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Integration

1.6 Volumes

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

1Stage 1Conceptual

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

Q1Stage 1

Consider a right circular cone.

Figure from prob_s1.6, line 2

Figure from prob_s1.6, line 2

What shape are horizontal cross-sections? Are the vertical cross-sections the same?

Hint

The horizontal cross-sections were discussed in Example 1.6.1 of the CLP-2 text.

Answer

The horizontal cross-sections are circles, but the vertical cross-sections are not.

Full solution

If we take a horizontal slice of a cone, we get a circle. If we take a vertical cross-section, the base is flat (it's a chord on the circular base of the cone), so we know right away it isn't a circle. Indeed, if we slice down through the very centre, we get a triangle. (Other vertical slices have a curvy top, corresponding to a class of curves known as hyperbolas.)

Q2Stage 1

Two potters start with a block of clay hh units tall, and identical square cookie cutters. They form columns by pushing the square cookie cutter straight down over the clay, so that its cross-section is the same square as the cookie cutter. Potter A pushes their cookie cutter down while their clay block is sitting motionless on a table; Potter B pushes their cookie cutter down while their clay block is rotating on a potter's wheel, so their column looks twisted. Which column has greater volume?

Figure from prob_s1.6, line 3

Figure from prob_s1.6, line 3

Figure from prob_s1.6, line 11

Figure from prob_s1.6, line 11

Hint

What are the dimensions of the cross-sections?

Answer

The columns have the same volume.

Full solution

The columns have the same volume. We can see this by chopping up the columns into horizontal cross-sections. Each cross-section has the same area as the cookie cutter, AA, and height dy\dee{y}. Then in both cases, the volume of the column is

∫0hA dy=hA cubic units\int_{0}^h A~\dee{y} = hA \text{ cubic units}
Q3Stage 1

Let RR be the region bounded above by the graph of y=f(x)y=f(x) shown below and bounded below by the xx-axis, from x=0x=0 to x=6x=6. Sketch the washers that are formed by rotating RR about the yy-axis. In your sketch, label all the radii in terms of yy, and label the thickness.

Figure from prob_s1.6, line 2

Figure from prob_s1.6, line 2

Hint

There are two different kinds of washers.

Answer

  • If y>1y>1, then our washer has inner radius 2+23y2+\frac{2}{3}y, outer radius 6−23y6-\frac{2}{3}y, and height dy\dee{y}.

    Figure from prob_s1.6, line 2

    Figure from prob_s1.6, line 2

  • When 0≤y<10 \le y < 1, we have a “double washer," two concentric rings. The inner washer has inner radius r1=yr_1=y and outer radius R1=2−yR_1=2-y. The outer washer has inner radius r2=2+23yr_2=2+\frac{2}{3}y and outer radius R2=6−23yR_2=6-\frac{2}{3}y. The thickness of the washers is dy\dee{y}.

    Figure from prob_s1.6, line 2

    Figure from prob_s1.6, line 2

Full solution

Notice f(x)f(x) is a piecewise linear function, so we can find explicit equations for each of its pieces from the graph. The radii will be determined by the xx-values, so below we give the xx-values as functions of yy.

Figure from prob_s1.6, line 2

Figure from prob_s1.6, line 2

If we imagine rotating the region from the picture about the yy-axis, there will be two kinds of washers formed: when y<1y<1, we have a “double washer," two concentric rings. When y>1y>1, we have a single ring.

  • If y>1y>1, then our washer has inner radius 2+23y2+\frac{2}{3}y, outer radius 6−23y6-\frac{2}{3}y, and height dy\dee{y}.

    Figure from prob_s1.6, line 2

    Figure from prob_s1.6, line 2

  • When 0≤y<10 \le y < 1, we have a “double washer," two concentric rings corresponding to the two “humps" in the function. The inner washer has inner radius r1=yr_1=y and outer radius R1=2−yR_1=2-y. The outer washer has inner radius r2=2+23yr_2=2+\frac{2}{3}y and outer radius R2=6−23yR_2=6-\frac{2}{3}y. The thickness of the washers is dy\dee{y}.

    Figure from prob_s1.6, line 2

    Figure from prob_s1.6, line 2

Q4Stage 1Past exam · 2000D

Write down definite integrals that represent the following quantities. Do not evaluate the integrals explicitly.

  1. The volume of the solid obtained by rotating around the xx–axis the region between the xx–axis and y=x ex2y=\sqrt{x}\, e^{x^2} for 0≤x≤30\le x\le 3.

  2. The volume of the solid obtained by revolving the region bounded by the curves y=x2y=x^2 and y=x+2y=x+2 about the line x=3x=3.

Hint

Draw sketches. The mechanically easiest way to answer part (b) uses the method of cylindrical shells, which is in the optional section 1.6 of the CLP-2 text. The method of washers also works, but requires you to have more patience and also to have a good idea what the specified region looks like. Look at your sketch very careful when identifying the ends of your horizontal strips.

Answer

(a) π∫03xe2x2 dx\pi\displaystyle\int_{0}^{3} xe^{2x^2}\ \dee{x}

(b) ∫01π[(3+y)2−(3−y)2]dy+∫14π[(5−y)2−(3−y)2]dy\displaystyle\int _0^1 \pi\big[\big(3+\sqrt{y}\big)^2-\big(3-\sqrt{y}\big)^2\big]\dee{y} +\displaystyle\int _ 1^4 \pi\big[\big(5-y\big)^2-\big(3-\sqrt{y}\big)^2\big]\dee{y}

Full solution

(a) When the strip shown in the figure

Figure from prob_s1.6, line 295

Figure from prob_s1.6, line 295

is rotated about the xx–axis, it forms a thin disk of radius xex2\sqrt{x}e^{x^2} and thickness dx\dee{x} and hence of cross sectional area πxe2x2\pi xe^{2x^2} and volume πxe2x2 dx\pi xe^{2x^2}\,\dee{x} So the volume of the solid is

π∫03xe2x2 dx\begin{align*} \pi\int_{0}^{3} xe^{2x^2}\ \dee{x} \end{align*}

(b) The curves intersect at (−1,1)(-1,1) and (2,4)(2,4).

Figure from prob_s1.6, line 295

Figure from prob_s1.6, line 295

We'll use horizontal washers as in Example 1.6.5 of the CLP-2 text.

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

  • When we rotate about the line x=3x=3, each strip sweeps out a thin washer

    • whose inner radius is rin=3−yr_{in}=3-\sqrt{y}, and

    • whose outer radius is rout=3−(y−2)=5−yr_{out}=3-(y-2)=5-y when y≥1y\ge 1 (see the red strip in the figure on the right above), and whose outer radius is rout=3−(−y)=3+yr_{out}= 3-(-\sqrt{y})=3+\sqrt{y} when y≤1y\le 1 (see the blue strip in the figure on the right above) and

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

    • whose volume is π(rout2−rin2)dy=π[(5−y)2−(3−y)2]dy\pi(r_{out}^2 - r_{in}^2)\dee{y} = \pi\big[\big(5-y\big)^2-\big(3-\sqrt{y}\big)^2\big]\dee{y} when y≥1y\ge 1 and whose volume is π(rout2−rin2)dy=π[(3+y)2−(3−y)2]dy\pi(r_{out}^2 - r_{in}^2)\dee{y} =\pi\big[\big(3+\sqrt{y}\big)^2-\big(3-\sqrt{y}\big)^2\big]\dee{y} when y≤1y\le 1 and

  • As our bottommost strip is at y=0y=0 and our topmost strip is at y=4y=4, the total volume is

    ∫01π[(3+y)2−(3−y)2]dy+∫14π[(5−y)2−(3−y)2]dy\begin{align*} \int _0^1 \pi\big[\big(3+\sqrt{y}\big)^2-\big(3-\sqrt{y}\big)^2\big]\dee{y} +\int _ 1^4 \pi\big[\big(5-y\big)^2-\big(3-\sqrt{y}\big)^2\big]\dee{y} \end{align*}
Q5Stage 1Past exam · 2001A,M121 2001A

Write down definite integrals that represent the following quantities. Do not evaluate the integrals explicitly.

  1. The volume of the solid obtained by rotating the finite plane region bounded by the curves y=1−x2y=1-x^2 and y=4−4x2y=4-4x^2 about the line y=−1y=-1.

  2. The volume of the solid obtained by rotating the finite plane region bounded by the curve y=x2−1y=x^2-1 and the line y=0y=0 about the line x=5x=5.

Hint

Draw sketchs.

