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Multiple Integrals

3.2 Double Integrals in Polar Coordinates

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

1Stage 1Conceptual

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

Q1Stage 1

Consider the points

(x1,y1)=(3,0)(x2,y2)=(1,1)(x3,y3)=(0,1)(x4,y4)=(−1,1)(x5,y5)=(−2,0)\begin{align*} (x_1,y_1) &= (3,0) & (x_2,y_2) &= (1,1) & (x_3,y_3) &= (0,1) \\ (x_4,y_4) &= (-1,1) & (x_5,y_5) &= (-2,0) \end{align*}

For each 1≤i≤51\le i\le 5,

  • sketch, in the xyxy-plane, the point (xi,yi)(x_i,y_i) and

  • find polar coordinates rir_i and θi\theta_i, with ri>0r_i>0 and 0≤θi<2π0\le\theta_i<2\pi, for the point (xi,yi)(x_i,y_i).

Answer

The left hand sketch below contains the points, (x1,y1)(x_1,y_1), (x3,y3)(x_3,y_3), (x5,y5)(x_5,y_5), that are on the axes. The right hand sketch below contains the points, (x2,y2)(x_2,y_2), (x4,y4)(x_4,y_4), that are not on the axes.

Figure from prob_s3.2, line 39

Figure from prob_s3.2, line 39

Figure from prob_s3.2, line 39

Figure from prob_s3.2, line 39

r1=3r_1 = 3, θ1=0\theta_1=0 r2=2r_2 = \sqrt{2}, θ2=π4\theta_2=\frac{\pi}{4} r3=1r_3 = 1, θ3=π2\theta_3=\frac{\pi}{2} r4=2r_4 = \sqrt{2}, θ4=3π4\theta_4=\frac{3\pi}{4}
r5=2r_5 = 2, θ5=π\theta_5=\pi

Full solution
Q2Stage 1

For each of the following points (xi,yi)(x_i,y_i),

  • find all pairs (ri,θi)(r_i,\theta_i) such that (xi,yi)=(ricos⁡θi , risin⁡θi)(x_i,y_i) = \big(r_i\cos\theta_i\,,\,r_i\sin\theta_i\big), and

  • in particular, find a pair (ri,θi)(r_i,\theta_i) with ri<0r_i<0 and 0≤θi<2π0\le\theta_i<2\pi such that (xi,yi)=(ricos⁡θi , risin⁡θi)(x_i,y_i) = \big(r_i\cos\theta_i\,,\,r_i\sin\theta_i\big)

  1. (x1,y1)=(−2,0)(x_1,y_1) = (-2,0)

  2. (x2,y2)=(1,1)(x_2,y_2) = (1,1)

  3. (x3,y3)=(−1,−1)(x_3,y_3) = (-1,-1)

  4. (x4,y4)=(3,0)(x_4,y_4) = (3,0)

  5. (x5,y5)=(0,1)(x_5,y_5) = (0,1)

Hint

rr is allowed to be negative.

Answer

(a) (r1=2 , θ1=nπ, n odd integer )\big(r_1=2\,,\,\theta_1= n\pi,\ n\text{ odd integer }\big) or (r1=−2 , θ1=nπ, n even integer )\big(r_1=-2\,,\,\theta_1= n\pi,\ n\text{ even integer }\big). In particular, (r1=−2 , θ1=0)\big(r_1=-2\,,\,\theta_1= 0\big) has r1<0r_1<0 and 0≤θ1<2π0\le\theta_1<2\pi.

(b) (r2=2 , θ2=π ⁣/4+2nπ)\big(r_2=\sqrt{2}\,,\,\theta_2= \nicefrac{\pi}{4} + 2n\pi\big) or (r2=−2 , θ2=5π ⁣/4+2nπ)\big(r_2=-\sqrt{2}\,,\,\theta_2= \nicefrac{5\pi}{4} + 2n\pi\big), with nn integer. In particular, (r2=−2 , θ2=5π ⁣/4)\big(r_2=-\sqrt{2}\,,\, \theta_2= \nicefrac{5\pi}{4}\big) has r2<0r_2<0 and 0≤θ2<2π0\le\theta_2<2\pi.

