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Multiple Integrals

3.6 Triple Integrals in Cylindrical Coordinates

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

1Stage 1Conceptual

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

Q1Stage 1

Use (r,θ,z)(r,\theta,z) to denote cylindrical coordinates.

  1. Draw r=0r=0.

  2. Draw r=1r=1.

  3. Draw θ=0\theta=0.

  4. Draw θ=π ⁣/4\theta=\nicefrac{\pi}{4}.

Answer

(a), (b)

Figure from prob_s3.6, line 30

Figure from prob_s3.6, line 30

Figure from prob_s3.6, line 30

Figure from prob_s3.6, line 30

(c), (d)

Figure from prob_s3.6, line 30

Figure from prob_s3.6, line 30

Figure from prob_s3.6, line 30

Figure from prob_s3.6, line 30

Full solution

(a), (b) Since the cylindrical coordinate r(x,y,z)r(x,y,z) of a point (x,y,z)(x,y,z) is the distance, x2+y2\sqrt{x^2+y^2}, from (x,y,z)(x,y,z) to the zz-axis, the sets

{ (x,y,z) ∣ r(x,y,z)=0 }={ (x,y,z) ∣ x2+y2=0 }={ (x,y,z) ∣ x=y=0 }=the z-axis{ (x,y,z) ∣ r(x,y,z)=1 }={ (x,y,z) ∣ x2+y2=1 }=the cylinder of radius 1 centred on the z-axis\begin{align*} \Set{(x,y,z)}{r(x,y,z)=0} &=\Set{(x,y,z)}{x^2+y^2=0} =\Set{(x,y,z)}{x=y=0} \\ &=\text{the }z\text{-axis} \\ \Set{(x,y,z)}{r(x,y,z)=1} &=\Set{(x,y,z)}{x^2+y^2=1} \\ &=\text{the cylinder of radius }1\text{ centred on the }z\text{-axis} \end{align*}

Figure from prob_s3.6, line 30

Figure from prob_s3.6, line 30

Figure from prob_s3.6, line 30

Figure from prob_s3.6, line 30

(c), (d) Since the cylindrical coordinate θ(x,y,z)\theta(x,y,z) of a point (x,y,z)(x,y,z) is the angle between the positive xx-axis and the line from (0,0,0)(0,0,0) to (x,y,0)(x,y,0), the sets

{ (x,y,z) ∣ θ(x,y,z)=0 }=the half of the xz-plane with x>0{ (x,y,z) ∣ θ(x,y,z)=π ⁣/4 }=the half of the plane y=x with x>0\begin{align*} \Set{(x,y,z)}{\theta(x,y,z)=0} &=\text{the half of the }xz\text{-plane with }x>0 \\ \Set{(x,y,z)}{\theta(x,y,z)=\nicefrac{\pi}{4}} &=\text{the half of the plane }y=x\text{ with }x>0 \end{align*}

Figure from prob_s3.6, line 30

Figure from prob_s3.6, line 30

Figure from prob_s3.6, line 30

Figure from prob_s3.6, line 30

Q2Stage 1

Sketch the points with the specified cylindrical coordinates.

  1. r=1r=1, θ=0\theta=0, z=0z=0

  2. r=1r=1, θ=π4\theta=\frac{\pi}{4}, z=0z=0

  3. r=1r=1, θ=π2\theta=\frac{\pi}{2}, z=0z=0

  4. r=0r=0, θ=π\theta=\pi, z=1z=1

  5. r=1r=1, θ=π4\theta=\frac{\pi}{4}, z=1z=1

Answer

Figure from prob_s3.6, line 101

Figure from prob_s3.6, line 101

Full solution

The sketch is below. To help build up this sketch, it is useful to recall the following facts.

  • The cylindrical coordinate rr is the distance of the point from the zz-axis. In particular all points with r=0r=0 lie on the zz-axis (for all values of θ\theta).

  • The cylindrical coordinate zz is the distance of the point from the xyxy-plane. In particular all points with z=0z=0 lie on the xyxy-plane.

Figure from prob_s3.6, line 101

Figure from prob_s3.6, line 101

Q3Stage 1

Convert from cylindrical to Cartesian coordinates.