Answer

(a) ∫−11π[(5−4x2)2−(2−x2)2] dx\displaystyle\int_{-1}^{1}\pi\big[{(5-4x^2)}^2-{(2-x^2)}^2\big]\,\dee{x} (b) ∫−10π[(5+y+1)2−(5−y+1)2] dy\displaystyle\int _{-1}^0 \pi\big[\big(5+\sqrt{y+1}\big)^2-\big(5-\sqrt{y+1}\big)^2\big]\,\dee{y}

Full solution

(a) The curves intersect at (1,0)(1,0) and (−1,0)(-1,0). When the strip shown in the figure

Figure from prob_s1.6, line 381

Figure from prob_s1.6, line 381

is rotated about the line y=−1y=-1, it forms a thin washer with:

  • inner radius (1−x2)−(−1)=2−x2(1-x^2)-(-1)=2-x^2,

  • outer radius (4−4x2)−(−1)=5−4x2(4-4x^2)-(-1)=5-4x^2 and

  • thickness dx\dee{x} ; so, it has

  • cross sectional area π[(5−4x2)2−(2−x2)2]\pi\big[{(5-4x^2)}^2-{(2-x^2)}^2\big] and

  • volume π[(5−4x2)2−(2−x2)2] dx\pi\big[{(5-4x^2)}^2-({2-x^2)}^2\big]\,\dee{x}.

So the volume of the solid is

∫−11π[(5−4x2)2−(2−x2)2] dx\begin{align*} \int_{-1}^{1}\pi\big[{(5-4x^2)}^2-{(2-x^2)}^2\big]\,\dee{x} \end{align*}

(b) The curve y=x2−1y=x^2-1 intersects y=0y=0 at (1,0)(1,0) and (−1,0)(-1,0).

Figure from prob_s1.6, line 381

Figure from prob_s1.6, line 381

We'll use horizontal washers.

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

  • When we rotate about the line x=5x=5, each strip sweeps out a thin washer

    • whose inner radius is rin=5−y+1r_{in}=5-\sqrt{y+1}, and

    • whose outer radius is rout=5−(−y+1)=5+y+1r_{out}= 5-(-\sqrt{y+1})=5+\sqrt{y+1} and

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

    • whose volume is π(rout2−rin2) dy=π[(5+y+1)2−(5−y+1)2] dy\pi(r_{out}^2 - r_{in}^2)\,\dee{y} = \pi\big[\big(5+\sqrt{y+1}\big)^2-\big(5-\sqrt{y+1}\big)^2\big]\,\dee{y}

  • As our topmost strip is at y=0y=0 and our bottommost strip is at y=−1y=-1 (when x=0x=0), the total volume is

    ∫−10π[(5+y+1)2−(5−y+1)2] dy\begin{align*} \int _{-1}^0 \pi\left[\big(5+\sqrt{y+1}\big)^2-\big(5-\sqrt{y+1}\big)^2\right]\,\dee{y} \end{align*}
Q6Stage 1Past exam · 2001D

Write down a definite integral that represents the volume of the solid obtained by rotating around the line y=−1y=-1 the region between the curves y=x2y=x^2 and y=8−x2y=8-x^2. Do not evaluate the integrals explicitly.

Hint

Draw a sketch.

Answer

π∫−22[(9−x2)2−(x2+1)2] dx\pi\displaystyle\int_{-2}^{2}\big[{(9-x^2)}^2-{(x^2+1)}^2\big]\ \dee{x}

Full solution

The curves intersect at (−2,4)(-2,4) and (2,4)(2,4). When the strip shown in the figure

Figure from prob_s1.6, line 461

Figure from prob_s1.6, line 461

is rotated about the line y=−1y=-1, it forms a thin washer (punctured disc) of

  • inner radius x2+1x^2+1,

  • outer radius 9−x29-x^2 and

  • thickness dx\dee{x} and hence of

  • cross sectional area π[(9−x2)2−(x2+1)2]\pi\big[{(9-x^2)}^2-{(x^2+1)}^2\big] and

  • volume π[(9−x2)2−(x2+1)2] dx\pi\big[{(9-x^2)}^2-{(x^2+1)}^2\big]\,\dee{x}.

So the volume of the solid is

π∫−22[(9−x2)2−(x2+1)2] dx\begin{align*} \pi\int_{-2}^{2}\big[{(9-x^2)}^2-{(x^2+1)}^2\big]\ \dee{x} \end{align*}
Q7Stage 1

A tetrahedron is a three-dimensional shape with four faces, each of which is an equilateral triangle. (You might have seen this shape as a 4-sided die; think of a pyramid with a triangular base.) Using the methods from this section, calculate the volume of a tetrahedron with side-length ℓ\ell. You may assume without proof that the height of a tetrahedron with side-length ℓ\ell is 23ℓ\sqrt{\frac{2}{3}}\ell.

Figure from prob_s1.6, line 2

Figure from prob_s1.6, line 2

Hint

If you take horizontal slices (parallel to one face), they will all be equilateral triangles.

Be careful not to confuse the height of a triangle with the height of the tetrahedron.

Answer

212ℓ3\dfrac{\sqrt{2}}{12}\ell^3

Full solution

We'll make horizontal slices, parallel to one of the faces of the tetrahedron. Then our slices will be equilateral triangles, of varying sizes.

Figure from prob_s1.6, line 2

Figure from prob_s1.6, line 2

For the sake of ease, as in Example 1.6.1 of the CLP-2 text, we picture the tetrahedron perched on a tip, one base horizontal on top.

Figure from prob_s1.6, line 2

Figure from prob_s1.6, line 2

Notice our slice forms the horizontal top of a smaller tetrahedron. The horizontal top of the full tetrahedron has side length ℓ\ell, which is 32\sqrt{\frac{3}{2}} times the height of the full tetrahedron. Our slice is the horizontal top of a tetrahedron of height yy and so has side length 32y\sqrt{\frac{3}{2}}y. An equilateral triangle with side length LL has base LL and height 32L\frac{\sqrt{3}}{2}L, and hence area 34L2\frac{\sqrt{3}}{4}L^2. So, the area of our slice with side length 32y\sqrt{\frac{3}{2}}y is

A=34(32y)2=338y2A = \frac{\sqrt{3}}{4}\left(\sqrt{\frac{3}{2}}y\right)^2 = \frac{3\sqrt{3}}{8}y^2

So, the volume of a tetrahedron with side length ℓ\ell is:

Volume=∫023ℓ338y2 dy=38⋅(23ℓ)3=212ℓ3\begin{align*} \text{Volume}&=\int_0^{\sqrt{\frac{2}{3}}\ell}\frac{3\sqrt{3}}{8}y^2\ \dee{y}\\ &=\frac{\sqrt{3}}{8}\cdot \left(\sqrt{\frac{2}{3}}\ell\right)^3=\frac{\sqrt{2}}{12}\ell^3 \end{align*}

You were given the height of a tetrahedron, but for completeness we calculate it here.

Draw a line starting at one tip, and dropping straight down to the middle of the opposite face. It forms a right triangle with one edge of the tetrahedron, and a line from the middle of the face to the corner.

Figure from prob_s1.6, line 2

Figure from prob_s1.6, line 2

We know the length of the hypotenuse of this right triangle (it's ℓ\ell), so if we know the length of its base (labeled AcAc in the diagram), we can figure out its third side, the height of our tetrahedron. Note by using the Pythagorean theorem, we see that the height of an equilateral triangle with edge length ℓ\ell is 32ℓ\sqrt{\frac{3}{2}}\ell.

Here is a sketch of the base of the pyramid:

Figure from prob_s1.6, line 2

Figure from prob_s1.6, line 2

The triangles ABCABC and AbcAbc are similar (since bb and BB are right angles, and also AA has the same angle in both). Therefore,

AcAb=ACABAcℓ/2=ℓ3ℓ/2Ac=13ℓ\begin{align*} \frac{{Ac}}{{Ab}}&=\frac{{AC}}{{AB}}\\ \frac{Ac}{\ell/2}&=\frac{\ell}{\sqrt{3}\ell/2}\\ Ac&=\frac{1}{\sqrt{3}}\ell \end{align*}

With this in our pocket, we can find the height of the tetrahedron: ℓ2−(13ℓ)2=23ℓ\sqrt{\ell^2 - \left(\frac{1}{\sqrt{3}}\ell\right)^2} =\sqrt{\frac{2}{3}}\ell.

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q8Stage 2Past exam · 2016Q3

Let a>0a>0 be a constant. Let RR be the finite region bounded by the graph of y=1+xex2y=1+\sqrt{x}e^{x^2}, the line y=1y=1, and the line x=ax=a. Using vertical slices, find the volume generated when RR is rotated about the line y=1y=1.

Hint

Sketch the region.

Answer

π4(e2a2−1)\displaystyle\frac{\pi}{4}\Big(e^{2a^2}-1\Big)

Full solution

Let f(x)=1+xex2f(x)=1+\sqrt{x}e^{x^2}. On the vertical slice a distance xx from the yy-axis, sketched in the figure below, yy runs from 11 to f(x)f(x). Upon rotation about the line y=1y=1, this thin slice sweeps out a thin disk of thickness dx\dee{x} and radius f(x)−1f(x)-1 and hence of volume π[f(x)−1]2 dx\pi[f(x)-1]^2\,\dee{x}. The full volume generated (for any fixed a>0a>0) is

∫0aπ[f(x)−1]2 dx=π∫0axe2x2 dx.\begin{align*} \int_0^a\pi[f(x)-1]^2\,\dee{x} =\pi\int_0^axe^{2x^2}\,\dee{x}. \end{align*}

Using the substitution u=2x2u=2x^2, so that du=4x dx\dee{u}=4x\,\dee{x}:

Volume=π∫02a2eu du4=π4eu∣02a2=π4(e2a2−1)\includegraphics\begin{align*} \text{Volume} = \pi\int_0^{2a^2}e^u\,\frac{\dee{u}}{4} =\frac{\pi}{4}e^u\Big|_0^{2a^2} =\frac{\pi}{4}\Big(e^{2a^2}-1\Big) \qquad\qquad\smash{{\includegraphics{OQ16_3_4b}}} \end{align*}

Remark: we spent a good deal of time last semester developing highly accurate but time-consuming methods for sketching common functions. For the purposes of questions like this, we don't need a detailed picture of a function–broad outlines suffice. Notice that x>0\sqrt{x} > 0 whenever x>0x>0, and ex2>0e^{x^2}>0 for all xx. Therefore, xex2\sqrt{x}e^{x^2} is nonnegative over its entire domain, and so the graph y=1+xex2y=1+\sqrt{x}e^{x^2} is always the top function, above the bottom function y=1y=1. That is the only information we needed to perform our calculation.

Q9Stage 2Past exam · 2014A

Find the volume of the solid generated by rotating the finite region bounded by y=1/xy = 1/x and 3x+3y=103x + 3y = 10 about the xx–axis.

Hint

Sketch the region first.

Answer

π[383−51434]=π51281\pi\left[\dfrac{38}{3}-\dfrac{514}{3^4}\right] = \pi\dfrac{512}{81}

Full solution

The curves y=1/xy=1/x and 3x+3y=103x+3y=10, i.e. y=103−xy =\frac{10}{3}-x intersect when

1x=103−x  ⟺  3=10x−3x2  ⟺  3x2−10x+3=0  ⟺  (3x−1)(x−3)=0  ⟺  x=3 , 13\begin{align*} \frac{1}{x} = \frac{10}{3}-x &\iff 3 = 10x-3x^2 \iff 3x^2-10x+3=0 \\ &\iff(3x-1)(x-3)=0 \\ &\iff x=3\,,\,\frac{1}{3} \end{align*}

Figure from prob_s1.6, line 825

Figure from prob_s1.6, line 825

When the region is rotated about the xx–axis, the vertical strip in the figure above sweeps out a washer with thickness dx\dee{x}, outer radius T(x)=103−xT(x)=\frac{10}{3}-x and inner radius B(x)=1xB(x)=\frac{1}{x}. This washer has volume

π(T(x)2−B(x)2) dx=π(1009−203x+x2−1x2) dx\begin{equation*} \pi\big(T(x)^2- B(x)^2\big)\,\dee{x} = \pi\Big(\frac{100}{9}-\frac{20}{3}x+x^2-\frac{1}{x^2}\Big)\,\dee{x} \end{equation*}

Hence the volume of the solid is

π∫1/33(1009−203x+x2−1x2) dx=π[100x9−103x2+13x3+1x]1/33=π[383−51434]=π51281\begin{align*} \pi\int_{1/3}^3\Big(\frac{100}{9}-\frac{20}{3}x+x^2-\frac{1}{x^2}\Big)\,\dee{x} &=\pi\Big[\frac{100x}{9}-\frac{10}{3}x^2+\frac{1}{3}x^3 +\frac{1}{x}\Big]_{1/3}^3 \\ &=\pi\Big[\frac{38}{3}-\frac{514}{3^4}\Big] = \pi\frac{512}{81} \end{align*}
Q10Stage 2Past exam · 2015A

Let RR be the region inside the circle x2+(y−2)2=1x^2 + (y-2)^2=1. Let SS be the solid obtained by rotating RR about the xx-axis.

  1. Write down an integral representing the volume of SS.

  2. Evaluate the integral you wrote down in part (a).

Hint

You can save yourself quite a bit of work by interpreting the integral as the area of a known geometric figure.

Answer

(a) 8π∫−111−x2 dx8\pi\int_{-1}^1\sqrt{1-x^2}\,\dee{x} (b) 4π24\pi^2

Full solution

(a) The top and the bottom of the circle have equations y=T(x)=2+1−x2y=T(x)=2+\sqrt{1-x^2} and y=B(x)=2−1−x2y=B(x) = 2-\sqrt{1-x^2}, respectively.

Figure from prob_s1.6, line 882

Figure from prob_s1.6, line 882

When RR is rotated about the xx–axis, the vertical strip of RR in the figure above sweeps out a washer with thickness dx\dee{x}, outer radius T(x)T(x) and inner radius B(x)B(x). This washer has volume

π(T(x)2−B(x)2) dx=π(T(x)+B(x))(T(x)−B(x)) dx=π×4×21−x2 dx\begin{equation*} {}\hskip0.5in\pi\big(T(x)^2- B(x)^2\big)\,\dee{x} = \pi\big(T(x)+ B(x)\big)\big(T(x)- B(x)\big)\,\dee{x} = \pi\times 4\times 2\sqrt{1-x^2}\,\dee{x} \end{equation*}

Hence the volume of the solid is

8π∫−111−x2 dx\begin{equation*} 8\pi\int_{-1}^1\sqrt{1-x^2}\,\dee{x} \end{equation*}

(b) Since y=1−x2y=\sqrt{1-x^2} is equivalent to x2+y2=1x^2+y^2=1, y≥0y\ge 0, the integral is 8π8\pi times the area of the upper half of the circle x2+y2=1x^2+y^2=1 and hence is 8π×12π12=4π28\pi\times \frac{1}{2}\pi 1^2 = 4\pi^2.

Q11Stage 2Past exam · 1996D

The region RR is the portion of the first quadrant which is below the parabola y2=8xy^2=8x and above the hyperbola y2−x2=15y^2-x^2=15.

  1. Sketch the region RR.

  2. Find the volume of the solid obtained by revolving RR about the xx axis.

Hint

See Example 1.6.3 in the

CLP-2 text.

Answer

(a) The region RR is the region between the blue and red curves, with 3≤x≤53\le x\le 5, in the figures below.

Figure from prob_s1.6, line 932

Figure from prob_s1.6, line 932

Figure from prob_s1.6, line 932

Figure from prob_s1.6, line 932

(b) 43π≈4.19\frac{4}{3}\pi\approx 4.19

Full solution

(a) The two curves intersect when xx obeys 8x=x2+158x=x^2+15 or x2−8x+15=(x−5)(x−3)=0x^2-8x+15=(x-5)(x-3)=0. The points of intersection, in the first quadrant, are (3,24)(3,\sqrt{24}) and (5,40)(5, \sqrt{40}). The region RR is the region between the blue and red curves, with 3≤x≤53\le x\le 5, in the figures below.

Figure from prob_s1.6, line 932

Figure from prob_s1.6, line 932

Figure from prob_s1.6, line 932

Figure from prob_s1.6, line 932

(b) The part of the solid with xx coordinate between xx and x+dxx+\dee{x} is a “washer” shaped region with inner radius x2+15\sqrt{x^2+15}, outer radius 8x\sqrt{8x} and thickness dx\dee{x}. The surface area of the washer is π(8x)2−π(x2+15)2=π(8x−x2−15)\pi(\sqrt{8x})^2 -\pi(\sqrt{x^2+15})^2=\pi(8x-x^2-15) and its volume is π(8x−x2−15) dx\pi(8x-x^2-15)\,\dee{x}. The total volume is

∫35π(8x−x2−15) dx=π[4x2−13x3−15x]35=π[100−1253−75−36+9+45]=43π≈4.19\begin{align*} \int_3^5 \pi(8x-x^2-15)\,\dee{x} &=\pi\Big[4x^2-\frac{1}{3}x^3-15 x\Big]_3^5 =\pi\Big[100-\frac{125}{3}-75-36+9+45\Big] \\ &=\frac{4}{3}\pi\approx 4.19 \end{align*}
Q12Stage 2Past exam · 1996D

The region RR is bounded by y=log⁡xy=\log x, y=0y=0, x=1x=1 and x=2x=2. (Recall that we are using log⁡x\log x to denote the logarithm of xx with base ee. In other courses it is often denoted ln⁡x\ln x.)

  1. Sketch the region RR.

  2. Find the volume of the solid obtained by revolving this region about the yy axis.

Hint

See Example 1.6.5 in the

CLP-2 text.

Answer

(a) The region RR is sketched below.

Figure from prob_s1.6, line 989

Figure from prob_s1.6, line 989

(b) π[4log⁡2−32]≈3.998\pi\Big[4\log 2 - \frac{3}{2}\Big] \approx 3.998

Full solution

(a) The region RR is sketched in the figure on the left below. (The bound y=0y=0 renders the bound x=1x=1 unnecessary, since the graph y=log⁡xy=\log x hits the xx-axis when x=1x=1.)

Figure from prob_s1.6, line 989

Figure from prob_s1.6, line 989

Figure from prob_s1.6, line 999

Figure from prob_s1.6, line 999

(b) We'll use horizontal washers as in Example 1.6.5 of the CLP-2 text.

  • We cut RR into thin horizontal strips of height dy\dee{y} as in the figure on the right above.

  • When we rotate RR about the yy–axis, i.e. about the line x=0x=0, each strip sweeps out a thin washer

    • whose inner radius is rin=eyr_{in}= e^y and outer radius is rout=2r_{out}=2, and

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

    • whose volume π(rout2−rin2)dy=π(4−e2y)dy\pi(r_{out}^2 - r_{in}^2)\dee{y} = \pi\big(4-e^{2y}\big)\dee{y}.

  • As our bottommost strip is at y=0y=0 and our topmost strip is at y=log⁡2y=\log 2 (since at the top x=2x=2 and x=eyx=e^y), the total

    Volume=∫0log⁡2π(4−e2y) dy=π[4y−e2y/2]0log⁡2=π[4log⁡2−2+12]=π[4log⁡2−32]\begin{align*} \text{Volume} &= \int _0^{\log 2} \pi\big(4-e^{2y}\big)\ \dee{y} =\pi\big[4y -e^{2y}/2\big]_0^{\log 2} =\pi\Big[4\log 2 - 2 +\frac{1}{2}\Big] \\ &=\pi\Big[4\log 2 - \frac{3}{2}\Big] \end{align*}

    Using a calculator, we see this is approximately 3.9983.998.

Q13Stage 2Past exam · 2016Q3

The finite region between the curves y=cos⁡(x2)y = \cos(\frac x2) and y=x2−π2y = x^2 - \pi^2 is rotated about the line y=−π2y=-{\pi^2}. Using vertical slices (disks and/or washers), find the volume of the resulting solid.

Hint

Sketch the region. To find where the curves intersect, look at where cos⁡(x2)\cos(\frac x2) and x2−π2x^2 - \pi^2 both have roots.

Answer

π2+8π3+8π65\pi^2 + 8\pi^3 + \frac{8\pi^6}{5}

Full solution

Here is a sketch of the curves y=cos⁡(x2)y = \cos(\frac x2) and y=x2−π2y = x^2 - \pi^2.

Figure from prob_s1.6, line 1052

Figure from prob_s1.6, line 1052

By inspection, the curves meet at x=±πx = \pm {\pi} where both cos⁡(x2)\cos(\frac x2) and x2−π2x^2 - \pi^2 take the value zero. We'll use vertical washers as specified in the question.

  • We cut the specified region into thin vertical strips of width dx\dee{x} as in the figure above.

  • When we rotate about the line y=−π2y=-\pi^2, each strip sweeps out a thin washer

    • whose inner radius is rin=(x2−π2)−(−π2)=x2r_{in}= (x^2 - {\pi^2} ) - ( {-} {\pi^2} )=x^2 and outer radius is rout=cos⁡(x2)−(−π2)=cos⁡(x2)+π2r_{out}=\cos(\frac x2) - ( {-} {\pi^2}) =\cos(\frac x2) +\pi^2, and

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

    • whose volume π(rout2−rin2)dx=π((cos⁡(x2)+π2)2−(x2)2)dx\pi(r_{out}^2 - r_{in}^2)\dee{x} = \pi\big( {(\cos(\frac x2) +\pi^2)}^2 - {(x^2)}^2\big)\dee{x}.

  • As our leftmost strip is at x=−πx=-\pi and our rightmost strip is at x=πx=\pi,

the total volume is

π∫−ππ(cos⁡2(x2)+2π2cos⁡(x2)+π4−x4) dx=π∫−ππ(1+cos⁡(x)2+2π2cos⁡(x2)+π4−x4) dx\begin{align*}&\pi \int_{-\pi}^{\pi} \left( \cos^2 (\tfrac x2) +2{\pi^2}\cos (\tfrac x2) +{\pi^4} -x^4\right)\,\dee{x} \\ &\hskip0.5in = \pi \int_{-\pi}^{\pi} \left(\frac{1+\cos(x)}{2} +2{\pi^2}\cos (\tfrac x2) +{\pi^4} -x^4\right)\,\dee{x}\end{align*}

Because the integrand is even,

=2π∫0π(1+cos⁡(x)2+2π2cos⁡(x2)+π4−x4) dx=2π[12x+12sin⁡(x)+4π2sin⁡(x2)+π4x−15x5]0π=2π[π2+0+4π2+π5−π55]=π2+8π3+8π65\begin{align*}&\hskip0.5in = 2\pi \int_0^{\pi} \left(\frac{1+\cos(x)}{2} +2{\pi^2}\cos (\tfrac x2) +{\pi^4} -x^4\right)\,\dee{x} \\ &\hskip0.5in = {2\pi \left[ \frac{1}{2}x + \frac{1}{2}\sin(x) + 4{\pi^2}\sin (\tfrac x2) +{\pi^4} x -\frac{1}{5}x^5 \right]} _0^\pi\\ &\hskip0.5in = 2\pi \left[\frac{\pi}{2} + 0 + 4{\pi^2} +{\pi^5} - \frac{\pi^5}{5} \right] \\ &\hskip0.5in = {\pi^2} + 8\pi^3 + \frac{8\pi^6}{5}\end{align*}

We used the fact that the integrand is an even function and the interval of integration [−π,π][-\pi, \pi] is symmetric, but one can also compute directly.

Q14Stage 2Past exam · 1997D

The solid VV is 2 meters high and has square horizontal cross sections. The length of the side of the square cross section at height xx meters above the base is 21+x\frac{2}{1+x} m. Find the volume of this solid.

Hint

See Example 1.6.6 in the

CLP-2 text.

Answer

83\dfrac{8}{3}

Full solution

As in Example 1.6.6 of the

CLP-2 text notes, we slice VV into thin horizontal “square pancakes”.

  • We are told that the pancake at height xx is a square of side 21+x\frac{2}{1+x} and so

  • has cross-sectional area (21+x)2\big(\frac{2}{1+x}\big)^2 and thickness dx\dee{x} and hence

  • has volume (21+x)2dx\big(\frac{2}{1+x}\big)^2\dee{x}.

Hence the volume of VV is

∫02[21+x]2 dx=∫134u2 du=4u−1−1∣13=−4[13−1]=83\int_0^2{\Big[\frac{2}{1+x}\Big]}^2\,\dee{x} =\int_1^3\frac{4}{u^2}\,\dee{u} =4\frac{u^{-1}}{-1}\bigg|_1^3 =-4\Big[\frac{1}{3}-1\Big] =\frac{8}{3}

We made the change of variables u=1+xu=1+x, du=dx\dee{u}=\dee{x}.

Q15Stage 2Past exam · 1998A

Consider a solid whose base is the finite portion of the xyxy–plane bounded by the curves y=x2y=x^2 and y=8−x2y=8-x^2. The cross–sections perpendicular to the xx–axis are squares with one side in the xyxy–plane. Compute the volume of this solid.

Hint

See Example 1.6.6 in the

CLP-2 text. Imagine cross-sections with shadow parallel to the yy-axis, sticking straight out of the xyxy-plane.

Answer

256×815=136.53˙\dfrac{256\times 8}{15}=136.5\dot3

Full solution

Here is a sketch of the base region.

Figure from prob_s1.6, line 1172

Figure from prob_s1.6, line 1172

Consider the thin vertical cross–section resting on the heavy red line in the figure above. It has thickness dx\dee{x}. Its face is a square whose side runs from y=x2y=x^2 to y=8−x2y=8-x^2, a distance of 8−2x28-2x^2. So the face has area (8−2x2)2{(8-2x^2)}^2 and the slice has volume (8−2x2)2 dx{(8-2x^2)}^2\,\dee{x}. The two curves cross when x2=8−x2x^2=8-x^2, i.e. when x2=4x^2=4 or x=±2x=\pm 2. So xx runs from −2-2 to 22 and the total volume is

∫−22(8−2x2)2 dx=2∫024(4−x2)2 dx=8∫02[16−8x2+x4] dx=8[16×2−8323+1525]=256×815=136.53˙\begin{align*} \int_{-2}^{2}{(8-2x^2)}^2\,\dee{x}&=2\int_0^2 4{(4-x^2)}^2\,\dee{x} =8\int_0^2\big[16-8x^2+x^4\big]\,\dee{x}\cr &=8\Big[16\times 2-\frac{8}{3}2^3+\frac{1}{5}2^5\Big] =\frac{256\times 8}{15}=136.5\dot3 \end{align*}

In the first simplification step, we used the fact that our integrand was even, but we also could have finished our computation without this step.

Q16Stage 2Past exam · 2001D

A frustum of a right circular cone (as shown below) has height hh. Its base is a circular disc with radius 44 and its top is a circular disc with radius 22. Calculate the volume of the frustum.

Figure from prob_s1.6, line 1196

Figure from prob_s1.6, line 1196

Hint

See Example 1.6.1 in the

CLP-2 text.

Answer

283πh\dfrac{28}{3}\pi h

Full solution

Slice the frustum into horizontal discs. When the disc is a distance tt from the top of the frustum it has radius 2+2t/h2+2t/h. Note that as tt runs from 00 (the top of the frustum) to t=ht=h (the bottom of the frustum) the radius 2+2t/h2+2t/h increases linearly from 22 to 44.

Figure from prob_s1.6, line 1215

Figure from prob_s1.6, line 1215

Thus the disk has volume π(2+2t/h)2dt\pi \big(2+2t/h\big)^2 \dee{t}. The total volume of the frustum is

π∫0h(2+2t/h)2dt=4π∫0h(1+t/h)2dt=4π[(1+t/h)33/h]0h=43πh×7=283πh\begin{align*} \pi\int_0^h \big(2+2t/h\big)^2 \dee{t} =4\pi\int_0^h \big(1+t/h\big)^2 \dee{t} =4\pi\left[\frac{(1+t/h)^3}{3/h}\right]_0^h =\frac{4}{3}\pi h\times 7 =\frac{28}{3}\pi h \end{align*}

Remark: we could also solve this problem using the formula for the volume of a cone. Using similar triangles, the frustum in question is shaped like a right circular cone of height 2h2h and base radius 4 (and hence of volume 13π(42)(2h)\dfrac{1}{3}\pi(4^2)(2h)), but missing its top, which is a right circular cone of height hh and base radius 22 (and hence volume 13π(22)h\dfrac{1}{3}\pi(2^2)h). So, the volume of the frustum is 13π(42)(2h)−13π(22)h=283πh\dfrac{1}{3}\pi(4^2)(2h) - \dfrac{1}{3}\pi(2^2)h = \dfrac{28}{3}\pi h.

3Stage 3Application

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

Q17Stage 3

The shape of the earth is often approximated by an oblate spheroid, rather than a sphere. An oblate spheroid is formed by rotating an ellipse about its minor axis (its shortest diameter).

  1. Find the volume of the oblate spheroid obtained by rotating the upper (positive) half of the ellipse (ax)2+(by)2=1(ax)^2+(by)^2=1 about the xx-axis, where aa and bb are positive constants with a≥ba \geq b.

  2. Suppose (Earth Fact Sheet, NASA, https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html, accessed 2 July 2017) the earth has radius at the equator of 6378.137 km, and radius at the poles of 6356.752 km. If we model the earth as an oblate spheroid formed by rotating the upper half of the ellipse (ax)2+(by)2=1(ax)^2+(by)^2=1 about the xx-axis, what are aa and bb?

  3. What is the volume of this model of the earth? (Use a calculator.)

  4. Suppose we had calculated the volume of the earth by modelling it as a sphere with radius 6378.1376378.137 km. What would our absolute and relative errors be, compared to our oblate spheroid calculation?

Hint

(a) Don't be put off by phrases like “rotating an ellipse about its minor axis." This is the same kind of volume you've been calculating all section.
(b) Hopefully, you sketched the ellipse in part (a). What was its smallest radius? Its largest? These correspond to the polar and equitorial radii, respectively.
(c) Combine your answers from (a) and (b).
(d) Remember that the absolute error is the absolute difference of your two results–that is, you subtract them and take the absolute value. The relative error is the absolute error divided by the actual value (which we're taking, for our purposes, to be your answer from (c)). When you take the relative error, lots of terms will cancel, so it's easiest to not use a calculator till the end.

Answer

(a) 4π3b2a\dfrac{4\pi}{3b^2a} cubic units (b) a=16356.752a = \dfrac{1}{6356.752} and b=16378.137b=\dfrac{1}{6378.137}

(c) Approximately 1.08321×1012 km31.08321\times 10^{12} ~\mathrm{km}^3, or 1.08321×1021 m31.08321\times 10^{21}~ \mathrm{m}^3

(d) Absolute error is about 3.64×109 km33.64\times 10^{9}~ \mathrm{km}^3, and relative error is about 0.003360.00336, or 0.336%0.336\%.

Full solution

(a)

We'll want to start by graphing the upper half of the ellipse (ax)2+(by)2=1(ax)^2+(by)^2=1. Its intercepts will be enough to get us an idea: (0,1b)(0,\frac{1}{b}) and (±1a,0)(\pm\frac{1}{a},0):

Figure from prob_s1.6, line 2

Figure from prob_s1.6, line 2

We note a few things at the outset: first, since a≥ba \geq b, then 1a≤1b\frac{1}{a} \leq \frac{1}{b}, so indeed the xx-axis is the minor axis. That is, we're rotating about the proper axis to create an oblate spheroid.

Second, if we solve our equation for yy, we get y=1b1−(ax)2y=\frac{1}{b}\sqrt{1-(ax)^2}. (Since we only want the upper half of the ellipse, we only need to consider the positive square root.)

Now, we have a standard volume-of-revolution problem. We make vertical slices, of width dx\dee{x} and height y=1b1−(ax)2y=\frac{1}{b}\sqrt{1-(ax)^2}. When we rotate these slices about the xx-axis, they form thin disks of volume π[1b1−(ax)2]2dx\pi\left[\frac{1}{b}\sqrt{1-(ax)^2}\right]^2\dee{x} . Since xx runs from −1a-\frac{1}{a} to 1a\frac{1}{a}, the volume of our oblate spheroid is:

Volume=∫−1a1aπ[1b1−(ax)2]2dx=πb2∫−1a1a1−(ax)2 dx=2πb2∫01a1−(ax)2 dx(even function)=2πb2[x−a2x33]01a=2πb2[1a−13a]=4π3b2a\begin{align*} \text{Volume}&=\int_{-\frac{1}{a}}^{\frac{1}{a}} \pi \left[\frac{1}{b}\sqrt{1-(ax)^2}\right]^2\dee{x}\\ &=\frac{\pi}{b^2}\int_{-\frac{1}{a}}^{\frac{1}{a}} 1-(ax)^2\ \dee{x}\\ &=\frac{2\pi}{b^2}\int_{0}^{\frac{1}{a}} 1-(ax)^2\ \dee{x}&\text{(even function)}\\ &=\frac{2\pi}{b^2}\left[x - \frac{a^2x^3}{3}\right]_{0}^{\frac{1}{a}}\\ &=\frac{2\pi}{b^2}\left[\frac{1}{a} - \frac{1}{3a} \right] = \frac{4\pi}{3b^2a} \end{align*}

(b) As we saw in the sketch from part (a), the shortest radius of the ellipse is 1a\frac{1}{a}, while the largest is 1b\frac{1}{b}. So, 1a=6356.752\frac{1}{a} = 6356.752, and 1b=6378.137\frac{1}{b} = 6378.137. That is, a=16356.752a = \dfrac{1}{6356.752} and b=16378.137b=\dfrac{1}{6378.137}.

Note a≥ba \geq b, as specified in part (a).

(c) Combining our answers from (a) and (b), the volume of an oblate spheroid with approximately the same dimensions as the earth is:

4π3b2a=4π3(1b)2(1a)=4π3(6378.137)2(6356.752)≈1.08321×1012km3≈1.08321×1021m3\begin{align*} \frac{4\pi}{3b^2a} &= \frac{4\pi}{3}\left(\frac{1}{b}\right)^2\left(\frac{1}{a}\right)\\ &=\frac{4\pi}{3}\left(6378.137\right)^2\left(6356.752\right)\\ &\approx 1.08321\times 10^{12} \quad\mathrm{km}^3\\ &\approx 1.08321\times 10^{21} \quad\mathrm{m}^3 \end{align*}

(d) A sphere of radius 6378.137 has volume

43π(6378.137)3\begin{align*}&\dfrac{4}{3}\pi\left(6378.137\right)^3\end{align*}

So, our absolute error is:

∣4π3(6378.137)2(6356.752)−43π(6378.137)3∣=4π3(6378.137)2∣6356.752−6378.137∣=4π3(6378.137)2(21.385)≈ 3.64×109 km3\begin{align*}&\left|\frac{4\pi}{3}\left(6378.137\right)^2\left(6356.752\right) - \dfrac{4}{3}\pi\left(6378.137\right)^3 \right|\\ =&\frac{4\pi}{3}\left(6378.137\right)^2\big|6356.752 - 6378.137\big|\\ =&\frac{4\pi}{3}\left(6378.137\right)^2(21.385)\\ \approx&~3.64 \times 10^{9}~\mathrm{km}^3\end{align*}

And our relative error is:

abs erroractual value=4π3(6378.137)2∣6356.752−6378.137∣4π3(6378.137)2(6356.752)=∣6356.752−6378.137∣6356.752=6378.1376356.752−1≈ 0.00336\begin{align*}\frac{\text{abs error}}{\text{actual value}}&=\frac{\frac{4\pi}{3}\left(6378.137\right)^2\big|6356.752 - 6378.137\big|}{\frac{4\pi}{3}\left(6378.137\right)^2\left(6356.752\right)}\\ &=\frac{\big|6356.752 - 6378.137\big|}{6356.752}\\ &=\frac{6378.137}{6356.752}-1\\ &\approx ~0.00336\end{align*}

That is, about 0.336%0.336\%, or about one-third of one percent.

Q18Stage 3Past exam · 2012A

Let RR be the bounded region that lies between the curve y=4−(x−1)2y = 4 - (x - 1)^2 and the line y=x+1y = x + 1.

  1. Sketch RR and find its area.

  2. Write down a definite integral giving the volume of the region obtained by rotating RR about the line y=5y = 5. Do not evaluate this integral.

Hint

To find the points of intersection, set 4−(x−1)2=x+14-(x-1)^2=x+1.

Answer

(a) 92\dfrac{9}{2} (b) π∫−12[(4−x)2−(1+(x−1)2)2] dx\pi\displaystyle\int_{-1}^2 \big[{\big(4-x\big)}^2-{\big(1+(x-1)^2\big)}^2\big]\,\dee{x}

Full solution

(a) The curve y=4−(x−1)2y = 4 - (x - 1)^2 is an “upside down parabola” and line y=x+1y = x + 1 has slope 1. They intersect at points (x,y)(x,y) which satisfy both y=x+1y=x+1 and y=4−(x−1)2y=4-(x-1)^2. That is, when xx obeys

x+1=4−(x−1)2x+1=4−x2+2x−1x2−x−2=0(x−2)(x+1)=0x=−1orx=2\begin{align*} x+1&=4-(x-1)^2\\ x+1 &= 4 -x^2+2x-1\\ x^2-x-2&=0 \\ (x-2)(x+1)&=0\\ x&=-1 \quad\text{or}\quad x=2 \end{align*}

Thus the intersection points are (−1,0)(-1,0) and (2,3)(2,3). Here is a sketch of RR:

Figure from prob_s1.6, line 1353

Figure from prob_s1.6, line 1353

The red strip in the sketch above runs from y=x+1y=x+1 to y=4−(x−1)2y=4-(x-1)^2 and so has area [4−(x−1)2−(x+1)] dx=[2+x−x2] dx[4-(x-1)^2 -(x+1)]\,\dee{x} = [2+x-x^2]\,\dee{x}. All together RR has

Area=∫−12[2+x−x2] dx=[2x+x22−x33]−12=6+32−93=92\begin{align*} \text{Area} &= \int_{-1}^2 \big[2+x-x^2\big]\ \dee{x} \\ &=\bigg[2x+\frac{x^2}{2}-\frac{x^3}{3}\bigg]_{-1}^2 \\ &=6+\frac{3}{2}-\frac{9}{3}=\frac{9}{2} \end{align*}

(b) We'll use vertical washers as in Example 1.6.3 of the CLP-2 text. Note that the highest point achieved by y=4−(x−1)2y=4-(x-1)^2 is y=4y=4, so rotating around the line y=5y=5 causes no unexpected problems.

Figure from prob_s1.6, line 1353

Figure from prob_s1.6, line 1353

  • We cut RR into thin vertical strips of width dx\dee{x} like the red strip in the figure above.

  • When we rotate RR about the horizontal line y=5y=5, each strip sweeps out a thin washer

    • whose inner radius is rin=5−[4−(x−1)2]=1+(x−1)2r_{in}=5-[4-(x-1)^2]=1+(x-1)^2, and

    • whose outer radius is rout=5−[x+1]=4−xr_{out}= 5-[x+1]=4-x and

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

    • whose volume is π[rout2−rin2] dx=π[(4−x)2−(1+(x−1)2)2] dx\pi\big[r_{out}^2-r_{in}^2\big]\,\dee{x} =\pi\big[{\big(4-x\big)}^2-{\big(1+(x-1)^2\big)}^2\big]\,\dee{x}

  • As our leftmost strip is at x=−1x=-1 and our rightmost strip is at x=2x=2, the total

    Volume=π∫−12[(4−x)2−(1+(x−1)2)2] dx\begin{align*} {\rm Volume} &= \pi\int_{-1}^2 \big[{\big(4-x\big)}^2-{\big(1+(x-1)^2\big)}^2\big]\,\dee{x} \end{align*}
Q19Stage 3Past exam · M121 1999A

Let R={(x,y) : (x−1)2+y2≤1 and x2+(y−1)2≤1 }\cR=\big\{(x,y)\ :\ (x-1)^2+y^2\le 1\text{ and } x^2+(y-1)^2\le 1\ \big\}.

  1. Sketch R\cR and find its area.

  2. If R\cR rotates around the yy–axis, what volume is generated?

Hint

You can somewhat simplify your calculations in part (a) (but not part (b)) by using the fact that R\cR is symmetric about the line y=xy=x.

When you're solving an equation for xx, be careful about your signs: x−1x-1 is negative.

Answer

(a) π2−1\dfrac{\pi}{2}-1 (b) π22−π≈1.793\dfrac{\pi^2}{2}-\pi\approx 1.793

Full solution

(a) The curves (x−1)2+y2=1(x-1)^2+y^2 = 1 and x2+(y−1)2=1x^2+(y-1)^2 = 1 are circles of radius 11 centered on (1,0)(1,0) and (0,1)(0,1) respectively. Both circles pass through (0,0)(0,0) and (1,1)(1,1). They are sketched below.

Figure from prob_s1.6, line 1443

Figure from prob_s1.6, line 1443

The region R\cR is symmetric about the line y=xy=x, so the area of R\cR is twice the area of the part of R\cR to the left of the line y=xy=x. The red strip in the sketch above runs from the edge of the lower circle to x=yx=y. So, given a value of yy in [0,1][0,1], we need to find the corresponding value of xx along the circle. We solve (x−1)2+y2=1(x-1)^2+y^2=1 for xx, keeping in mind that 0≤x≤10 \leq x\leq 1:

(x−1)2+y2=1(x−1)2=1−y2∣x−1∣=1−y21−x=1−y2x=1−1−y2\begin{align*}(x-1)^2+y^2&=1\\ (x-1)^2&=1-y^2\\ |x-1|&=\sqrt{1-y^2}\\ 1-x&=\sqrt{1-y^2}\\ x&=1-\sqrt{1-y^2}\end{align*}

Now, we calculate:

Area=2∫01[y−(1−1−y2)] dy=2{∫01y−1 dy+∫011−y2 dy}=2{[y22−y]01+∫011−y2 dy}=π2−1\begin{align*}\text{Area} &= 2\int_0^1 \big[y-\big(1-\sqrt{1-y^2}\big)\big]\ \dee{y} \\ &=2\left\{ \int_0^1 y-1\ \dee{y} + \int_0^1 \sqrt{1-y^2}\ \dee{y} \right\}\\ &=2\Big\{\Big[\frac{y^2}{2}-y\Big]_0^1 +\int_0^1\sqrt{1-y^2}\ \dee{y}\Big\} \\ &=\frac{\pi}{2}-1\end{align*}

Here the integral ∫011−y2 dy\int_0^1\sqrt{1-y^2}\ \dee{y} was evaluated simply as the area of one quarter of a cicular disk of radius 11. It can also be evaluated by substituting y=sin⁡θy=\sin\theta, a technique we'll learn more about in Section 1.9 of the CLP-2 text.

(b) We'll use horizontal washers as in Example 1.6.5 of the in the CLP-2 text.

  • We cut R\cR into thin horizontal strips of width dy\dee{y} like the blue strip in the figure above.

  • When we rotate R\cR about the yy–axis, each strip sweeps out a thin washer

    • whose inner radius is rin=1−1−y2r_{in}=1-\sqrt{1-y^2}, and

    • whose outer radius is rout=1−(y−1)2r_{out}= \sqrt{1-(y-1)^2} and

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

    • whose volume is

      π[(1−(y−1)2)2−(1−1−y2 )2] dy=π[1−(y−1)2−1+21−y2−(1−y2)]=2π[1−y2+y−1] dy\begin{align*} &\pi\big[{\big(\sqrt{1-(y-1)^2}\big)}^2-{\big(1-\sqrt{1-y^2}\,\big)}^2\big]\,\dee{y}\\ =&\pi \big[ 1 - (y-1)^2 -1 + 2\sqrt{1-y^2} - (1-y^2) \big]\\ =&2\pi\big[\sqrt{1-y^2}+y-1\big]\,\dee{y}\end{align*}

  • As our bottommost strip is at y=0y=0 and our topmost strip is at y=1y=1, the total

    Volume=2π∫01[1−y2+y−1] dy=2π[π4+12−1]=π22−π≈1.793\begin{align*} {\rm Volume} &= 2\pi\int_{0}^1\big[\sqrt{1-y^2}+y-1\big]\ \dee{y} = 2\pi\Big[\frac{\pi}{4}+\frac{1}{2}-1\Big]\\ &=\frac{\pi^2}{2}-\pi\approx 1.793 \end{align*}

    Here, we again used that ∫011−y2 dy\int_{0}^1 \sqrt{1-y^2}\ \dee{y} is the area of a quarter circle of radius one, and we used a calculator to approximate the final answer.

Q20Stage 3Past exam · 1997A

Let R\cR be the plane region bounded by x=0, x=1, y=0x=0,\ x=1,\ y=0 and y=c1+x2y=c\sqrt{1+x^2}, where c≥0c\ge 0 is a constant.

  1. Find the volume V1V_1 of the solid obtained by revolving R\cR about the xx–axis.

  2. Find the volume V2V_2 of the solid obtained by revolving R\cR about the yy–axis.

  3. If V1=V2V_1=V_2, what is the value of cc?

Hint

The mechanically easiest way to answer part (b) uses the method of cylindrical shells, which we have not covered. The method of washers also works, but requires you have enough patience and also to have a good idea what R\cR looks like. So it is crucial to first sketch R\cR. Then be very careful in identifying the left end of your horizontal strips.

Answer

(a) V1=43πc2V_1=\dfrac{4}{3}\pi c^2 (b) V2=π c3[42−2]V_2 =\dfrac{\pi\,c}{3}\big[4\sqrt{2}-2 \big] (c) c=0 or c=2−12c=0\text{ or }c=\sqrt{2}-\frac{1}{2}

Full solution

Before we start, it will be useful to have a reasonable sketch of the graph y=c1+x2y=c\sqrt{1+x^2} over the interval [0,1][0,1]. Its endpoints are (0,c)(0,c) and (1,c2)(1,c\sqrt{2}). The function is entirely above the xx-axis, which we need to know for part (a). For part (b), we need to know whether it is always increasing or not: when we're drawing horizontal strips, we need to know their endpoints, and if the function has “humps," the right endpoint will not be simply the line x=1x=1.

If you're comfortable noticing that 1+x21+x^2 increases as xx increases because we only consider nonnegative values of xx, then you can also be confident that 1+x2\sqrt{1+x^2} is simply increasing. Alternately, we can consider the derivative:

ddx{c1+x2}=c⋅121+x2⋅2x=cx1+x2\begin{align*} \diff{}{x}\left\{c\sqrt{1+x^2}\right\}&=c\cdot \dfrac{1}{2\sqrt{1+x^2}}\cdot 2x = \dfrac{cx}{\sqrt{1+x^2}} \end{align*}

Since we only consider positive values of xx, this derivative is never negative, so the function is never decreasing. This gives us the following basic sketch:

Figure from prob_s1.6, line 2

Figure from prob_s1.6, line 2

The figures in the solution below use a slightly more detailed rendering of our function, but so much accuracy is not necessary.

(a) Let V1\cV_1 be the solid obtained by revolving R\cR about the xx–axis. The portion of V1\cV_1 with xx–coordinate between xx and x+dxx+\dee{x} is obtained by rotating the red vertical strip in the figure on the left below about the xx–axis. That portion is a disk of radius c1+x2c\sqrt{1+x^2} and thickness dx\dee{x}. The volume of this disk is π(c1+x2)2dx=πc2(1+x2) dx\pi(c\sqrt{1+x^2})^2\dee{x}=\pi c^2 (1+x^2)\,\dee{x}. So the total volume of V1\cV_1 is

V1=∫01πc2(1+x2) dx=πc2[x+x33]01=43πc2\begin{align*} V_1=\int_0^1 \pi c^2 (1+x^2)\,\dee{x} =\pi c^2\Big[x+\frac{x^3}{3}\Big]_0^1 =\frac{4}{3}\pi c^2 \end{align*}

Figure from prob_s1.6, line 1543

Figure from prob_s1.6, line 1543

Figure from prob_s1.6, line 1543

Figure from prob_s1.6, line 1543

(b) We'll use horizontal washers as in Example 1.6.5 of the in the CLP-2 text.

  • We cut R\cR into thin horizontal strips of width dy\dee{y} as in the figure on the right above.

  • When we rotate R\cR about the yy–axis, i.e. about the line x=0x=0, each strip sweeps out a thin washer

    • whose outer radius is rout=1r_{out}=1, and

    • whose inner radius is rin=y2c2−1r_{in}= \sqrt{\frac{y^2}{c^2}-1} when y≥c1+02=cy\ge c\sqrt{1+0^2}=c (see the red strip in the figure on the right above), and whose inner radius is rin=0r_{in}= 0 when y≤cy\le c (see the blue strip in the figure on the right above) and

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

    • whose volume is π(rout2−rin2)dy=π(2−y2c2)dy\pi(r_{out}^2 - r_{in}^2)\dee{y} = \pi\big(2-\frac{y^2}{c^2}\big)\dee{y} when y≥cy\ge c and whose volume is π(rout2−rin2)dy=π dy\pi(r_{out}^2 - r_{in}^2)\dee{y} = \pi\,\dee{y} when y≤cy\le c and

  • As our bottommost strip is at y=0y=0 and our topmost strip is at y=2 cy=\sqrt{2}\,c (since at the top x=1x=1 and y=c1+x2y= c\sqrt{1+x^2}), the total

    V2=∫c2 cπ(2−y2c2)dy+∫0cπ dy=π[2y−y33c2]c2 c+πc=π c[423−53]+πc=π c3[42−2]\begin{align*} V_2 &= \int _c^{\sqrt{2}\,c} \pi\Big(2-\frac{y^2}{c^2}\Big)\dee{y} +\int _ 0^c \pi\,\dee{y} \\ &=\pi{\Big[2y -\frac{y^3}{3c^2}\Big]}_c^{\sqrt{2}\,c} +\pi c\\ &=\pi\,c\Big[\frac{4\sqrt{2}}{3}-\frac{5}{3} \Big]+\pi c \\[0.05in] &=\frac{\pi\,c}{3}\big[4\sqrt{2}-2 \big] \end{align*}

(c) We have V1=V2V_1=V_2 if and only if

43πc2=π c3[42−2]4c2=c(42−2)4c2−c(42−2)=04c(c−(2−12))=0c=0orc=2−12\begin{align*} \frac{4}{3}\pi c^2&=\frac{\pi\,c}{3}\big[4\sqrt{2}-2 \big] \\ 4c^2&=c\left(4\sqrt{2}-2\right)\\ 4c^2-c\left(4\sqrt{2}-2\right)&=0\\ 4c\left(c - \left(\sqrt{2}-\frac{1}{2}\right)\right)&=0\\ c=0 \quad\text{or}\quad c&=\sqrt{2}-\frac{1}{2} \end{align*}
Q21Stage 3Past exam · 2013A

The graph below shows the region between y=4+πsin⁡xy = 4 + \pi \sin x and y=4+2π−2xy = 4 + 2\pi - 2x.

Figure from prob_s1.5, line 652

Figure from prob_s1.5, line 652

The region is rotated about the line y=−1y = -1. Express in terms of definite integrals the volume of the resulting solid. Do not evaluate the integrals.

Hint

Note that the curves cross. The area of this region was found in Problem 14 of Section 1.5. It would be useful to review that problem.

Answer

∫π/2ππ[(5+πsin⁡x)2−(5+2π−2x)2] dx+∫π3π/2π[(5+2π−2x)2−(5+πsin⁡x)2] dx\displaystyle\int_{\pi/2}^\pi \pi\big[(5 + \pi \sin x)^2-(5 + 2\pi - 2x)^2\big]\ \dee{x} +\displaystyle\int^{3\pi/2}_\pi \pi\big[(5 + 2\pi - 2x)^2-(5 + \pi \sin x)^2\big]\ \dee{x}

Full solution

We will compute the volume by rotating thin vertical strips as in the sketch

Figure from prob_s1.5, line 671

Figure from prob_s1.5, line 671

about the line y=−1y=-1 to generate thin washers. We need to know when the line y=4+2π−2xy = 4 + 2\pi - 2x intersects the curve y=4+πsin⁡xy = 4 + \pi \sin x. Looking at the graph, it appears to be at π2\frac{\pi}{2}, π\pi, and 3π2\frac{3\pi}{2}. By plugging in these values of xx to both functions, we see they are indeed the points of intersection.

  • When π2≤x≤π\frac{\pi}{2} \le x \le \pi, the top of the strip is at y=4+πsin⁡xy = 4 + \pi \sin x and the bottom of the strip is at y=4+2π−2xy = 4 + 2\pi - 2x. When the strip is rotated, we get a thin washer with outer radius R1(x)=1+4+πsin⁡x=5+πsin⁡xR_1(x)= 1+ 4 + \pi \sin x=5 + \pi \sin x and inner radius r1(x)=1+4+2π−2x=5+2π−2xr_1(x) = 1+4 + 2\pi - 2x=5 + 2\pi - 2x.

  • When π≤x≤3π2\pi \le x \le \frac{3\pi}{2}, the top of the strip is at y=4+2π−2xy = 4 + 2\pi - 2x and the bottom of the strip is at y=4+πsin⁡xy = 4 + \pi \sin x. When the strip is rotated, we get a thin washer with outer radius R2(x)=1+4+2π−2x=5+2π−2xR_2(x) = 1+4 + 2\pi - 2x=5 + 2\pi - 2x and inner radius r2(x)=1+4+πsin⁡x=5+πsin⁡xr_2(x) = 1+ 4 + \pi \sin x=5 + \pi \sin x.

So, the total

Volume=∫π/2ππ[R1(x)2−r1(x)2] dx+∫π3π/2π[R2(x)2−r2(x)2] dx=∫π/2ππ[(5+πsin⁡x)2−(5+2π−2x)2] dx+∫π3π/2π[(5+2π−2x)2−(5+πsin⁡x)2] dx\begin{align*} \hbox{Volume} &= \int_{\pi/2}^\pi \pi\big[R_1(x)^2-r_1(x)^2\big]\ \dee{x} +\int^{3\pi/2}_\pi \pi\big[R_2(x)^2-r_2(x)^2\big]\ \dee{x}\\ &= \int_{\pi/2}^\pi \pi\big[(5 + \pi \sin x)^2-(5 + 2\pi - 2x)^2\big]\ \dee{x} \\ &\hskip0.5in +\int^{3\pi/2}_\pi \pi\big[(5 + 2\pi - 2x)^2-(5 + \pi \sin x)^2\big]\ \dee{x} \end{align*}
Q22Stage 3

On a particular, highly homogeneous (This is clearly a simplified model: air density changes all the time, and depends on lots of complicated factors aside from altitude. However, the equation we're using is not so far off from an idealized model of the earth's atmosphere, taken from Pressure and the Gas Laws by H.P. Schmid, http://www.indiana.edu/ geog109/topics/10_Forces&Winds/GasPressWeb/PressGasLaws.html, accessed 3 July 2017.) planet, we observe that the density of the atmosphere hh kilometres above the surface is given by the equation ρ(h)=c2−h/6kgm3\rho(h) = c2^{-h/6}\quad \frac{\mathrm{kg}}{\mathrm{m^3}}, where cc is the density on the planet's surface.

  1. What is the mass of the atmosphere contained in a vertical column with radius one metre, sixty kilometres high?

  2. What height should a column be to contain 3000cπlog⁡2\dfrac{3000c\pi}{\log 2} kilograms of air?

Hint

You can use ideas from this section to answer the question. If you take a very thin slice of the column, the density is almost constant, so you can find the mass. Then you can add up all your little slices. It's the same idea as volume, only applied to mass.

Do be careful about units: in the problem statement, some are given in metres, others in kilometres.

If you're having a hard time with the antiderivative, try writing the exponential function with base ee. Remember 2=elog⁡22 = e^{\log 2}.

Answer

(a) 6000cπlog⁡2(1−1210)\dfrac{6000c\pi}{\log 2}\left(1-\dfrac{1}{2^{10}}\right), which is close to 6000cπlog⁡2\dfrac{6000c\pi}{\log 2}.

(b) 6km: that is, there is roughly the same mass of air in the lowest 6 km of the column as there is in the remaining 54 km.

Full solution

(a)

We use the same ideas for volume, and apply them to mass. We want to take slices of the column, approximate their mass, then add them up. To reconcile our units, let k=1000hk=1000h, so kk is the height in metres. Then the density of air at height kk is c2−k/6000 kgm3c2^{-k/6000} ~\frac{\mathrm{kg}}{\mathrm{m}^3}.

A horizontal slice of the column is a circular disk with height dk\dee{k} and radius 11 m. So, its volume is π dk m3\pi~ \dee{k}~\mathrm{m^3}. What we're interested in, though, is its mass. At height kk, its mass is

(volume)×(density)=(π dk m3)×(c2−k/6000 kgm3)=cπ2−k/6000 dkkg\begin{align*}(\mathrm{volume})\times (\mathrm{density})&=\left(\pi ~\dee{k} ~\mathrm{m^3}\right)\times \left(c2^{-k/6000}~\frac{\mathrm{kg}}{\mathrm{m^3}}\right)\\ &=c\pi2^{-k/6000}~\dee{k}\quad\mathrm{kg}\end{align*}

Since kk runs from 0 to 60,00060,000, the total mass is given by

∫060000cπ2−k/6000 dk=cπ∫0600002−k/6000 dk\begin{align*}\int_0^{60000} c\pi2^{-k/6000}~\dee{k}&=c\pi\int_0^{60000} 2^{-k/6000}~\dee{k}\end{align*}

To facilitate integration, we can write our exponential function in terms of ee, then use the substitution u=−k6000log⁡2u=-\frac{k}{6000}\log 2, du=−16000log⁡2 dk\dee{u} = -\frac{1}{6000}\log 2~\dee{k}.

=cπ∫060000(elog⁡2)−k/6000dk=cπ∫060000e−k6000log⁡2dk=−6000cπlog⁡2∫0−10log⁡2eudu=6000cπlog⁡2∫−10log⁡20eudu=6000cπlog⁡2(1−1210)\begin{align*}&=c\pi\int_0^{60000}\left(e^{\log 2}\right)^{-k/6000}\dee{k}\\ &=c\pi\int_0^{60000}e^{-\tfrac{k}{6000}\log 2}\dee{k}\\ &=-\frac{6000c\pi}{\log 2}\int_0^{-10\log 2}e^{u}\dee{u}\\ &=\frac{6000c\pi}{\log 2}\int_{-10\log 2}^0e^{u}\dee{u}\\ &=\frac{6000c\pi}{\log 2}\left(1-\frac{1}{2^{10}}\right)\end{align*}

We note this is fairly close to 6000cπlog⁡2\dfrac{6000c\pi}{\log 2}.

We also remark that this is a demonstration of the usefulness of integrals. We wanted to know how much of something there was, but the amount of that something was different everywhere: more in some places, less in others. Integration allowed us to account for this gradient. You've seen this behaviour exploited to find distances travelled, areas, volumes, and now mass. In your studies, you will doubtless learn to use it to find still more quantities, and we will discuss other applications in Chapter 2 of the CLP-2 text.

(b) We want to find the value of kk that gives a mass of 3000cπlog⁡2\dfrac{3000c\pi}{\log 2}. By following our reasoning above, the mass of air in the column from the ground to height kk is

6000cπlog⁡2(1−12k/6000)\begin{align*}\frac{6000c\pi}{\log 2}\left(1-\frac{1}{2^{k/6000}}\right)&\end{align*}

So, we set this equal to the mass we want, and solve for kk.

6000cπlog⁡2(1−12k/6000)=3000cπlog⁡22(1−12k/6000)=11=22k/60002k/6000=21k=6000h=6\begin{align*}\frac{6000c\pi}{\log 2}\left(1-\frac{1}{2^{k/6000}}\right)&=\frac{3000c\pi}{\log 2}\\ 2\left(1-\frac{1}{2^{k/6000}}\right)&=1\\ 1&=\frac{2}{2^{k/6000}}\\ 2^{k/6000}&=2^1\\ k&=6000\\ h&=6\end{align*}

This means that there is roughly the same mass of air in the lowest 6 km of the column as there is in the remaining 54 km.

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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.