(c) (r3=2 , θ3=5π ⁣/4+2nπ)\big(r_3=\sqrt{2}\,,\,\theta_3= \nicefrac{5\pi}{4} + 2n\pi\big) or (r3=−2 , θ3=π ⁣/4+2nπ)\big(r_3=-\sqrt{2}\,,\,\theta_3= \nicefrac{\pi}{4} + 2n\pi\big), with nn integer. In particular, (r3=−2 , θ3=π ⁣/4)\big(r_3=-\sqrt{2}\,,\, \theta_3= \nicefrac{\pi}{4}\big) has r3<0r_3<0 and 0≤θ3<2π0\le\theta_3<2\pi.

(d) (r4=3 , θ4=0+2nπ)\big(r_4=3\,,\,\theta_4= 0 + 2n\pi\big) or (r4=−3 , θ4=π+2nπ)\big(r_4=-3\,,\,\theta_4= \pi + 2n\pi\big), with nn integer. In particular, (r4=−3 , θ4=π)\big(r_4=-3\,,\, \theta_4= \pi\big) has r4<0r_4<0 and 0≤θ4<2π0\le\theta_4<2\pi.

(e) (r5=1 , θ5=π ⁣/2+2nπ)\big(r_5=1\,,\,\theta_5= \nicefrac{\pi}{2} + 2n\pi\big) or (r5=−1 , θ5=3π ⁣/2+2nπ)\big(r_5=-1\,,\,\theta_5= \nicefrac{3\pi}{2} + 2n\pi\big), with nn integer. In particular, (r5=−1 , θ5=3π ⁣/2)\big(r_5=-1\,,\, \theta_5= \nicefrac{3\pi}{2}\big) has r5<0r_5<0 and 0≤θ5<2π0\le\theta_5<2\pi.

Full solution
Q3Stage 1

Consider the points

(x1,y1)=(3,0)(x2,y2)=(1,1)(x3,y3)=(0,1)(x4,y4)=(−1,1)(x5,y5)=(−2,0)\begin{align*} (x_1,y_1) &= (3,0) & (x_2,y_2) &= (1,1) & (x_3,y_3) &= (0,1) \\ (x_4,y_4) &= (-1,1) & (x_5,y_5) &= (-2,0) \end{align*}

Also define, for each angle θ\theta, the vectors

e^r(θ)=cos⁡θ ı^+sin⁡θ ȷ^e^θ(θ)=−sin⁡θ ı^+cos⁡θ ȷ^\begin{align*} \he_r(\theta)=\cos\theta\ \hi + \sin\theta\ \hj\qquad \he_\theta(\theta) = -\sin\theta\ \hi + \cos\theta\ \hj \end{align*}
  1. Determine, for each angle θ\theta, the lengths of the vectors e^r(θ)\he_r(\theta) and e^θ(θ)\he_\theta(\theta) and the angle between the vectors e^r(θ)\he_r(\theta) and e^θ(θ)\he_\theta(\theta). Compute e^r(θ)×e^θ(θ)\he_r(\theta)\times\he_\theta(\theta) (viewing e^r(θ)\he_r(\theta) and e^θ(θ)\he_\theta(\theta) as vectors in three dimensions with zero k^\hk components).

  2. For each 1≤i≤51\le i\le 5, sketch, in the xyxy-plane, the point (xi,yi)=(ricos⁡θi , risin⁡θi)(x_i,y_i)= \big(r_i\cos\theta_i\,,\,r_i\sin\theta_i\big) and the vectors e^r(θi)\he_r(\theta_i) and e^θ(θi)\he_\theta(\theta_i). In your sketch of the vectors, place the tails of the vectors e^r(θi)\he_r(\theta_i) and e^θ(θi)\he_\theta(\theta_i) at (xi,yi)(x_i,y_i).

Hint

Compute, for each angle θ\theta, the dot product e^r(θ)⋅e^θ(θ)\he_r(\theta)\cdot\he_\theta(\theta).

Answer

(a) Both e^r(θ)\he_r(\theta) and e^θ(θ)\he_\theta(\theta) have length 1. The angle between them is π2\frac{\pi}{2}. The cross product is e^r(θ)×e^θ(θ)=k^\he_r(\theta) \times \he_\theta(\theta)=\hk.

(b) Here is a sketch of (xi,yi)(x_i,y_i), e^r(θi)\he_r(\theta_i), e^θ(θi)\he_\theta(\theta_i) for i=1,3,5i =1,3,5 (the points on the axes)

Figure from prob_s3.2, line 365

Figure from prob_s3.2, line 365

and here is a sketch (to a different scale) of (xi,yi)(x_i,y_i), e^r(θi)\he_r(\theta_i), e^θ(θi)\he_\theta(\theta_i) for i=2,4i =2,4 (the points off the axes).

Figure from prob_s3.2, line 365

Figure from prob_s3.2, line 365

Full solution
Q4Stage 1

Let <a,b>\llt a, b\rgt be a vector. Let rr be the length of <a,b>\llt a, b\rgt and θ\theta be the angle between <a,b>\llt a, b\rgt and the xx-axis.

  1. Express aa and bb in terms of rr and θ\theta.

  2. Let <A,B>\llt A, B\rgt be the vector gotten by rotating <a,b>\llt a, b\rgt by an angle φ\varphi about its tail. Express AA and BB in terms of aa, bb and φ\varphi.

Hint

Sketch <a,b>\llt a, b\rgt and <A,B>\llt A, B\rgt. The trigonometric addition formulas

sin⁡(θ+φ)=sin⁡θcos⁡φ+cos⁡θsin⁡φcos⁡(θ+φ)=cos⁡θcos⁡φ−sin⁡θsin⁡φ\begin{align*} \sin(\theta+\varphi) &=\sin\theta\cos\varphi+\cos\theta\sin\varphi \\ \cos(\theta+\varphi) &=\cos\theta\cos\varphi-\sin\theta\sin\varphi \end{align*}

will help.

Answer

(a) a=rcos⁡θa=r\cos\theta, b=rsin⁡θb=r\sin\theta

(b) A=acos⁡φ−bsin⁡φA=a\cos\varphi-b\sin\varphi, B=bcos⁡φ+asin⁡φB=b\cos\varphi+a\sin\varphi

Full solution
Q5Stage 1

For each of the regions R\cR sketched below, express ∬Rf(x,y) dx dy\dblInt_\cR f(x,y)\,\dee{x}\,\dee{y} as an iterated integral in polar coordinates in two different ways.

(a)

Figure from prob_s3.2, line 490

Figure from prob_s3.2, line 490

(b)

Figure from prob_s3.2, line 490

Figure from prob_s3.2, line 490

(c)

Figure from prob_s3.2, line 490

Figure from prob_s3.2, line 490

(d)

Figure from prob_s3.2, line 490

Figure from prob_s3.2, line 490

Answer

(a) ∬Rf(x,y) dx dy=∫0π ⁣/4dθ∫02dr r f(rcos⁡θ,rsin⁡θ)=∫02dr∫0π ⁣/4dθ r f(rcos⁡θ,rsin⁡θ)\displaystyle\dblInt_\cR f(x,y)\,\dee{x}\,\dee{y} =\int_0^{\nicefrac{\pi}{4}}\dee{\theta} \int_0^2\dee{r}\ r\ f(r\cos\theta,r\sin\theta) =\int_0^2\dee{r} \int_0^{\nicefrac{\pi}{4}}\dee{\theta} \ r\ f(r\cos\theta,r\sin\theta)

(b) ∬Rf(x,y) dx dy=∫0π ⁣/2dθ∫12dr r f(rcos⁡θ,rsin⁡θ)=∫12dr∫0π ⁣/2dθ r f(rcos⁡θ,rsin⁡θ)\displaystyle\dblInt_\cR f(x,y)\,\dee{x}\,\dee{y} =\int_0^{\nicefrac{\pi}{2}}\dee{\theta} \int_1^2\dee{r}\ r\ f(r\cos\theta,r\sin\theta) =\int_1^2\dee{r} \int_0^{\nicefrac{\pi}{2}}\dee{\theta} \ r\ f(r\cos\theta,r\sin\theta)