  1. r=1r=1, θ=0\theta=0, z=0z=0

  2. r=1r=1, θ=π4\theta=\frac{\pi}{4}, z=0z=0

  3. r=1r=1, θ=π2\theta=\frac{\pi}{2}, z=0z=0

  4. r=0r=0, θ=π\theta=\pi, z=1z=1

  5. r=1r=1, θ=π4\theta=\frac{\pi}{4}, z=1z=1

Answer

(a) (1,0,0)(1,0,0) (b) (12,12,0)\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right) (c) (0,1,0)(0,1,0) (d) (0,0,1)(0,0,1) (e) (12,12,1)\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},1\right)

Full solution

(a) When θ=0\theta=0, sin⁡θ=0\sin\theta=0 and cos⁡θ=1\cos\theta=1, so that the polar coordinates r=1r=1, θ=0\theta=0, z=0z=0 correspond to the Cartesian coordinates

(x,y,z)=(rcos⁡θ,rsin⁡θ,z)=(1×cos⁡0, 1×sin⁡0, 0)=(1,0,0)\begin{equation*} (x,y,z) = (r\cos\theta,r\sin\theta,z) = (1\times\cos 0,\ 1\times\sin 0,\ 0) =(1,0,0) \end{equation*}

(b) When θ=π4\theta=\frac{\pi}{4}, sin⁡θ=cos⁡θ=12\sin\theta=\cos\theta=\frac{1}{\sqrt{2}}, so that the polar coordinates r=1r=1, θ=π4\theta=\frac{\pi}{4}, z=0z=0 correspond to the Cartesian coordinates

(x,y,z)=(rcos⁡θ,rsin⁡θ,z)=(1×cos⁡π4, 1×sin⁡π4, 0)=(12,12,0)\begin{equation*} (x,y,z) = (r\cos\theta,r\sin\theta,z) = \left(1\times\cos \frac{\pi}{4},\ 1\times\sin \frac{\pi}{4},\ 0\right) = \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right) \end{equation*}

(c) When θ=π2\theta=\frac{\pi}{2}, sin⁡θ=1\sin\theta=1 and cos⁡θ=0\cos\theta=0, so that the polar coordinates r=1r=1, θ=π2\theta=\frac{\pi}{2}, z=0z=0 correspond to the Cartesian coordinates

(x,y,z)=(rcos⁡θ,rsin⁡θ,z)=(1×cos⁡π2, 1×sin⁡π2, 0)=(0,1,0)\begin{equation*} (x,y,z) = (r\cos\theta,r\sin\theta,z) = \left(1\times\cos \frac{\pi}{2},\ 1\times\sin \frac{\pi}{2},\ 0\right) = (0,1,0) \end{equation*}

(d) When θ=π\theta=\pi, sin⁡θ=0\sin\theta=0 and cos⁡θ=−1\cos\theta=-1, so that the polar coordinates r=0r=0, θ=π\theta=\pi, z=1z=1 correspond to the Cartesian coordinates

(x,y,z)=(rcos⁡θ,rsin⁡θ,z)=(0×cos⁡π, 0×sin⁡π, 1)=(0,0,1)\begin{equation*} (x,y,z) = (r\cos\theta,r\sin\theta,z) = \left(0\times\cos\pi,\ 0\times\sin\pi,\ 1\right) = (0,0,1) \end{equation*}

(e) When θ=π4\theta=\frac{\pi}{4}, sin⁡θ=cos⁡θ=12\sin\theta=\cos\theta=\frac{1}{\sqrt{2}}, so that the polar coordinates r=1r=1, θ=π4\theta=\frac{\pi}{4}, z=1z=1 correspond to the Cartesian coordinates

(x,y,z)=(rcos⁡θ,rsin⁡θ,z)=(1×cos⁡π4, 1×sin⁡π4, 1)=(12,12,1)\begin{equation*} (x,y,z) = (r\cos\theta,r\sin\theta,z) = \left(1\times\cos \frac{\pi}{4},\ 1\times\sin \frac{\pi}{4},\ 1\right) = \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},1\right) \end{equation*}
Q4Stage 1

Convert from Cartesian to cylindrical coordinates.

  1. (1,1,2)(1,1,2)

  2. (−1,−1,2)(-1,-1,2)

  3. (−1,3,0)(-1,\sqrt{3}, 0)

  4. (0,0,1)(0,0,1)

Answer

(a) r=2r=\sqrt{2}, z=2z=2, θ=π4\theta=\frac{\pi}{4} (plus possibly any integer multiple of 2π2\pi)

(b) r=2r=\sqrt{2}, z=2z=2, θ=5π4\theta=\frac{5\pi}{4} (plus possibly any integer multiple of 2π2\pi)

(c) r=2r=2, z=0z=0, θ=2π3\theta=\frac{2\pi}{3} (plus possibly any integer multiple of 2π2\pi)

(d) r=0r=0, z=1z=1, θ=arbitrary\theta=\text{arbitrary}

Full solution

(a) The cylindrical coordinates must obey

1=x=rcos⁡θ1=y=rsin⁡θ2=z\begin{equation*} 1=x=r\cos\theta\qquad 1=y=r\sin\theta\qquad 2=z \end{equation*}

So z=2z=2, r=12+12=2r=\sqrt{1^2+1^2}=\sqrt{2} and tan⁡θ=yx=11=1\tan\theta=\frac{y}{x} =\frac{1}{1}=1. Recall that tan⁡(π4+kπ)=1\tan\left(\frac{\pi}{4}+k\pi\right)=1 for all integers kk. As (x,y)=(1,1)(x,y)=(1,1) lies in the first quadrant, 0≤θ≤π20\le\theta\le\frac{\pi}{2}. So θ=π4\theta=\frac{\pi}{4} (plus possibly any integer multiple of 2π2\pi).

(b) The cylindrical coordinates must obey

−1=x=rcos⁡θ−1=y=rsin⁡θ2=z\begin{equation*} -1=x=r\cos\theta\qquad -1=y=r\sin\theta\qquad 2=z \end{equation*}

So z=2z=2, r=(−1)2+(−1)2=2r=\sqrt{(-1)^2+(-1)^2}=\sqrt{2} and tan⁡θ=yx=−1−1=1\tan\theta=\frac{y}{x} =\frac{-1}{-1}=1. Recall that tan⁡(π4+kπ)=1\tan\left(\frac{\pi}{4}+k\pi\right)=1 for all integers kk. As (x,y)=(−1,−1)(x,y)=(-1,-1) lies in the third quadrant, π≤θ≤3π2\pi\le\theta\le\frac{3\pi}{2}. So θ=5π4\theta=\frac{5\pi}{4} (plus possibly any integer multiple of 2π2\pi).

(c) The cylindrical coordinates must obey

−1=x=rcos⁡θ3=y=rsin⁡θ0=z\begin{equation*} -1=x=r\cos\theta\qquad \sqrt{3}=y=r\sin\theta\qquad 0=z \end{equation*}

So z=0z=0, r=(−1)2+(3)2=2r=\sqrt{(-1)^2+\big(\sqrt{3}\big)^2}=2 and tan⁡θ=yx=3−1=−3\tan\theta=\frac{y}{x} =\frac{\sqrt{3}}{-1}=-\sqrt{3}. Recall that tan⁡(2π3+kπ)=−3\tan\left(\frac{2\pi}{3}+k\pi\right)=-\sqrt{3} for all integers kk. As (x,y)=(−1,3)(x,y)=(-1,\sqrt{3}) lies in the second quadrant, π2≤θ≤π\frac{\pi}{2}\le\theta\le\pi. So θ=2π3\theta=\frac{2\pi}{3} (plus possibly any integer multiple of 2π2\pi).

(d) The cylindrical coordinates must obey

0=x=rcos⁡θ0=y=rsin⁡θ1=z\begin{equation*} 0=x=r\cos\theta\qquad 0=y=r\sin\theta\qquad 1=z \end{equation*}

So z=1z=1, r=02+02=0r=\sqrt{0^2+0^2}=0 and θ\theta is completely arbitrary.

Q5Stage 1

Rewrite the following equations in cylindrical coordinates.

  1. z=2xyz=2xy

  2. x2+y2+z2=1x^2+y^2+z^2=1

  3. (x−1)2+y2=1(x-1)^2 + y^2 =1

Answer

(a) z=r2sin⁡(2θ)z=r^2\sin(2\theta) (b) r2+z2=1r^2+z^2=1 (c) r=2cos⁡θr=2\cos\theta

Full solution

(a) As x=rcos⁡θx=r\cos\theta and y=rsin⁡θy=r\sin\theta,

z=2xy  ⟺  z=2r2cos⁡θsin⁡θ=r2sin⁡(2θ)\begin{equation*} z=2xy \iff z=2r^2\cos\theta\sin\theta = r^2\sin(2\theta) \end{equation*}

(b) As x=rcos⁡θx=r\cos\theta and y=rsin⁡θy=r\sin\theta,

x2+y2+z2=1  ⟺  r2cos⁡2θ+r2sin⁡2θ+z2=1  ⟺  r2+z2=1\begin{equation*} x^2+y^2+z^2=1 \iff r^2\cos^2\theta+r^2\sin^2\theta + z^2 =1 \iff r^2+z^2=1 \end{equation*}

(c) As x=rcos⁡θx=r\cos\theta and y=rsin⁡θy=r\sin\theta,

(x−1)2+y2=1  ⟺  (rcos⁡θ−1)2+(rsin⁡θ)2=1  ⟺  r2cos⁡2θ−2rcos⁡θ+1+r2sin⁡2θ=1  ⟺  r2=2rcos⁡θ  ⟺  r=2cos⁡θ or r=0  ⟺  r=2cos⁡θ\begin{align*} (x-1)^2 + y^2 =1 &\iff (r\cos\theta-1)^2 + (r\sin\theta)^2 =1 \\ &\iff r^2\cos^2\theta -2r\cos\theta +1 + r^2\sin^2\theta = 1 \\ &\iff r^2=2r\cos\theta \iff r=2\cos\theta\text{ or }r=0 \\ &\iff r=2\cos\theta \end{align*}

Note that the solution r=0r=0 is included in r=2cos⁡θr=2\cos\theta — just choose θ=π2\theta=\frac{\pi}{2}.

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q6Stage 2

Use cylindrical coordinates to evaluate the volumes of each of the following regions.

  1. Above the xyxy-plane, inside the cone z=2a−x2+y2z=2a-\sqrt{x^2+y^2} and inside the cylinder x2+y2=2ayx^2+y^2=2ay, where a>0a>0 is a constant.

  2. Above the xyxy-plane, under the paraboloid z=1−x2−y2z=1-x^2-y^2 and in the wedge −x≤y≤3x-x\le y\le \sqrt{3}x.

  3. Above the paraboloid z=x2+y2z=x^2+y^2 and below the plane z=2yz=2y.

Answer

(a) a3(2π−329)a^3\big(2\pi-\frac{32}{9}\big) (b) 748π\frac{7}{48}\pi (c) π2\frac{\pi}{2}

Full solution

(a) In cylindrical coordinates, the cone z=2a−x2+y2z=2a-\sqrt{x^2+y^2} is z=2a−rz=2a-r and the cylinder x2+y2=2ayx^2+y^2=2ay is r2=2arsin⁡θr^2=2ar\sin\theta or r=2asin⁡θr=2a\sin\theta. The figures below show the parts of the cone, the cylinder and the intersection, respectively, that are in the first octant.

Figure from prob_s3.6, line 365

Figure from prob_s3.6, line 365

Figure from prob_s3.6, line 365

Figure from prob_s3.6, line 365

Figure from prob_s3.6, line 365

Figure from prob_s3.6, line 365

The specified region is

V={ (rcos⁡θ , rsin⁡θ , z) ∣ 0≤θ≤π, r≤2asin⁡θ, 0≤z≤2a−r }\begin{equation*} V=\Set{(r\cos\theta\,,\,r\sin\theta\,,\,z)} {0\le\theta\le\pi,\ r\le 2a\sin\theta,\ 0\le z\le 2a-r} \end{equation*}

By symmetry under x→−xx\rightarrow -x, the full volume is twice the volume in the first octant.

So the

Volume=2∫0π2dθ∫02asin⁡θdr r∫02a−rdz[0]=2∫0π2dθ∫02asin⁡θdr r(2a−r)[0]=2∫0π2dθ [4a3sin⁡2θ−8a33sin⁡3θ][0]=8a3[∫0π2dθ 1−cos⁡(2θ)2+23∫10dt (1−t2)]where t=cos⁡θ=8a3[π4−23(1−13)]=a3(2π−329)\begin{align*} \text{Volume} &= 2\int_0^{{\pi\over 2}} \dee{\theta}\int_0^{2a\sin\theta}\dee{r}\ r \int_0^{2a -r}\dee{z} [0]\\ &= 2\int_0^{{\pi\over 2}} \dee{\theta}\int_0^{2a\sin\theta}\dee{r}\ r(2a-r) [0]\\ &= 2\int_0^{{\pi\over 2}} \dee{\theta}\ \left[4a^3\sin^2\theta-\frac{8a^3}{3}\sin^3\theta\right] [0]\\ &= 8a^3\left[\int_0^{{\pi\over 2}} \dee{\theta}\ \frac{1-\cos(2\theta)}{2} +\frac{2}{3}\int^0_1 \dee{t}\ \big(1-t^2\big)\right] \qquad\text{where }t=\cos\theta \\ &= 8a^3\left[\frac{\pi}{4}-\frac{2}{3}\left(1-\frac{1}{3}\right)\right] =a^3\big(2\pi-\frac{32}{9}\big) \end{align*}

For an efficient, sneaky, way to evaluate ∫0π2dθ sin⁡2θ\int_0^{{\pi\over 2}} \dee{\theta}\ \sin^2\theta, see Remark 3.3.5 in the CLP-3 text.

(b) The domain of integration is

V={ (x,y,z) ∣ −x≤y≤3x, 0≤z≤1−x2−y2 }\begin{equation*} V = \Set{(x,y,z)}{-x\le y\le \sqrt{3}x,\ 0\le z \le 1-x^2-y^2} \end{equation*}

Recall that in polar coordinates yx=tan⁡θ\frac{y}{x}=\tan\theta. So the boundaries of the wedge −x≤y≤3x-x\le y\le \sqrt{3}x, or equivalently −1≤yx≤3-1\le\frac{y}{x}\le\sqrt{3}, correspond, in polar coordinates, to θ=tan⁡−1(−1)=−π4\theta=\tan^{-1}(-1)=-\frac{\pi}{4} and θ=tan⁡−13=π3\theta=\tan^{-1}\sqrt{3}=\frac{\pi}{3}. In cylindrical coordinates, the paraboloid z=1−x2−y2z=1-x^2-y^2 becomes z=1−r2z=1-r^2. There are zz's that obey 0≤z≤1−r20\le z\le 1-r^2 if and only if r≤1r\le 1. So, in cylindrical coordinates,

V={ (rcos⁡θ,rsin⁡θ,z) ∣ −π4≤θ≤π3, 0≤r≤1, 0≤z≤1−r2 }\begin{equation*} V = \Set{(r\cos\theta,r\sin\theta,z)} {-\tfrac{\pi}{4}\le \theta\le \tfrac{\pi}{3},\ 0\le r\le 1,\ 0\le z \le 1-r^2} \end{equation*}

and

Volume=∫−π4π3dθ∫01dr r∫01−r2dz=(π3+π4)∫01dr r(1−r2)=712π(12−14)=748π\begin{align*} \text{Volume} &= \int_{-{\pi\over 4}}^{{\pi\over 3}} \dee{\theta}\int_0^1\dee{r}\ r \int_0^{1 -r^2}\dee{z} = \left(\frac{\pi}{3}+\frac{\pi}{4}\right)\int_0^1\dee{r}\ r(1-r^2) = \frac{7}{12}\pi\left(\frac{1}{2}-\frac{1}{4}\right) = \frac{7}{48}\pi \end{align*}

(c) The region is

V={ (x,y,z) ∣ x2+y2≤z≤2y }\begin{equation*} V = \Set{(x,y,z)}{x^2+y^2\le z\le 2y} \end{equation*}

There are zz's that obey x2+y2≤z≤2yx^2+y^2\le z\le 2y if and only if

x2+y2≤2y  ⟺  x2+y2−2y≤0  ⟺  x2+(y−1)2≤1\begin{align*} x^2+y^2\le 2y &\iff x^2+y^2-2y \le 0 \iff x^2 + (y-1)^2 \le 1 \end{align*}

This disk is sketched in the figure

Figure from prob_s3.6, line 365

Figure from prob_s3.6, line 365

In cylindrical coordinates,

  • the bottom, z=x2+y2z=x^2+y^2, is z=r2z=r^2,

  • the top, z=2yz=2y, is z=2rsin⁡θz=2r\sin\theta, and

  • the disk x2+y2≤2yx^2+y^2\le 2y is r2≤2rsin⁡θr^2\le 2r\sin\theta, or equivalently r≤2sin⁡θr\le2\sin\theta,

so that, looking at the figure above,

V={ (rcos⁡θ,rsin⁡θ,z) ∣ 0≤θ≤π, 0≤r≤2sin⁡θ, r2≤z≤2rsin⁡θ }\begin{equation*} V = \Set{(r\cos\theta,r\sin\theta,z)}{0\le\theta\le\pi, \ 0\le r\le 2\sin\theta,\ r^2\le z\le 2r\sin\theta} \end{equation*}

By symmetry under x→−xx\rightarrow -x, the full volume is twice the volume in the first octant so that

Volume=2∫0π2dθ∫02sin⁡θdr r∫r22rsin⁡θdz=2∫0π2dθ∫02sin⁡θdr r(2rsin⁡θ−r2)=2∫0π2dθ (243−244)sin⁡4θ\begin{align*} \text{Volume} &= 2\int_{0}^{{\pi\over 2}} \dee{\theta}\int_0^{2\sin\theta}\dee{r}\ r\int_{r^2}^{2r\sin\theta}\dee{z} = 2\int_{0}^{{\pi\over 2}} \dee{\theta}\int_0^{2\sin\theta}\dee{r}\ r(2r\sin\theta-r^2)\cr &= 2\int_{0}^{{\pi\over 2}} \dee{\theta}\ \left(\frac{2^4}{3}-\frac{2^4}{4}\right)\sin^4\theta \end{align*}

To integrate (For a general discussion of trigonometric integrals see §1.8 in the CLP-2 text. In particular the integral
∫cos⁡4x dx\int \cos^4 x\ \dee{x} is evaluated in Example 1.8.8 in the CLP-2 text.) sin⁡4θ\sin^4\theta, we use the double angle formulae sin⁡2x=1−cos⁡(2x)2\sin^2 x= \frac{1-\cos(2x)}{2} and cos⁡2x=1+cos⁡(2x)2\cos^2 x= \frac{1+\cos(2x)}{2} to write

sin⁡4θ=[1−cos⁡(2θ)2]2=14−12cos⁡(2θ)+14cos⁡2(2θ)=14−12cos⁡(2θ)+18(1+cos⁡(4θ))=38−12cos⁡(2θ)+18cos⁡(4θ)\begin{align*} \sin^4\theta &= \left[ \frac{1-\cos(2\theta)}{2} \right]^2 \\ &= \frac{1}{4} - \frac{1}{2} \cos(2\theta) + \frac{1}{4}\cos^2(2\theta)\\ &= \frac{1}{4} - \frac{1}{2} \cos(2\theta) + \frac{1}{8}\left(1 + \cos(4\theta) \right)\\ &= \frac{3}{8} - \frac{1}{2} \cos(2\theta) + \frac{1}{8}\cos(4\theta) \end{align*}

So

Volume=2 2412[38θ−14sin⁡(2θ)+132sin⁡(4θ)]0π2=2 2412 316π=π2\begin{align*} \text{Volume} &= 2\ \frac{2^4}{12} \left[ \frac{3}{8}\theta - \frac{1}{4} \sin(2\theta) + \frac{1}{32}\sin(4\theta) \right]_0^{\pi\over 2} = 2\ \frac{2^4}{12}\ \frac{3}{16}\pi =\frac{\pi}{2} \end{align*}
Q7Stage 2Past exam · M200 2008A

Let E be the region bounded between the parabolic surfaces z=x2+y2z = x^2 + y^2 and z=2−x2−y2z = 2 - x^2 - y^2 and within the cylinder x2+y2≤1x^2 + y^2 \le 1. Calculate the integral of f(x,y,z)=(x2+y2)3/2f(x,y,z) = {(x^2 + y^2)}^{3/2} over the region EE.

Answer

8π35\frac{8\pi}{35}

Full solution

Note that the paraboloids z=x2+y2z=x^2+y^2 and z=2−x2−y2z=2-x^2-y^2 intersect when z=x2+y2=1z=x^2+y^2=1. We'll use cylindrical coordinates. Then x2+y2=r2x^2+y^2=r^2, dV=r dr dθ dz\dee{V}=r \ \dee{r}\,\dee{\theta}\,\dee{z}, and

E={ (rcos⁡θ , rsin⁡θ , z)) ∣ 0≤r≤1, r2≤z≤2−r2, 0≤θ≤2π }\begin{equation*} E=\Set{(r\cos\theta\,,\,r\sin\theta\,,\,z))}{0\le r\le 1,\ r^2\le z\le 2-r^2,\ 0\le\theta\le 2\pi} \end{equation*}

so that

∭Ef(x,y,z) dV=∫01dr∫r22−r2dz∫02πdθ r r3⏞f=2π∫01dr r4(2−r2−r2)=2π[2155−2177]=8π35\begin{align*} \tripInt_E f(x,y,z)\ \dee{V} &=\int_0^1\dee{r}\int_{r^2}^{2-r^2}\dee{z}\int_0^{2\pi}\dee{\theta}\ r\ \overbrace{r^3}^{f} \\ &=2\pi \int_0^1\dee{r}\ r^4\big(2-r^2-r^2\big) \\ &=2\pi \left[2\frac{1^5}{5}-2\frac{1^7}{7}\right] \\ &=\frac{8\pi}{35} \end{align*}
Q8Stage 2Past exam · M200 2010D

Let EE be the region bounded above by the sphere x2+y2+z2=2x^2 + y^2 + z^2 = 2 and below by the paraboloid z=x2+y2z = x^2 + y^2. Find the centroid of EE.

Answer

(0,0,7162−14≈0.811)\big(0,0,\frac{7}{16\sqrt{2}-14} \approx 0.811\big)

Full solution

Observe that both the sphere x2+y2+z2=2x^2+y^2+z^2=2 and the paraboloid z=x2+y2z=x^2+y^2 are invariant under rotations around the zz–axis. So EE is invariant under rotations around the zz–axis and the centroid (centre of mass) of EE will lie on the zz–axis. Thus xˉ=yˉ=0\bar x=\bar y=0 and we just have to find

zˉ=∭Ez dV∭EdV\begin{align*} \bar z = \frac{\tripInt_E z\ \dee{V}}{\tripInt_E\dee{V}} \end{align*}

The surfaces z=x2+y2z = x^2 + y^2 and x2+y2+z2=2x^2 + y^2 + z^2 = 2 intersect when z=x2+y2z=x^2+y^2 and

z+z2=2  ⟺  z2+z−2=0  ⟺  (z+2)(z−1)=0\begin{equation*} z+z^2=2 \iff z^2+z-2=0 \iff(z+2)(z-1)=0 \end{equation*}

Since z=x2+y2≥0z=x^2+y^2\ge 0, the surfaces intersect on the circle z=1z=1, x2+y2=2x^2+y^2=2. So

E={ (x,y,z) ∣ x2+y2≤1, x2+y2≤z≤2−x2−y2 }\begin{equation*} E = \Set{(x,y,z)}{ x^2+y^2\le 1,\ x^2+y^2\le z\le \sqrt{2-x^2-y^2}} \end{equation*}

Here is a sketch of the y=0y=0 cross section of E.

Figure from prob_s3.6, line 540

Figure from prob_s3.6, line 540

Let's use cylindrical coordinates to do the two integrals. In cylindrical coordinates

  • E={ (rcos⁡θ,rsin⁡θ,z) ∣ 0≤r≤1, 0≤θ≤2π, r2≤z≤2−r2 }E = \Set{(r\cos\theta,r\sin\theta,z)}{ 0\le r\le 1,\ 0\le\theta\le 2\pi,\ r^2\le z\le \sqrt{2-r^2}}, and

  • dV\dee{V} is r dr dθ dzr\,\dee{r}\,\dee{\theta}\,\dee{z}

so, for n=0,1n=0,1 (we'll try to do both integrals at the same time)

∭Ezn dV=∫01dr∫02πdθ∫r22−r2dz r zn=2π∫01dr r{2−r2−r2if n=012(2−r2−r4)if n=1\begin{align*} \tripInt_E z^n\ \dee{V} &=\int_0^1\dee{r} \int_0^{2\pi}\dee{\theta} \int_{r^2}^{\sqrt{2-r^2}} \dee{z}\ r\ z^n \\ &=2\pi\int_0^1\dee{r} \ r\begin{cases} \sqrt{2-r^2}-r^2 & \text{if }n=0 \\ \frac{1}{2}\big(2-r^2-r^4\big) & \text{if }n=1 \end{cases} \end{align*}

Since

∫01dr r2−r2=[−13(2−r2)3/2]01=13(22−1)\begin{align*} \int_0^1 \dee{r}\ r\sqrt{2-r^2} &=\left[-\frac{1}{3}(2-r^2)^{3/2}\right]_0^1 =\frac{1}{3}\big(2\sqrt{2}-1\big) \end{align*}

we have

∭Ezn dV=2π{13(22−1)−14if n=012−18−112if n=1}=2π{232−712if n=0724if n=1\begin{align*} \tripInt_E z^n\ \dee{V} =2\pi\left.\begin{cases} \frac{1}{3}(2\sqrt{2}-1)-\frac{1}{4} &\text{if }n=0 \\ \frac{1}{2}-\frac{1}{8}-\frac{1}{12} &\text{if }n=1 \end{cases}\right\} =2\pi\begin{cases} \frac{2}{3}\sqrt{2}-\frac{7}{12} &\text{if }n=0 \\ \frac{7}{24} &\text{if }n=1 \end{cases} \end{align*}

and xˉ=yˉ=0\bar x=\bar y=0 and

zˉ=∭Ez dV∭EdV=724232−712=7162−14≈0.811\begin{align*} \bar z =\frac{\tripInt_E z\ \dee{V}}{\tripInt_E\dee{V}} =\frac{\frac{7}{24}}{\frac{2}{3}\sqrt{2}-\frac{7}{12}} =\frac{7}{16\sqrt{2}-14} \approx 0.811 \end{align*}
Q9Stage 2Past exam · M200 2012A

Let EE be the smaller of the two solid regions bounded by the surfaces z=x2+y2z = x^2 + y^2 and x2+y2+z2=6x^2 + y^2 + z^2 = 6. Evaluate ∭E(x2+y2) dV\tripInt_E (x^2+y^2)\ \dee{V} .

Answer

π[4856−32815]≈1.65π\pi\left[\frac{48}{5}\sqrt{6}-\frac{328}{15}\right] \approx 1.65\pi

Full solution

Note that both surfaces are invariant under rotations about the zz–axis. Here is a sketch of the y=0y=0 cross section of E.

Figure from prob_s3.6, line 628

Figure from prob_s3.6, line 628

The surfaces z=x2+y2z = x^2 + y^2 and x2+y2+z2=6x^2 + y^2 + z^2 = 6 intersect when z=x2+y2z=x^2+y^2 and

z+z2=6  ⟺  z2+z−6=0  ⟺  (z+3)(z−2)=0\begin{equation*} z+z^2=6 \iff z^2+z-6=0 \iff(z+3)(z-2)=0 \end{equation*}

Since z=x2+y2≥0z=x^2+y^2\ge 0, the surfaces intersect on the circle z=2z=2, x2+y2=2x^2+y^2=2. So

E={ (x,y,z) ∣ x2+y2≤2, x2+y2≤z≤6−x2−y2 }\begin{equation*} E = \Set{(x,y,z)}{ x^2+y^2\le 2,\ x^2+y^2\le z\le \sqrt{6-x^2-y^2}} \end{equation*}

Let's use cylindrical coordinates to do the integral. In cylindrical coordinates

  • E={ (rcos⁡θ,rsin⁡θ,z) ