T(x)=\dfrac{8}{x^3} and the equation of the bottom of the region is y=B(x)=0y=B(x)=0. Using vertical strips, as in the figure on the left above, the yy-coordinate of the centre of mass is

yˉ=12A∫1∞[T(x)2−B(x)2] dx=18∫1∞(8x3)2 dx=lim⁡t→∞(∫1t8x6 dx)=lim⁡t→∞[−85x5]1t=lim⁡t→∞[−85t5+85×15]=85\begin{align*} \bar y & = \frac{1}{2A}\int_1^\infty \big[T(x)^2-B(x)^2\big]\,\dee{x} \\ &= \frac{1}{8}\int_1^\infty \bigg( \frac{8}{x^3} \bigg)^2 \,\dee{x} \\ &= \lim_{t\to\infty} \bigg( \int_1^t \frac{8}{x^6}\,\dee{x} \bigg) \\ &= \lim_{t\to\infty} \bigg[ {-}\frac{8}{5x^5}\bigg]_1^t \\ &= \lim_{t\to\infty} \bigg[ {-}\frac{8}{5t^5} + \frac{8}{5\times1^5} \bigg] = \frac{8}{5} \end{align*}

Vertical strips:
Since y=8x3y=\dfrac{8}{x^3} is equivalent to x=8y3x=\sqrt[3]{\dfrac{8}{y}}, the equation of the right-hand side of the region is x=R(y)=2y1/3x=R(y)=\dfrac{2}{y^{1/3}} and the equation of the left hand side of the region is x=L(y)=1x=L(y)=1. The point at the top of the region is (1,8)(1,8). Thus yy runs from 00 to 88. So, using horizontal strips, as in the figure on the right above, the yy-coordinate of the centre of mass is

yˉ=1A∫08y[R(y)−L(y)] dy=14∫08y[2y−1/3−1] dy=14∫08[2y2/3−y] dy=14[65y5/3−y22]08=14[6×325−8×82]=8[65−1]=85\begin{align*} \bar y & = \frac{1}{A}\int_0^8 y \big[R(y)-L(y)\big]\,\dee{y} \\ &= \frac{1}{4}\int_0^8 y\big[ 2y^{-1/3} -1 \big] \,\dee{y} \\ &=\frac{1}{4}\int_0^8 \big[2y^{2/3}-y\big]\,\dee{y} \\ &= \frac{1}{4} \bigg[ \frac{6}{5}y^{5/3}-\frac{y^2}{2}\bigg]_0^8 \\ &= \frac{1}{4}\bigg[ \frac{6\times 32}{5} - \frac{8\times 8}{2} \bigg] = 8\bigg[ \frac{6}{5} - 1 \bigg] = \frac{8}{5} \end{align*}
Q25Stage 3Past exam · 2016A

Let AA be the region to the right of the yy-axis that is bounded by the graphs of y=x2y=x^2 and y=6−xy = 6-x.

  1. Find the centroid of AA, assuming it has constant density ρ=1\rho=1. The area of AA is 223\dfrac{22}{3} (you don't have to show this).

  2. Write down an expression, using horizontal slices (disks), for the volume obtained when the region AA is rotated around the yy-axis. Do not evaluate any integrals; simply write down an expression for the volume.

Hint

Draw a sketch. In part (b) be careful about the equation of the right hand boundary of AA.

Answer

(a) xˉ=811\displaystyle \bar x = \frac{8}{11}, yˉ=16655\displaystyle\bar y = \frac{166}{55} (b) π∫04y dy+π∫46(6−y)2 dy\displaystyle\pi \int_0^4 y\,\dee{y} + \pi \int_4^6 (6-y)^2\,\dee{y}

Full solution

(a) The two curves cross at points (x,y)(x,y) that satisfy both y=x2y=x^2 and y=6−xy = 6-x, and hence

x2=6−x  ⟺  x2+x−6=0  ⟺  (x+3)(x−2)=0\begin{align*} x^2 = 6-x \iff x^2+x-6=0 \iff (x+3)(x-2)=0 \end{align*}

So we see that the two curves intersect at x=2x=2 (as well as x=−3x=-3, which is to the left of the yy-axis and therefore irrelevant). Here is a sketch of AA.

Figure from prob_s2.3, line 1719

Figure from prob_s2.3, line 1719

The top of AA has equation y=T(x)=6−xy=T(x)=6-x, the bottom has equation y=B(x)=x2y=B(x)=x^2 and xx runs from 00 to 22. So, using vertical strips,

xˉ=1A∫02x[T(x)−B(x)] dx=122/3∫02x[(6−x)−x2] dx=322∫02(6x−x2−x3) dx=322[3x2−x33−x44]02=322[12−83−4]=322163=811\begin{align*} \bar x &=\frac{1}{A} \int_0^2 x \big[T(x)-B(x)\big]\,\dee{x} \\ &= \frac{1}{22/3} \int_0^2 x \big[ (6-x) - x^2 \big]\,\dee{x} = \frac{3}{22}\int_0^2 (6x-x^2-x^3) \,\dee{x} \\ &= \frac{3}{22} \bigg[ 3x^2 - \frac{x^3}{3} - \frac{x^4}{4} \bigg]_0^2 \\ &= \frac{3}{22}\Big[12 - \frac{8}{3} - 4\Big] = \frac{3}{22} \frac{16}{3} =\frac{8}{11} \end{align*}

and

yˉ=12A∫02[T(x)2−B(x)2] dx=12⋅122/3∫02((6−x)2−x4 ) dx=344[−(6−x)33−x55]02=344(−64−2163−325)=344⋅66415=16655\begin{align*} \bar y &=\frac{1}{2A} \int_0^2 \big[T(x)^2-B(x)^2\big]\,\dee{x} \\ &= \frac{1}{2}\cdot\frac{1}{22/3}\int_0^2 \big( (6-x)^2 - x^4\ \big) \,\dee{x} = \frac{3}{44} \bigg[ {-}\frac{(6-x)^3}{3} - \frac{x^5}{5} \bigg]_0^2 \\ &=\frac{3}{44} \bigg(-\frac{64-216}{3} - \frac{32}{5} \bigg) = \frac{3}{44}\cdot\frac{664}{15} =\frac{166}{55} \end{align*}

The integral was evaluated by guessing an antiderivative for the integrand. It could also be evaluated as

344∫02(36−12x+x2−x4 ) dx=344[36x−6x2+x33−x55]02=344(72−24+83−325)=34466415=16655\begin{align*} \frac{3}{44}\int_0^2 \big( 36-12x+x^2 - x^4\ \big) \,\dee{x} &= \frac{3}{44} \bigg[36x-6x^2+\frac{x^3}3 - \frac{x^5}5\bigg]_0^2 \\ &= \frac{3}{44} \bigg( 72-24+\frac83-\frac{32}5 \bigg) = \frac{3}{44}\frac{664}{15} = \frac{166}{55} \end{align*}

(b) The question specifies the use of horizontal slices (as in Example 1.6.5 of the CLP-2 text). The radius of the slice at height yy is the xx-value of the right-hand boundary of the region at that point. So, we start by converting both equations y=6−xy=6-x and y=x2y=x^2 into equations of the form x=f(y)x=f(y). To do so we solve for xx in both equations, yielding x=yx=\sqrt y and x=6−yx=6-y.

Figure from prob_s2.3, line 1719

Figure from prob_s2.3, line 1719

  • We use thin horizontal strips of width dy\dee{y} as in the figure above.

  • When we rotate about the yy–axis, each strip sweeps out a thin disk

    • whose radius is r=6−yr=6-y when 4≤y≤64\le y\le 6 (see the blue strip in the figure above), and whose radius is r=yr=\sqrt{y} when 0≤y≤40\le y\le 4 (see the red strip in the figure above) and

    • whose thickness is dy\dee{y} and hence

    • whose volume is πr2 dy=π(6−y)2 dy\pi r^2\,\dee{y} = \pi(6-y)^2\,\dee{y} when 4≤y≤64\le y\le 6 and whose volume is πr2 dy=πy dy\pi r^2\,\dee{y} =\pi y\,\dee{y} when 0≤y≤40\le y\le 4.

  • As our bottommost strip is at y=0y=0 and our topmost strip is at y=6y=6, the total volume is

    π∫04y dy+π∫46(6−y)2 dy\begin{align*} \pi \int_0^4 y\,\dee{y} + \pi \int_4^6 (6-y)^2\,\dee{y} \end{align*}
Q26Stage 3Past exam · 2014D

(a) Find the yy–coordinate of the centroid of the region bounded by y=exy = e^x, x=0x = 0, x=1x = 1, and y=−1y = -1.

(b) Calculate the volume of the solid generated by rotating the region from part (a) about the line y=−1y = -1.

Hint

Draw a sketch. Rotating about a horizontal line is similar to rotating about the xx-axis, but for the radius of a slice, you'll need to know ∣y−(−1)∣|y-(-1)|: the distance from the outer edge of the region (the boundary function's yy-value) to y=−1y=-1.

Answer

(a) yˉ=e4−34e\displaystyle\bar y = \frac{e}{4} - \frac{3}{4e} (b) π(e22+2e−32)\displaystyle\pi\left(\frac{e^{2}}{2}+2e -\frac{3}{2}\right)

Full solution

(a) Here is a sketch of the specified region, which we shall call RR.

Figure from prob_s2.3, line 1828

Figure from prob_s2.3, line 1828

The top of RR has equation y=T(x)=exy=T(x)=e^x, the bottom has equation y=B(x)=−1y=B(x)=-1 and xx runs from 00 to 11. So, using vertical strips, we see that RR has area

A=∫01[T(x)−B(x)] dx=∫01[ex−(−1)] dx=∫01[ex+1] dx=[ex+x]01=e\begin{align*} A & = \int_0^1 \big[T(x)-B(x)\big]\,\dee{x} = \int_0^1 \big[e^x-(-1)\big]\,\dee{x} = \int_0^1 \big[e^x+1\big]\,\dee{x} = \big[e^x+x\big]_0^1 = e \end{align*}

and

yˉ=12A∫01[T(x)2−B(x)2] dx=12e∫01[e2x−1 ] dx=12e[e2x2−x]01=12e(e22−1−12)=e4−34e\begin{align*} \bar y &=\frac{1}{2A} \int_0^1 \big[T(x)^2-B(x)^2\big]\,\dee{x} \\ &= \frac{1}{2e}\int_0^1 \big[ e^{2x} - 1\ \big] \,\dee{x} = \frac{1}{2e} \bigg[\frac{e^{2x}}{2} - x \bigg]_0^1 \\ &=\frac{1}{2e} \bigg(\frac{e^2}{2} - 1 - \frac{1}{2} \bigg) = \frac{e}{4} - \frac{3}{4e} \end{align*}

(b) To compute the volume when RR is rotated about the line y=−1y=-1

  • we use thin vertical strips of width dx\dee{x} as in the figure above.

  • When we rotate about the line y=−1y=-1, each strip sweeps out a thin disk

    • whose radius is r=T(x)−B(x)=ex+1r=T(x)-B(x)=e^x+1 and

    • whose thickness is dx\dee{x} and hence

    • whose volume is πr2 dx=π(ex+1)2 dx\pi r^2\,\dee{x} = \pi(e^x+1)^2\,\dee{x}.

  • As our leftmost strip is at x=0x=0 and our rightmost strip is at x=1x=1, the total volume is

    π∫01(ex+1)2 dx=π∫01(e2x+2ex+1) dx=π[e2x2+2ex+x]01=π[(e22+2e+1)−(12+2+0)]=π(e22+2e−32)\begin{align*} \pi \int_0^1 (e^x+1)^2\,\dee{x} &=\pi \int_0^1 (e^{2x}+2e^x+1)\,\dee{x} =\pi\bigg[\frac{e^{2x}}{2}+2e^x + x\bigg]_0^1 \\ &=\pi\bigg[\Big(\frac{e^{2}}{2}+2e + 1\Big) -\Big(\frac{1}{2}+2 + 0\Big)\bigg] \\ &=\pi\Big(\frac{e^{2}}{2}+2e -\frac{3}{2}\Big) \end{align*}
Q27Stage 3

Suppose a rectangle has width 4 m, height 3 m, and its density xx metres from its left edge is x2x^2 kg/m2^2. Find the centre of mass of the rectangle.

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

Hint

Go back to the derivation of Equation 2.3.3 in the CLP-2 text (centroid for a region) to figure out what to do when your surface does not have uniform density. We will consider a rod RR that reaches from x=0x=0 to x=4x=4, and the mass of the section of the rod along [a,b][a,b] is equal to the mass of the strip of our rectangle along [a,b][a,b].

Answer

(3,1.5)(3,1.5)

Full solution

By symmetry, yˉ=1.5\bar y = 1.5. We can't immediately use Equation 2.3.3 in the CLP-2 text to find xˉ\bar x, because the density is not constant. Instead, we'll go through the derivation of Equation 2.3.3, to figure out what to do with a non-constant density. (This is a good time to review Questions 9 and 10 in this section.)

Our model is that we're making a rod RR that reaches from x=0x=0 to x=4x=4, and the mass of the section of the rod along [a,b][a,b] is equal to the mass of the strip of our rectangle along [a,b][a,b]. If we have a formula ρ(x)\rho(x) for the density of RR, we can find the centre of mass of RR, which is also the xx-coordinate of the centre of mass of the rectangle.

A thin vertical strip of the rectangle with length dx\dee{x} at position xx has area 3dx3\dee{x} m2^2 and density x2x^2 kg/m2^2, so it has mass 3x2 dx3x^2\,\dee{x} kg. Therefore, a short section of RR at position xx with length dx\dee{x} ought to have mass 3x2 dx3x^2\,\dee{x} kg as well. Then its density at xx is ρ(x)=3x2 dx kgdx m=3x2\rho(x) = \frac{3x^2\,\dee{x}\text{ kg}}{\dee{x} \text{ m}} = 3x^2 kg/m.

Now, we can use Equation 2.3.2 in the CLP-2 text to find the centre of mass of the rod, which is also the xx-coordinate of the centre of mass of our rectangle:

xˉ=∫04xρ(x) dx∫04ρ(x) dx=∫043x3 dx∫043x2 dx=[34x4]04[34x3]04=3⋅4343=3\begin{align*} \bar x &= \frac{\int_0^4 x\rho(x)\,\dee{x}}{\int_0^4 \rho(x)\,\dee{x}} = \frac{\int_0^4 3x^3\,\dee{x}}{\int_0^4 3x^2\,\dee{x}} = \frac{\left[\frac{3}{4}x^4\right]_0^4}{\left[\vphantom{\frac34}x^3\right]_0^4}=\frac{3\cdot 4^3}{4^3}=3 \end{align*}

The centre of mass of our rectangle is (3,1.5)(3,1.5).

Q28Stage 3

Suppose a circle of radius 3 m has density (2+y)(2+y) kg/m2^2 at any point yy metres above its bottom. Find the centre of mass of the circle.

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

Hint

Horizontal slices will help you, where symmetry doesn't, to set up a rod RR whose centre of mass is the same as one coordinate of the centre of mass of the circle. When you're integrating, trigonometric substitutions are sometimes the easiest way, and sometimes not.

The equation of a circle of radius 3, centred at (0,3)(0,3), is x2+(y−3)2=9x^2+(y-3)^2=9.

Answer

(0,3.45)(0,3.45)

Full solution

By symmetry, the xx-coordinate of the centre of mass will be xˉ=0\bar x =0; that is, exactly in the middle, horizontally. To find the yy-coordinate of the centre of mass, we need to consider the origin of Equation 2.3.3 in the CLP-2 text.

We can make vertical strips or horizontal strips. A vertical strip of the circle has a density that varies from the bottom of the strip to the top, but a horizontal strip has a constant density (assuming the strip is very thin). So it seems that horizontal strips in this case will be the easier route.

Following the derivation of Equation 2.3.3 in the CLP-2 text, we model our circle as a vertical rod RR, filling the yy-interval [0,6][0, 6]. A portion of the rod with a≤y≤ba \le y \le b should have the same mass as the portion of the circle with a≤y≤ba \le y \le b. To achieve this, we slice the circle into thin horizontal strips of thickness dy\dee{y}, calculate their mass, then use that to find ρ(y)\rho(y), the density of RR.

First, let's find a formula for the mass of a thin horizontal strip of the circle at position yy with height dy\dee{y}.

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

The circle with radius 3 centred at (0,3)(0,3) has equation x2+(y−3)2=9x^2+(y-3)^2=9. So, the right half of the circle has equation x=9−(y−3)2x=\sqrt{9-(y-3)^2}, and the left half of the circle has equation x=−9−(y−3)2x=-\sqrt{9-(y-3)^2}. So, the width of a strip at height yy is 29−(y−3)22\sqrt{9-(y-3)^2} m. Its height is dy\dee{y} m, so its area is 29−(y−3)2 dy2\sqrt{9-(y-3)^2}\,\dee{y} m2^2. Its density is 2+y kgm22+y~\frac{\text{kg}}{\text{m}^2} , so its mass is 2(2+y)9−(y−3)2 dy2(2+y)\sqrt{9-(y-3)^2}\,\dee{y} kg.

Now we can find ρ(y)\rho(y), the density of RR at position yy. The mass of the section of RR at position yy with length dy\dee{y} is 2(2+y)9−(y−3)2 dy2(2+y)\sqrt{9-(y-3)^2}\,\dee{y} kg (the mass of the strip in the paragraph above), so its density is 2(2+y)9−(y−3)2 dy kgdy m=2(2+y)9−(y−3)2 kgm=ρ(y)\frac{2(2+y)\sqrt{9-(y-3)^2}\,\dee{y}~\text{kg}}{\dee{y}~\text{m}} = 2(2+y)\sqrt{9-(y-3)^2}~\frac{\text{kg}}{\text{m}} = \rho(y).

Now, Equation 2.3.2 in the CLP-2 text will tell us the centre of mass of RR, which is also the yy-coordinate of the centre of mass of the circle.

yˉ=∫abyρ(y) dy∫abρ(y) dy=∫06y×2(2+y)9−(y−3)2 dy∫062(2+y)9−(y−3)2 dy=∫06y(2+y)9−(y−3)2 dy∫06(2+y)9−(y−3)2 dy\begin{align*}\bar y &= \frac{\int_a^b y\rho(y)\,\dee{y}}{\int_a^b \rho(y)\,\dee{y}} = \frac{\int_0^6 y\times 2(2+y)\sqrt{9-(y-3)^2}\,\dee{y}}{\int_0^6 2(2+y)\sqrt{9-(y-3)^2}\,\dee{y}}\\ &= \frac{\int_0^6 y(2+y)\sqrt{9-(y-3)^2}\,\dee{y}}{\int_0^6 (2+y)\sqrt{9-(y-3)^2}\,\dee{y}}\end{align*}

To make things look a little cleaner, we use the substitution u=y−3u=y-3, du=dy\dee{u}=\dee{y}. Then the limits of integration become −3-3 and 33, respectively, and y=u+3y=u+3. (Geometrically, we're re-centring the circle at the origin, instead of at the point (0,3).)

=∫−33(u+3)(2+u+3)9−u2 du∫−33(2+u+3)9−u2 du=∫−33(u2+8u+15)9−u2 du∫−33(u+5)9−u2 du=ND\begin{align*}&=\frac{\int_{-3}^3 (u+3)(2+u+3)\sqrt{9-u^2}\,\dee{u}}{\int_{-3}^3 \big(2+u+3\big)\sqrt{9-u^2}\,\dee{u}} \\&=\frac{\int_{-3}^3 \big(u^2+8u+15\big)\sqrt{9-u^2}\,\dee{u}}{\int_{-3}^3 \big(u+5\big)\sqrt{9-u^2}\,\dee{u}} = \frac{N}{D}\tag{$*$}\end{align*}

Let's start by finding DD, the integral of the denominator. If we break it into two pieces, we can use symmetry and geometry to evaluate it.

D=∫−33u9−u2 du+5∫−339−u2 du\begin{align*}D&=\int_{-3}^3 u\sqrt{9-u^2}\,\dee{u}+5\int_{-3}^3 \sqrt{9-u^2}\,\dee{u}\end{align*}

The left integrand is odd, so its integral over a symmetric interval is 0. (You can also evaluate this using the substitution w=9−u2w=9-u^2, dw=−2u du\dee{w}=-2u\,\dee{u}.) The right integral represents the area underneath half a circle of radius 3, centred at the origin.

D=0+5⋅12π⋅32=452π\begin{align*}D&=0+5\cdot\frac{1}{2}\pi\cdot 3^2 = \frac{45}{2}\pi\end{align*}

Now, let's evaluate our numerator integral from (∗*), N=∫−33(u2+8u+15)9−u2 duN=\int_{-3}^3 \big(u^2+8u+15\big)\sqrt{9-u^2}\,\dee{u}. If we break it into three pieces, we can simplify the integration somewhat.

N=∫−33u29−u2 du+8∫−33u9−u2 du+15∫−339−u2 du\begin{align*}N&=\int_{-3}^3 u^2\sqrt{9-u^2}\,\dee{u}+ 8\int_{-3}^3 u\sqrt{9-u^2}\,\dee{u}+ 15\int_{-3}^3 \sqrt{9-u^2}\,\dee{u}\end{align*}

The first integrand is even, with a symmetric interval of integration, so we can simplify its limits of integration a little bit. The middle integrand is odd, so its integral over the symmetric interval [−3,3][-3,3] is zero. The last integral is the area of half a circle of radius 3.

N=2∫03u29−u2 du+0+15⋅π⋅32=1352π+2∫03u29−u2 du\begin{align*}N&=2\int_{0}^3 u^2\sqrt{9-u^2}\,\dee{u}+0+15\cdot\pi\cdot3^2\\ &=\frac{135}{2}\pi+2\int_{0}^3 u^2\sqrt{9-u^2}\,\dee{u}\end{align*}

The remaining integral has a quadratic function underneath a square root with no obvious substitution, so we use a trigonometric substitution. Let u=3sin⁡θu=3\sin\theta, du=3cos⁡θ dθ\dee{u}=3\cos\theta\,\dee{\theta}. Note 3sin⁡(0)=03\sin(0)=0 and 3sin⁡(π/2)=33\sin(\pi/2)=3, so the limits of integration become 0 and π2\frac{\pi}{2}.

N=1352π+2∫0π/2(3sin⁡θ)29−(3sin⁡θ)2⋅3cos⁡θ dθ=1352π+2∫0π/29sin⁡2θ⋅9−9sin⁡2θ⋅3cos⁡θ dθ=1352π+54∫0π/2sin⁡2θ⋅9cos⁡2θ⋅cos⁡θ dθ=1352π+54∫0π/2sin⁡2θ⋅3cos⁡θ⋅cos⁡θ dθ=1352π+162∫0π/2sin⁡2θ⋅cos⁡2θ dθ\begin{align*}N&=\frac{135}{2}\pi+2\int_{0}^{\pi/2} \big(3\sin\theta\big)^2\sqrt{9-\big(3\sin\theta\big)^2}\cdot 3\cos\theta\,\dee{\theta}\\ &=\frac{135}{2}\pi+2\int_{0}^{\pi/2} 9\sin^2\theta \cdot \sqrt{9-9\sin^2\theta}\cdot 3\cos\theta\,\dee{\theta}\\ &=\frac{135}{2}\pi+54\int_{0}^{\pi/2} \sin^2\theta \cdot \sqrt{9\cos^2\theta}\cdot \cos\theta\,\dee{\theta}\\ &=\frac{135}{2}\pi+54\int_{0}^{\pi/2} \sin^2\theta \cdot 3\cos \theta\cdot \cos\theta\,\dee{\theta}\\ &=\frac{135}{2}\pi+162\int_{0}^{\pi/2} \sin^2\theta \cdot \cos^2 \theta\,\dee{\theta}\end{align*}

Using the identity sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta)=2\sin\theta\cos\theta, we see sin⁡2θcos⁡2θ=(sin⁡θcos⁡θ)2=14sin⁡2(2θ)\sin^2\theta\cos^2\theta =\big(\sin\theta\cos\theta\big)^2 = \frac{1}{4}\sin^2(2\theta)

N=1352π+162∫0π/214sin⁡2(2θ) dθ\begin{align*}N&=\frac{135}{2}\pi+162\int_{0}^{\pi/2} \frac{1}{4}\sin^2(2\theta)\,\dee{\theta}\end{align*}

Now, we use the identity sin⁡2x=12(1−cos⁡(2x))\sin^2 x = \frac{1}{2}(1-\cos(2x)), with x=2θ.x=2\theta.

N=1352π+162∫0π/218(1−cos⁡(4θ)) dθ=1352π+814∫0π/21−cos⁡(4θ) dθ=1352π+814[θ−14sin⁡(4θ)]0π/2=1352π+814(π2)=6218π\begin{align*}N&=\frac{135}{2}\pi+162\int_{0}^{\pi/2} \frac{1}{8}\big(1-\cos(4\theta)\big)\,\dee{\theta}\\ &=\frac{135}{2}\pi+\frac{81}{4}\int_{0}^{\pi/2} 1-\cos(4\theta)\,\dee{\theta}\\ &=\frac{135}{2}\pi+\frac{81}{4}\left[\theta-\frac{1}{4}\sin(4\theta)\right]_{0}^{\pi/2} \\ &=\frac{135}{2}\pi+\frac{81}{4}\left(\frac{\pi}{2}\right)\\ &=\frac{621}{8}\pi\end{align*}

Now, using equation (∗*), we find yˉ\bar y:

yˉ=ND=6218π452π=6920=3.45\begin{align*}\bar y &=\frac{N}{D} = \frac{\frac{621}{8}\pi}{\frac{45}{2}\pi} = \frac{69}{20}=3.45\end{align*}

Let's quickly check that this makes sense: if the circle has uniform density, its centre of mass would lie at (0,3)(0,3). Since it's denser at the top, the centre of mass should be higher, and indeed 3.453.45 is higher than 3 (without being so high it's above the entire circle).

Q29Stage 3

A right circular cone of uniform density has base radius rr m and height hh m. We want to find its centre of mass. By symmetry, we know that the centre of mass will occur somewhere along the straight vertical line through the tip of the cone and the centre of its base. The only question is the height of the centre of mass.

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

We will model the cone as a rod RR with height hh, such that the mass of the section of the rod from position aa to position bb is the same as the volume of the cone from height aa to height bb. (You can imagine that the cone is an umbrella, and we've closed it up to look like a cane. (This analogy isn't exact: if the cone were an umbrella, closing it would move the outside fabric vertically. A more accurate, but less familiar, image might be vacuum-wrapping an umbrella, watching it shrivel towards the middle but not move vertically.))

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

  1. Using this model, calculate how high above the base of the cone its centre of mass is.

  2. If we cut off the top h−kh-k metres of the cone (leaving an object of height kk), how high above the base is the new centre of mass?

Hint

The model in the question gives you the setup to solve this problem. You know how to find the centre of mass of a rod–that's Equation 2.3.2 in the CLP-2 text — so all you need to find is ρ(y)\rho(y), the density of the rod at position yy. To find this, consider a thin slice of the cone at position yy with thickness dy\dee{y}. Its volume V(y)V(y) is the same as the mass of the small section of the rod at position yy with thickness dy\dee{y}. So, the density of the rod at position yy is ρ(y)=V(y)dy\rho(y)=\frac{V(y)}{\dee{y}}.

Answer

(a) h4\dfrac{h}{4} (b) 12h2k−23hk2+14k3h2−hk+13k2\displaystyle\frac{\frac{1}{2}h^2k - \frac{2}{3}hk^2+\frac{1}{4}k^3}{h^2-hk+\frac{1}{3}k^2}

Full solution

  1. To find the centre of mass of the rod RR, we need to know its density at height yy, ρ(y)\rho(y). Since the mass of a section of RR is the same as the volume of a section of the cone, let's find the volume of a thin horizontal slice of the cone at height yy, with thickness dy.\dee{y}. To find its radius ss, we use similar triangles. The diagram below represents a vertical cross-section of the cone.

    Figure from prob_s2.3, line 2

    Figure from prob_s2.3, line 2

    Since rh=sh−y\frac{r}{h}=\frac{s}{h-y}, the radius of our slice at height yy is s=rh(h−y)s=\frac{r}{h}(h-y). Then the volume of the slice is πs2dy=π(rh(h−y))2 dy\pi s^2\dee{y}=\pi\left(\frac{r}{h}(h-y)\right)^2\,\dee{y}. Correspondingly, the mass of the piece of the rod at position yy with length dy\dee{y} is π(rh(h−y))2 dy\pi\left(\frac{r}{h}(h-y)\right)^2\,\dee{y}, so its density is

    ρ(y)=π(rh(h−y))2 dydy=π(rh(h−y))2.\rho(y) = \frac{\pi\left(\frac{r}{h}(h-y)\right)^2\,\dee{y}}{\dee{y}}=\pi\left(\frac{r}{h}(h-y)\right)^2.

    Now, we can find the centre of mass of RR:

    yˉ=∫0hyρ(y) dy∫0hρ(y) dy=∫0hyπ(rh(h−y))2 dy∫0hπ(rh(h−y))2 dy=r2h2π∫0hy(h−y)2 dyr2h2π∫0h(h−y)2 dy=∫0h(h2y−2hy2+y3) dy∫0h(h2−2hy+y2) dy=[h22y2−2h3y3+14y4]0h[h2y−hy2+13y3]0h=h42−2h43+h44h3−h3+13h3=h4h3⋅12−23+1413=h4\begin{align*} \bar y &= \frac{\int_0^h y\rho(y)\,\dee{y}}{\int_0^h \rho(y)\,\dee{y}} = \frac{\int_0^h y\pi\left(\frac{r}{h}(h-y)\right)^2\,\dee{y}}{\int_0^h \pi\left(\frac{r}{h}(h-y)\right)^2\,\dee{y}}\\ & = \frac{\frac{r^2}{h^2}\pi\int_0^h y\left(h-y\right)^2\,\dee{y}}{\frac{r^2}{h^2}\pi\int_0^h \left(h-y\right)^2\,\dee{y}}\\& = \frac{\int_0^h \left(h^2y-2hy^2+y^3\right)\,\dee{y}}{\int_0^h \left(h^2-2hy+y^2\right)\,\dee{y}}\\ &=\frac{\left[\frac{h^2}{2}y^2-\frac{2h}{3}y^3+\frac{1}{4}y^4\right]_0^h} {\left[h^2y-hy^2+\frac{1}{3}y^3\right]_0^h} \\&=\frac{\frac{h^4}{2}-\frac{2h^4}{3}+\frac{h^4}{4}} {h^3-h^3+\frac{1}{3}h^3}\\ &=\frac{h^4}{h^3}\cdot\frac{\frac{1}{2}-\frac{2}{3}+\frac{1}{4}}{\frac{1}{3}}=\frac{h}{4} \end{align*}

    So, the centre of mass of the cone occurs h4\frac{h}{4} metres above its base.

    Remark: it is quite interesting that the centre of mass does not depend on the radius of the cone!

  2. To find the centre of mass of a truncated cone, we simply consider a truncated rod. If the top h−kh-k metres are missing, then the height of the cone (and also the rod) is kk. Then the centre of mass has height:

    yˉ=∫0kyρ(y) dy∫0hρ(y) dy=∫0kyπ(rh(h−y))2 dy∫0hπ(rh(h−y))2 dy=r2h2π∫0ky(h−y)2 dyr2h2π∫0k(h−y)2 dy=∫0k(h2y−2hy2+y3) dy∫0k(h2−2hy+y2) dy=[h22y2−2h3y3+14y4]0k[h2y−hy2+13y3]0k=12h2k2−23hk3+14k4h2k−hk2+13k3=12h2k−23hk2+14k3h2−hk+13k2\begin{align*} \bar y &= \frac{\int_0^{k} y\rho(y)\,\dee{y}}{\int_0^h \rho(y)\,\dee{y}} = \frac{\int_0^{k} y\pi\left(\frac{r}{h}(h-y)\right)^2\,\dee{y}}{\int_0^h \pi\left(\frac{r}{h}(h-y)\right)^2\,\dee{y}}\\ & = \frac{\frac{r^2}{h^2}\pi\int_0^{k} y\left(h-y\right)^2\,\dee{y}}{\frac{r^2}{h^2}\pi\int_0^{k} \left(h-y\right)^2\,\dee{y}}\\& = \frac{\int_0^{k} \left(h^2y-2hy^2+y^3\right)\,\dee{y}}{\int_0^{k} \left(h^2-2hy+y^2\right)\,\dee{y}}\\ &=\frac{\left[\frac{h^2}{2}y^2-\frac{2h}{3}y^3+\frac{1}{4}y^4\right]_0^{k}} {\left[h^2y-hy^2+\frac{1}{3}y^3\right]_0^{k}}\\ &=\frac{\frac{1}{2}h^2k^2 - \frac{2}{3}hk^3+\frac{1}{4}k^4}{h^2k-hk^2+\frac{1}{3}k^3}\\ &=\frac{\frac{1}{2}h^2k - \frac{2}{3}hk^2+\frac{1}{4}k^3}{h^2-hk+\frac{1}{3}k^2} \end{align*}
Q30Stage 3

An hourglass is shaped like two identical truncated cones attached together. Their base radius is 5 cm, the height of the entire hourglass is 18 cm, and the radius at the thinnest point is .5.5 cm. The hourglass contains sand that fills up the bottom 6 cm when it's settled, with mass 600600 grams and uniform density. We want to know the work done flipping the hourglass smoothly, so the sand settles into a truncated, inverted-cone shape before it starts to fall down.

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

Using the methods of Section 2.1 to calculate the work done would be quite tedious. Instead, we will model the sand as a point of mass 0.60.6 kg, being lifted from the centre of mass of its original position to the centre of mass of its upturned position. Using the results of Question 29, how much work was done on the sand?

To simplify your calculation, you may assume that the height of the upturned sand (that is, the distance from the skinniest part of the hourglass to the top of the sand) is 8.8 cm. (Actually, it's 9373−1≈8.7854\sqrt[3]{937}-1\approx 8.7854 cm.) So, the top 0.2 cm of the hourglass is empty.

Hint

Use similar triangles to show that the shape of the lower (also upper) half of the hourglass is a truncated cone, where the untruncated cone would have had a height 10 cm.

To calculate the centre of mass of the upturned sand using the result of Question 29, you should find h=9.8h=9.8 (not h=10h=10–think carefully about our model from Question 29) and k=8.8k=8.8. For the centre of mass of the sand before turning, h=10h=10 and k=6k=6.

Answer

about 0.833 N

Full solution

To use the result of Question 29, we need to know the dimensions of the cone that was truncated to make the hourglass. The bottom (or top) half of our hourglass has base radius 5 cm, height 9 cm, and top radius 0.5 cm. Imagine extending it to a full cone. Let tt be the distance from the top of the half hourglass to the tip of the full cone.

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

Using similar triangles,

t0.5=t+95so 5t=12(t+9)4.5t=4.5t=1\begin{align*}\dfrac{t}{0.5} &= \dfrac{t+9}{5}\\ \text{so }\qquad 5t&=\frac{1}{2}(t+9)\\ 4.5t&=4.5\\ t&=1 \end{align*}

Then the height of the full cone (that we imagined truncating to make half of the hourglass) is h=10h=10 cm.

Before the hourglass is turned over, the sand forms a truncated cone of height 66 cm. So, it's the bottom k=6k=6 cm of a cone of height h=10h=10 cm. Using the result of Question 29, its centre of mass is at height:

12h2k−23hk2+14k3h2−hk+13k2=12102⋅6−2310⋅62+1463102−10⋅6+1362=5726≈2.2\begin{align*} \frac{\frac{1}{2}h^2k - \frac{2}{3}hk^2+\frac{1}{4}k^3}{h^2-hk+\frac{1}{3}k^2}&= \frac{\frac{1}{2}10^2\cdot 6 - \frac{2}{3}10\cdot 6^2+\frac{1}{4}6^3}{10^2-10\cdot 6+\frac{1}{3}6^2}=\textcolor{red}{\frac{57}{26}}\approx2.2 \end{align*}

Next, let's find the centre of mass of the sand after it's been rotated. We have to be a little careful with our vocabulary here: usually we imagine a cone sitting on its base, with its tip pointing up. The upturned sand is in the opposite configuration. When we say the “base" of the cone, we mean the larger horizontal face–the top of the sand as it sits in the hourglass.

The formula we have from Question 29 gives us our centre of mass as a distance from the base of the truncated cone (that is, the distance from the top of the upturned sand). If kk is the height the sand actually occupies, then we were told we may assume k=8.8k=8.8 cm. It's missing its “tip" of height 1 cm, so hh, the height of the “untruncated" cone, is 9.89.8 cm. Using our model from Question 29, we don't care about the empty, uppermost piece of the hourglass. The shape of the sand is of a cone of height 9.8 cm (not 10 cm), with a tip of height 11 cm chopped off.

Figure from prob_s2.3, line 1

Figure from prob_s2.3, line 1

12h2k−23hk2+14k3h2−hk+13k2=129.82⋅8.8−239.8⋅8.82+148.839.82−9.8⋅8.8+138.82≈2.443\begin{align*} \frac{\frac{1}{2}h^2k - \frac{2}{3}hk^2+\frac{1}{4}k^3}{h^2-hk+\frac{1}{3}k^2}&= \frac{\frac{1}{2}9.8^2\cdot 8.8 - \frac{2}{3}9.8\cdot 8.8^2+\frac{1}{4}8.8^3}{9.8^2-9.8\cdot 8.8+\frac{1}{3}8.8^2}\approx 2.443 \end{align*}

That is, the centre of mass of the upturned sand is about 2.4432.443 centimetres below its top, which is at height 8.8+10=18.88.8+10=18.8 cm above the very bottom of the hourglass. So, the centre of mass of the upturned sand is at height y=18.8−2.443=16.357y=18.8-2.443= \textcolor{blue}{16.357} cm.

Now, we have our model: the sand, viewed as a point mass, is moved from y=5726y=\textcolor{red}{\frac{57}{26}} to y=16.357y=\textcolor{red}{16.357} cm. That is, it moved about 14.16514.165 cm, or about 0.141650.14165 m. It has a mass of 0.60.6 kg, so the force required to lift it against gravity is

(0.6 kg)×(9.8msec2)×(0.14165 m)≈0.833 newtons\left(0.6~\text{kg}\right)\times \left(9.8 \frac{\text{m}}{\text{sec}^2}\right)\times\left(0.14165~\text{m}\right)\approx 0.833~\text{newtons}
Q31Stage 3

Tank AA is in the shape of half a sphere of radius 1 metre, with its flat face resting on the ground, and is completely filled with water. Tank BB is empty and rectangular, with a square base of side length 1 m and a height of 3 m.

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

  1. To pump the water from Tank AA to Tank BB, we need to pump all the water from Tank AA to a height of 3 m. How much work is done to pump all the water from Tank AA to a height of 3 m? You may model the water as a point mass, originally situated at the centre of mass of the full Tank AA.

  2. Suppose we could move the water from Tank AA directly to its final position in Tank BB without going over the top of Tank BB. (For example, maybe tank AA is elastic, and Tank BB is just Tank AA after being smooshed into a different form.) How much work is done pumping the water? (That is, how much work is done moving a point mass from the centre of mass of Tank AA to the centre of mass of Tank BB?)

  3. What percentage of work from part (a) was “wasted" by pumping the water over the top of Tank BB, instead of moving it directly to its final position?

You may assume that the only work done is against the acceleration due to gravity, g=9.8g=9.8 m/sec2^2, and that the density of water is 1000 kg/m3^3.

Remark: the answer from (b) is what you might think of as the net work involved in pumping the water from Tank AA to Tank BB. When work gets “wasted," the pump does some work pumping water up, then gravity does equal and opposite work bringing the water back down.

Hint

The techniques of Section 2.1 get pretty complicated here, so it's easiest to use the techniques we developed in Questions 6, 29, and 30 in this section. That is, (1) find the height of the centre of mass of the water in its starting and ending positions, and then (2) model the work done as the work moving a point mass with the weight of the water from the first centre of mass to the second.

The height change of the centre of mass is all that matters to calculate the work done against gravity, so you only have to worry about the height of the centres of mass.

Answer

(a) 17,150π\pi J (b) 24509π(8π−9)≈13,797 J\displaystyle\frac{2450}{9}\pi\left(8\pi-9\right)\approx 13,797\text{ J} (c) about 74%

Full solution

The techniques of Section 2.1 get pretty complicated here, so we will use the techniques we developed in Questions 6, 29 and 30 in this section. That is, (1) find the centre of mass of the water in its starting and ending positions, and then (2) compute the work done as the work moving a point mass with the weight of the water from the first centre of mass to the second. For the centre of mass, all we need to know is the height–for one thing, we could find the other coordinates by symmetry, but we don't need them. The height moved by the water is all that matters if we're calculating the work done opposing gravity.

Let's start by calculating the volume of the water. The volume of a sphere of radius 1 is 43π⋅13\frac{4}{3}\pi\cdot 1^3, so the volume of water is 23π\frac{2}{3}\pi m3^3.

Then the mass of the water is 20003π\frac{2000}{3}\pi kg.

Next, we calculate the centre of mass of Tank AA, and the work done to pump the water out of Tank AA to a height of 3 metres. Symmetry alone won't tell us the height of the centre of mass. We'll show you two ways to go about this.

  • As in Question 29, we'll model the tank of water as a vertical rod, along the yy–axis spanning the interval [0,1][0,1], such that the mass of a piece of the rod along [a,b][a,b] is the same as the mass of the water from height y=ay=a to height y=by=b. Then, the centre of mass of the rod will be the same as the centre of mass of the water.

    Consider a horizontal slice of water at height yy, with thickness dy\dee{y}. If the radius of this slice is r(y)r(y), then the volume of the slice is πr(y)2 dy\pi r(y)^2\,\dee{y} m3^3, so its mass is 1000πr(y)2 dy1000\pi r(y)^2\,\dee{y} kg. Then the mass of the slice of the rod at position yy with length dy\dee{y} is 1000πr(y)2 dy1000\pi r(y)^2\,\dee{y} kg, so its density ρ(y)\rho(y) is

    ρ(y)=1000πr(y)2 dy kgdy m=1000πr(y)2 kgm.\rho(y) = \frac{1000\pi r(y)^2\,\dee{y}\text{ kg}}{\dee{y}\text{ m}} = 1000\pi r(y)^2 ~\frac{\text{kg}}{\text{m}}.

    So, let's find r(y)r(y), the radius of the slice of water at height yy.

    Figure from prob_s2.3, line 2

    Figure from prob_s2.3, line 2

    Using the Pythagorean Theorem, r=1−y2r = \sqrt{1-y^2}. Therefore,

    ρ(y)=1000π(1−y2)\rho(y)= 1000\pi(1-y^2)

    We use Equation 2.3.2 in the CLP-2 text to calculate the centre of mass of the rod, which is the height of the centre of mass of Tank AA:

    yˉA=∫01yρ(y) dy∫01ρ(y) dy=∫011000πy(1−y2) dy∫011000π(1−y2) dy=∫01(y−y3) dy∫01(1−y2) dy=[12y2−14y4]01[y−13y3]01=12−141−13=38 m\begin{align*} \bar y_A&=\frac{\int_0^1 y\rho(y)\,\dee{y}}{\int_0^1 \rho(y)\,\dee{y}} = \frac{\int_0^1 1000\pi y(1-y^2)\,\dee{y}}{\int_0^1 1000\pi(1-y^2)\,\dee{y}}\\ &=\frac{\int_0^1(y-y^3)\,\dee{y}}{\int_0^1(1-y^2)\,\dee{y}} = \frac{\left[\frac{1}{2}y^2 - \frac{1}{4}y^4\right]_0^1}{\left[y-\frac{1}{3}y^3\right]_0^1} = \frac{\frac{1}{2}-\frac{1}{4}}{1-\frac{1}{3}}=\frac{3}{8}~\text{m} \end{align*}

    From here, we can find the work done moving pumping the water to a height of 3 metres. We've moved the centre of mass from yˉA=38\bar y_A = \frac{3}{8} metres to 3 metres.

    W=(20003π kg)×(3−38 m)×(9.8 msec2)=17,150π J\begin{align*} W&=\left(\frac{2000}{3}\pi~\text{kg}\right)\times\left(3 - \frac{3}{8}~\text{m}\right)\times\left(9.8 ~\frac{\text{m}}{\text{sec}^2}\right)\\ &=17,150\pi~\text{J} \end{align*}

  • We can use the techniques of Section 2.1 in the CLP-2 text to calculate the amount of work it takes to pump the water from tank AA to a height of 3 metres. That solves part (a), and we can use the amount of work to figure out the centre of gravity of the water in Tank AA to help us solve part (b).

    At height yy, a horizontal layer of water in Tank AA forms a disk with thickness dy\dee{y} and radius 1−y2\sqrt{1-y^2}. (The radius comes from the Pythagorean Theorem–see the diagram below.)

    Figure from prob_s2.3, line 1

    Figure from prob_s2.3, line 1

    The volume of the layer at height yy is π(1−y2)2 dy=π(1−y2) dy\pi \left(\sqrt{1-y^2}\right)^2\,\dee{y} = \pi(1-y^2)\,\dee{y} m3^3, so its mass is 1000π(1−y2) dy1000\pi(1-y^2)\,\dee{y} kg.

    The layer at height yy needs to be pumped a distance of 3−y3-y metres. So, the work involved pumping the layer at height yy is:

    dW=(1000π(1−y2) dy kg)×(3−y m)×(9.8 m/sec2)=9800π(y3−3y2−y+3) dy J\begin{align*}\dee{W}&=\left(1000\pi(1-y^2)\,\dee{y}~\text{kg}\right)\times\left(3-y~\text{m}\right)\times\left(9.8~\text{m}/\text{sec}^2\right)\\ &=9800\pi(y^3-3y^2-y+3)\,\dee{y}\text{ J}\end{align*}

    Then the work involved pumping out the entire tank to a height of 3 metres is:

    W=∫019800π(y3−3y2−y+3) dy=9800π[14y4−y3−12y2+3y]01=17,150π J\begin{align*}W&=\int_0^1 9800\pi(y^3-3y^2-y+3)\,\dee{y}\\ &=9800\pi\left[\frac{1}{4}y^4 - y^3-\frac{1}{2}y^2+3y\right]_0^1\\ &=17,150\pi~\text{J}\end{align*}

    This gives us an answer to part (a). To find the centre of mass of the water in Tank AA, note that the work done is equivalent to moving a point mass from the centre of mass of the tank to a height of 3 metres. We know the water in Tank AA has mass 20003π\frac{2000}{3}\pi kg. So, if yˉA\bar y_A is the centre of mass of the water in Tank AA:

    W=(20003π kg)×(3−yˉA m)×(9.8 m/sec2)17,150π=(20003π)(3−yˉA)(9.8)218=3−yˉAyˉA=38 m\begin{align*} W&=\left(\frac{2000}{3}\pi~\text{kg}\right)\times\left(3-\bar y_A~\text{m}\right)\times\left(9.8~\text{m}/\text{sec}^2\right)\\ 17,150\pi&=\left(\frac{2000}{3}\pi\right)\left(3-\bar y_A\right)(9.8)\\ \frac{21}{8}&=3-\bar y_A\\ \bar y_A &= \frac{3}{8}~\text{m} \end{align*}

Next let's calculate the centre of mass of the water in Tank BB. Since the volume of the water in Tank BB is 23π\frac{2}{3}\pi m3^3, and the base of Tank BB has area 1 m2^2, the height of the water in Tank BB is 23π\frac{2}{3}\pi m. Since the water is of uniform density, and Tank BB has uniform horizontal cross-sections, by symmetry the centre of mass of the water in Tank BB is at

yˉB=13π m.\bar y_B = \frac{1}{3}\pi\text{ m}.

Now, we can calculate the work done by moving the water directly from Tank AA to its final position in Tank BB. The work done moving a point mass of 20003π\frac{2000}{3}\pi kg a distance of yˉB−yˉA=13π−38\bar y_B - \bar y_A = \frac{1}{3}\pi-\frac{3}{8} m against the gravity, g=9.8g=9.8 m/sec2^2, is:

W=(20003π kg)×(13π−38 m)×(9.8m/sec2)=24509π(8π−9)≈13,797 J\begin{align*}W &= \left(\frac{2000}{3}\pi\text{ kg}\right)\times\left( \frac{1}{3}\pi-\frac{3}{8} \text{ m}\right)\times\left(9.8 \text{m/sec}^2\right)\\ &=\frac{2450}{9}\pi\left(8\pi-9\right)\approx 13,797\text{ J} \end{align*}

Finally, the “wasted" work is:

ΔW=17,150π−24509π(8π−9)=2450π(7−8π−99)=2450π(8−8π9)=19,600π(1−π9)\begin{align*}\Delta W &=17,150\pi -\frac{2450}{9}\pi\left(8\pi-9\right)\\ &=2450\pi\left(7-\frac{8\pi-9}{9}\right)\\ &=2450\pi\left(8-\frac{8\pi}{9}\right)\\ &=19,600\pi\left(1-\frac{\pi}{9}\right)\end{align*}

As a percentage of 17,150π\pi, this is:

waste=(19,600π(1−π9)17,150π)×100=87(1−π9)×100≈74%\begin{align*}\text{waste}&= \left(\frac{19,600\pi\left(1-\frac{\pi}{9}\right)}{17,150\pi}\right)\times 100\\ &=\frac{8}{7}\left(1-\frac{\pi}{9}\right)\times 100\approx 74\%\end{align*}
Q32Stage 3

Let RR be the region bounded above by y=2xsin⁡(x2)y=2x\sin (x^2) and below by the xx-axis, 0≤x≤π20 \le x \le \sqrt{\frac{\pi}{2}}. Give an approximation of the xx-value of the centroid of RR with error no more than 1100\frac{1}{100}.

You may assume without proof that ∣d4dx4{2x2sin⁡(x2)}∣≤415\left|\ddiff{4}{}{x}\left\{2x^2\sin(x^2)\right\} \right| \leq 415 over the interval [0,π2]\left[0,\sqrt{\frac{\pi}{2}}\right].

Hint

The area of RR is precisely one, so the error in your approximation is the error involved in approximating ∫0π/22x2sin⁡(x2) dx\int_0^{\sqrt{\pi/2}}2x^2\sin(x^2)\,\dee{x}.

Answer

xˉ=π162π2[sin⁡(π72)+2sin⁡(π18)+9sin⁡(π8)+8sin⁡(2π9)+25sin⁡(25π72)+9]≈0.976\displaystyle \bar x =\frac{\pi}{162}\sqrt{\frac{\pi}{2}}\bigg[ \sin\left(\frac{\pi}{72}\right)+2\sin\left(\frac{\pi}{18}\right)+9\sin\left(\frac{\pi}{8}\right)+ 8\sin\left(\frac{2\pi}{9}\right)+25\sin\left(\frac{25\pi}{72}\right) +9\bigg]\approx 0.976

Full solution

Using Equation 2.3.3 in the CLP-2 text with T(x)=2xsin⁡(x2)T(x) = 2x\sin (x^2) and B(x)=0B(x)=0,

xˉ=∫0π/22x2sin⁡(x2) dx∫0π/22xsin⁡(x2) dx\bar x = \frac{\displaystyle\int_0^{\sqrt{\pi/2}} 2x^2\sin(x^2)\,\dee{x}}{\displaystyle\int_0^{\sqrt{\pi/2}} 2x\sin(x^2)\,\dee{x}}

We can evaluate the bottom integral exactly with the substitution u=x2u=x^2, du=2xdx\dee{u}=2x\dee{x}. When x=0x=0, u=0u=0, and when x=π/2x=\sqrt{\pi/2}, u=π/2u=\pi/2.

∫0π/22xsin⁡(x2) dx=∫0π/2sin⁡u du=[−cos⁡u]0π/2=1\begin{align*} \int_0^{\sqrt{\pi/2}}2x\sin(x^2)\,\dee{x}&=\int_0^{\pi/2} \sin u\,\dee{u} =\Big[-\cos u\Big]_0^{\pi/2}=1\\ \end{align*}

So,

xˉ=∫0π/22x2sin⁡(x2) dx\bar x = \int_0^{\sqrt{\pi/2}} 2x^2\sin(x^2)\,\dee{x}

Evaluating the integral ∫x2sin⁡(x2) dx\int x^2\sin(x^2)\,\dee{x} is not so simple (Indeed, the antiderivative of 2x2sin⁡(x2)2x^2\sin(x^2) is not expressible as an elementary function.), so we use a numerical approximation. Since we're given an upper bound on the fourth derivative, we decide to use Simpson's rule. The error involved using Simpson's rule with nn intervals is at most L(b−a)5180n4\frac{L(b-a)^5}{180n^4}. For our approximation, a=0a=0 and b=π/2b=\sqrt{\pi/2}. According to the information given in the problem statement, ∣d4dx4{2x2sin⁡(x2)}∣≤415\left|\ddiff{4}{}{x}\left\{2x^2\sin(x^2)\right\} \right| \leq 415 over the interval [0,π2]\left[0,\sqrt{\frac{\pi}{2}}\right], so we set L=415L=415.

We want our final error to be no more than 1100\frac{1}{100}, so we want to find an even nn such that:

415(π2−0)5180n4≤1100n4≥415⋅100(π2)5/2180=2075π5/2362n≥2075π5/23624≈5.17\begin{align*} \frac{415\left(\sqrt{\frac{\pi}{2}}-0\right)^5}{180n^4} &\leq \frac{1}{100}\\ n^4 &\geq \frac{415\cdot 100\left(\frac{\pi}{2}\right)^{5/2}}{180} = \frac{2075\pi^{5/2}}{36\sqrt{2}}\\ n&\geq \sqrt[4]{\frac{2075\pi^{5/2}}{36\sqrt{2}}}\approx 5.17 \end{align*}

So, n=6n=6 intervals suffices. Then Δx=b−a6=16π2\De x = \frac{b-a}{6} = \frac{1}{6}\sqrt{\frac{\pi}{2}} and our grid points are x0=0x_0=0, x1=16π2x_1=\frac{1}{6}\sqrt{\frac{\pi}{2}}, x2=13π2x_2=\frac{1}{3}\sqrt{\frac{\pi}{2}}, x3=12π2x_3=\frac{1}{2}\sqrt{\frac{\pi}{2}}, x4=23π2x_4=\frac{2}{3}\sqrt{\frac{\pi}{2}}, x5=56π2x_5=\frac{5}{6}\sqrt{\frac{\pi}{2}}, and , x6=π2x_6=\sqrt{\frac{\pi}{2}}.

Figure from prob_s2.3, line 2

Figure from prob_s2.3, line 2

Following Equation 1.11.9 in the CLP-2 text, the Simpson's rule approximation of ∫0π/22x2sin⁡(x2) dx\displaystyle\int_0^{\sqrt{\pi/2}}2x^2\sin(x^2)\,\dee{x} is:

Δx3[f(x0)+4f(x1)+2f(x2)+4f(x3)+2f(x4)+4f(x5)+f(x6)]=16π2⋅13[0+4×2π72sin⁡(π72)+2×2π18sin⁡(π18)+4×2π8sin⁡(π8)+2×8π18sin⁡(4π18)+4×50π72sin⁡(25π72)+2π2sin⁡(π2)]=118π2[π9sin⁡(π72)+2π9sin⁡(π18)+πsin⁡(π8)+8π9sin⁡(2π9)+25π9sin⁡(25π72)+π]=π18π2[19sin⁡(π72)+29sin⁡(π18)+sin⁡(π8)+89sin⁡(2π9)+259sin⁡(25π72)+1]=π162π2[sin⁡(π72)+2sin⁡(π18)+9sin⁡(π8)+8sin⁡(2π9)+25sin⁡(25π72)+9]≈0.976\begin{align*} &\frac{\De x}{3}\Big[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+4f(x_5)+f(x_6)\Big]\\ =&\frac{1}{6}\sqrt{\frac{\pi}{2}}\cdot\frac{1}{3}\bigg[ 0+4\times \frac{2\pi}{72}\sin\left(\frac{\pi}{72}\right)+2\times\frac{2\pi}{18}\sin\left(\frac{\pi}{18}\right)+4\times\frac{2\pi}{8}\sin\left(\frac{\pi}{8}\right)+ 2\times \frac{8\pi}{18}\sin\left(\frac{4\pi}{18}\right)\\ &+4\times \frac{50\pi}{72}\sin\left(\frac{25\pi}{72}\right) +\frac{2\pi}{2}\sin\left(\frac{\pi}{2}\right) \bigg]\\=&\frac{1}{18}\sqrt{\frac{\pi}{2}}\bigg[ \frac{\pi}{9}\sin\left(\frac{\pi}{72}\right)+\frac{2\pi}{9}\sin\left(\frac{\pi}{18}\right)+\pi\sin\left(\frac{\pi}{8}\right)+ \frac{8\pi}{9}\sin\left(\frac{2\pi}{9}\right)\\ &+\frac{25\pi}{9}\sin\left(\frac{25\pi}{72}\right) +\pi\bigg]\\=&\frac{\pi}{18}\sqrt{\frac{\pi}{2}}\bigg[ \frac{1}{9}\sin\left(\frac{\pi}{72}\right)+\frac{2}{9}\sin\left(\frac{\pi}{18}\right)+\sin\left(\frac{\pi}{8}\right)+ \frac{8}{9}\sin\left(\frac{2\pi}{9}\right)+\frac{25}{9}\sin\left(\frac{25\pi}{72}\right) +1\bigg]\\=&\frac{\pi}{162}\sqrt{\frac{\pi}{2}}\bigg[ \sin\left(\frac{\pi}{72}\right)+2\sin\left(\frac{\pi}{18}\right)+9\sin\left(\frac{\pi}{8}\right)+ 8\sin\left(\frac{2\pi}{9}\right)+25\sin\left(\frac{25\pi}{72}\right) +9\bigg]\\ &\approx 0.976 \end{align*}

The absolute error in our answer is at most:

L(b−a)5180n4=415×π25180×64=82π51866242≈0.005\begin{align*} \frac{L(b-a)^5}{180n^4}&=\frac{415\times\sqrt{\frac{\pi}{2}}^5}{180\times 6^4} = \frac{82\sqrt{\pi}^5}{186624\sqrt{2}}\approx 0.005 \end{align*}

Remark: combining the error with our approximation, we see the actual value of xˉ\bar x is in the interval

[0.976−0.005,0.976+0.005]=[0.971,.981]\left[0.976-0.005,0.976+0.005\right] = \left[0.971,.981\right]

A computer algebra system approximates xˉ\bar x as 0.977451.

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From the UBC Math 101 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.