(c) ∬Rf(x,y) dx dy=∫0π ⁣/2dθ∫02cos⁡θdr r f(rcos⁡θ,rsin⁡θ)=∫02dr∫0arccos⁡r2dθ r f(rcos⁡θ,rsin⁡θ)\displaystyle \dblInt_\cR f(x,y)\,\dee{x}\,\dee{y} =\int_0^{\nicefrac{\pi}{2}}\dee{\theta} \int_0^{2\cos\theta}\dee{r}\ r\ f(r\cos\theta,r\sin\theta) =\int_0^2\dee{r} \int_0^{\arccos\frac{r}{2}}\dee{\theta} \ r\ f(r\cos\theta,r\sin\theta)

(d)

∬Rf(x,y) dx dy=∫π ⁣/4π ⁣/2dθ∫02 ⁣/sin⁡θdr r f(rcos⁡θ,rsin⁡θ)=∫02dr∫π ⁣/4π ⁣/2dθ r f(rcos⁡θ,rsin⁡θ)+∫222dr∫π ⁣/4arcsin⁡2rdθ r f(rcos⁡θ,rsin⁡θ)\begin{align*} \dblInt_\cR f(x,y)\,\dee{x}\,\dee{y} &=\int_{\nicefrac{\pi}{4}}^{\nicefrac{\pi}{2}}\dee{\theta} \int_0^{\nicefrac{2}{\sin\theta}}\dee{r}\ r\ f(r\cos\theta,r\sin\theta)\\ &=\int_0^2\dee{r} \int_{\nicefrac{\pi}{4}}^{\nicefrac{\pi}{2}}\dee{\theta} \ r\ f(r\cos\theta,r\sin\theta) +\int_2^{2\sqrt{2}}\dee{r} \int_{\nicefrac{\pi}{4}}^{\arcsin\frac{2}{r}}\dee{\theta} \ r\ f(r\cos\theta,r\sin\theta) \end{align*}
Full solution
Q6Stage 1

Sketch the domain of integration in the xyxy-plane for each of the following polar coordinate integrals.

  1. ∫12dr∫−π ⁣/4π ⁣/4dθ r f(rcos⁡θ,rsin⁡θ)\displaystyle \int_1^2\dee{r} \int_{-\nicefrac{\pi}{4}}^{\nicefrac{\pi}{4}}\dee{\theta} \ r\ f(r\cos\theta,r\sin\theta)

  2. ∫0π ⁣/4dθ∫02sin⁡θ+cos⁡θdr r f(rcos⁡θ,rsin⁡θ)\displaystyle \int_0^{\nicefrac{\pi}{4}}\dee{\theta} \int_0^{\frac{2}{\sin\theta+\cos\theta}}\dee{r} \ r\ f(r\cos\theta,r\sin\theta)

  3. ∫02πdθ∫03cos⁡2θ+9sin⁡2θdr r f(rcos⁡θ,rsin⁡θ)\displaystyle \int_0^{2\pi}\dee{\theta} \int_0^{\frac{3}{\sqrt{\cos^2\theta+9\sin^2\theta}}}\dee{r} \ r\ f(r\cos\theta,r\sin\theta)

Answer

(a)

Figure from prob_s3.2, line 806

Figure from prob_s3.2, line 806

(b)

Figure from prob_s3.2, line 806

Figure from prob_s3.2, line 806

(c)

Figure from prob_s3.2, line 806

Figure from prob_s3.2, line 806

Full solution

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q7Stage 2

Use polar coordinates to evaluate each of the following integrals.

  1. ∬S(x+y)dx dy\displaystyle\dblInt_S (x+y)\dee{x}\,\dee{y} where SS is the region in the first quadrant lying inside the disc x2+y2≤a2x^2+y^2\le a^2 and under the line y=3xy=\sqrt{3}x.

  2. ∬Sx dx dy\displaystyle\dblInt_Sx\ \dee{x}\,\dee{y}, where SS is the disc segment x2+y2≤2, x≥1x^2+y^2\le 2,\ x\ge 1.

  3. ∬T(x2+y2)dx dy\displaystyle\dblInt_T (x^2+y^2)\dee{x}\,\dee{y} where TT is the triangle with vertices (0,0),(1,0)(0,0), (1,0) and (1,1)(1,1).

  4. ∬x2+y2≤1ln⁡(x2+y2) dx dy\displaystyle\dblInt_{x^2+y^2\le 1} \ln(x^2+y^2)\,\dee{x}\,\dee{y}

Answer

(a) a36[3+1]\frac{a^3}{6}\big[\sqrt{3}+1\big] (b) 23\frac{2}{3} (c) 13\frac{1}{3} (d) −π-\pi

Full solution
Q8Stage 2

Find the volume lying inside the sphere x2+y2+z2=2x^2+y^2+z^2=2 and above the paraboloid z=x2+y2z=x^2+y^2.

Answer

π[432−76]≈2.26\pi \left[\frac{4}{3}\sqrt{2}-\frac{7}{6}\right] \approx 2.26

Full solution
Q9Stage 2

Let a>0a>0. Find the volume lying inside the cylinder x2+(y−a)2=a2x^2+(y-a)^2=a^2 and between the upper and lower halves of the cone z2=x2+y2z^2=x^2+y^2.

Answer

649a3\frac{64}{9}a^3

Full solution
Q10Stage 2

Let a>0a>0. Find the volume common to the cylinders x2+y2≤2axx^2+y^2\le 2ax and z2≤2axz^2\le 2ax.

Answer

12815a3\frac{128}{15}a^3

Full solution
Q11Stage 2Past exam · M200 2005D

Consider the region EE in 3–dimensions specified by the inequalities x2+y2≤2yx^2 + y^2 \le 2y and 0≤z≤x2+y20 \le z \le \sqrt{x^2 + y^2}.

  1. Draw a reasonably accurate picture of EE in 3–dimensions. Be sure to show the units on the coordinate axes.

  2. Use polar coordinates to find the volume of EE. Note that you will be “using polar coordinates” if you solve this problem by means of cylindrical coordinates.

    Hint:
    ∫sin⁡nu du=−1nsin⁡n−1u cos⁡u+n−1n∫sin⁡n−2u du\int \sin^n u\ \dee{u} =-\frac{1}{n}\sin^{n-1}u\ \cos u +\frac{n-1}{n}\int\sin^{n-2}u\ \dee{u}

Answer

(a)

Figure from prob_s3.2, line 1200

Figure from prob_s3.2, line 1200

(b) 329\frac{32}{9}

Full solution
Q12Stage 2Past exam · M200 2006A

Evaluate the iterated double integral

∫x=0x=2∫y=0y=4−x2(x2+y2)32 dy dx\begin{equation*} \int_{x=0}^{x=2}\int_{y=0}^{y=\sqrt{4-x^2}} {(x^2+y^2)}^{\frac{3}{2}}\ \dee{y}\,\dee{x} \end{equation*}
Answer

16π5\frac{16\pi}{5}

Full solution
Q13Stage 2Past exam · M200 2006D
  1. Sketch the region L\cL (in the first quadrant of the xyxy–plane) with boundary curves

    x2+y2=2, x2+y2=4, y=x, y=0.\begin{equation*} x^2 + y^2 = 2,\ x^2 + y^2 = 4,\ y = x,\ y = 0. \end{equation*}

    The mass of a thin lamina with a density function ρ(x,y)\rho(x,y) over the region L\cL is given by

    M=∬Lρ(x,y) dA\begin{equation*} M =\dblInt_\cL \rho(x,y)\,\dee{A} \end{equation*}
  2. Find an expression for MM as an integral in polar coordinates.

  3. Find M when

    ρ(x,y)=2xyx2+y2\begin{equation*} \rho(x,y) = \frac{2xy}{x^2+y^2} \end{equation*}
Answer

(a)

Figure from prob_s3.2, line 1376

Figure from prob_s3.2, line 1376

(b) M=∫0π/4dθ∫22dr r ρ(rcos⁡θ , rsin⁡θ)M = \int_0^{\pi/4}\dee{\theta}\int_{\sqrt{2}}^2\dee{r}\ r\, \rho(r\cos\theta\,,\,r\sin\theta) (c) 12\frac{1}{2}

Full solution
Q14Stage 2Past exam · M200 2009A

Evaluate ∬R21(1+x2+y2)2 dA\displaystyle \dblInt_{\bbbr^2} \frac{1}{{(1+x^2+y^2)}^2}\ \dee{A}.

Answer

π\pi

Full solution
Q15Stage 2Past exam · M200 2011D

Evaluate the double integral

∬Dyx2+y2 dA\begin{equation*} \dblInt_D y\sqrt{x^2+y^2}\,\dee{A} \end{equation*}

over the region D={ (x,y) ∣ x2+y2≤2, 0≤y≤x }D =\Set{(x,y)}{ x^2+y^2\le 2,\ 0\le y\le x}.

Answer

1−121-\frac{1}{\sqrt{2}}

Full solution
Q16Stage 2Past exam · M200 2013D

This question is about the integral

∫01∫3y4−y2ln⁡(1+x2+y2) dx dy\begin{equation*} \int_0^1 \int_{\sqrt{3}y}^{\sqrt{4-y^2}} \ln\big(1+x^2+y^2)\ \dee{x}\,\dee{y} \end{equation*}
  1. Sketch the domain of integration.

  2. Evaluate the integral by transforming to polar coordinates.

Answer

(a)

Figure from prob_s3.2, line 1532

Figure from prob_s3.2, line 1532

(b) π12[5ln⁡(5)−4]\frac{\pi}{12}\big[5\ln(5)-4\big]

Full solution
Q17Stage 2Past exam · M200 2014A

Let DD be the region in the xyxy–plane bounded on the left by the line x=2x = 2 and on the right by the circle x2+y2=16x^2 + y^2 = 16. Evaluate

∬D(x2+y2)−3/2 dA\begin{equation*} \dblInt_D\big(x^2+y^2\big)^{-3/2}\ \dee{A} \end{equation*}
Answer

32−π6\frac{\sqrt{3}}{2}-\frac{\pi}{6}

Full solution
Q18Stage 2Past exam · M200 2014D

In the xyxy–plane, the disk x2+y2≤2xx^2 + y^2 \le 2x is cut into 22 pieces by the line y=xy = x. Let DD be the larger piece.

  1. Sketch DD including an accurate description of the center and radius of the given disk. Then describe DD in polar coordinates (r,θ)(r, \theta).

  2. Find the volume of the solid below z=x2+y2z = \sqrt{x^2 + y^2} and above DD.

Answer

(a) D={ (rcos⁡θ , rsin⁡θ) ∣ −π ⁣/2≤θ≤π ⁣/4, 0≤r≤2cos⁡θ }D = \Set{(r\cos\theta\,,\,r\sin\theta)} {-\nicefrac{\pi}{2}\le\theta\le\nicefrac{\pi}{4},\ 0\le r\le 2\cos\theta}

(b) Volume=40182+169\text{Volume}=\frac{40}{18\sqrt{2}} +\frac{16}{9}

Full solution
Q19Stage 2Past exam · M200 2000A

Let DD be the shaded region in the diagram. Find the average distance of points in DD from the origin. You may use that ∫cos⁡n(x) dx=cos⁡n−1(x)sin⁡(x)n+n−1n∫cos⁡n−2(x) dx\int\cos^n(x)\,dx = \frac{\cos^{n-1}(x)\sin(x)}{n} +\frac{n-1}{n}\int\cos^{n-2}(x)\,dx for all natural numbers n≥2n\ge 2.

Figure from prob_s3.2, line 1687

Figure from prob_s3.2, line 1687

Answer

2π+44/9π+8≈1.4422\frac{\pi+44/9}{\pi+8}\approx 1.442

Full solution

3Stage 3Application

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

Q20Stage 3Past exam · M200 2010A

Let GG be the region in R2\bbbr^2 given by

x2+y2≤10≤x≤2yy≤2x\begin{gather*} x^2 + y^2 \le 1 \\ 0 \le x \le 2y \\ y \le 2x \end{gather*}
  1. Sketch the region GG.

  2. Express the integral ∬Gf(x,y) dA\dblInt_G f(x,y)\ \dee{A} a sum of iterated integrals ∬f(x,y) dxdy\dblInt f(x, y)\ \dee{x}\dee{y}.

  3. Express the integral ∬Gf(x,y) dA\dblInt_G f(x,y)\ \dee{A} as an iterated integral in polar coordinates (r,θ)(r, \theta) where x=rcos⁡(θ)x = r \cos(\theta) and y=rsin⁡(θ)y = r \sin(\theta).

Answer

(a)

Figure from prob_s3.2, line 1775

Figure from prob_s3.2, line 1775

(b)

∬Gf(x,y) dA=∫015dy∫y/22ydx f(x,y)+∫1525dy∫y/21−y2dx f(x,y)\begin{align*} \dblInt_G f(x,y)\ \dee{A} &=\int_0^{\frac{1}{\sqrt{5}}}\dee{y}\int_{y/2}^{2y}\dee{x}\ f(x,y) +\int_{\frac{1}{\sqrt{5}}}^{\frac{2}{\sqrt{5}}}\dee{y}\int_{y/2}^{\sqrt{1-y^2}} \dee{x}\ f(x,y) \end{align*}

(c)

∬Gf(x,y) dA=∫arctan⁡12arctan⁡2dθ∫01dr r f(rcos⁡θ,rsin⁡θ)\begin{align*} \dblInt_G f(x,y)\ \dee{A} &=\int_{\arctan\frac{1}{2}}^{\arctan 2}\dee{\theta} \int_0^1\dee{r}\ r\,f(r\cos\theta,r\sin\theta) \end{align*}
Full solution
Q21Stage 3Past exam · M200 2011A

Consider

J=∫02∫y4−y2yxex2+y2 dx dy\begin{equation*} J = \int_0^{\sqrt{2}} \int_y^{\sqrt{4-y^2}} \frac{y}{x} e^{x^2+y^2}\ \dee{x}\,\dee{y} \end{equation*}
  1. Sketch the region of integration.

  2. Reverse the order of integration.

  3. Evaluate JJ by using polar coordinates.

Answer

(a)

Figure from prob_s3.2, line 1912

Figure from prob_s3.2, line 1912

(b) J=∫02∫0xyxex2+y2 dy dx+∫22∫04−x2yxex2+y2 dy dxJ = \int_0^{\sqrt{2}}\int_0^x \frac{y}{x} e^{x^2+y^2}\ \dee{y}\,\dee{x} + \int_{\sqrt{2}}^2\int_0^{\sqrt{4-x^2}} \frac{y}{x} e^{x^2+y^2}\ \dee{y}\,\dee{x}

(c) 14[e4−1]ln⁡2\frac{1}{4}\left[e^4-1\right]\ln 2

Full solution
Q22Stage 3

Find the volume of the region in the first octant below the paraboloid

z=1−x2a2−y2b2z=1-\frac{x^2}{a^2}-\frac{y^2}{b^2}
Answer

π8ab\frac{\pi}{8}ab

Full solution
Q23Stage 3

A symmetrical coffee percolator holds 24 cups when full. The interior has a circular cross-section which tapers from a radius of 3" at the centre to 2" at the base and top, which are 12" apart. The bounding surface is parabolic. Where should the mark indicating the 6 cup level be placed?

Figure from prob_s3.2, line 2032

Figure from prob_s3.2, line 2032

Answer

About 3.5” above the bottom

Full solution
Q24Stage 3Past exam · M200 2003A

Consider the surface SS given by z=ex2+y2z=e^{x^2+y^2}.

  1. Compute the volume under SS and above the disk x2+y2≤9x^2+y^2\le 9 in the xyxy-plane.

  2. The volume under SS and above a certain region RR in the xyxy-plane is

    ∫01(∫0yex2+y2 dx)dy+∫12(∫02−yex2+y2 dx)dy\begin{equation*} \int_0^1\bigg(\int_0^y e^{x^2+y^2}\,\dee{x}\bigg)\dee{y} +\int_1^2\bigg(\int_0^{2-y} e^{x^2+y^2}\,\dee{x}\bigg)\dee{y} \end{equation*}

    Sketch RR and express the volume as a single iterated integral with the order of integration reversed. Do not compute either integral in part (b).

Answer

(a) π(e9−1)≈25,453\pi\big(e^9-1\big)\approx 25,453

(b) ∫01dx∫x2−xdy ex2+y2\int_0^1 \dee{x}\int_x^{2-x}\dee{y}\ e^{x^2+y^2}

Figure from prob_s3.2, line 2116

Figure from prob_s3.2, line 2116

Full solution

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From the UBC Math 200 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.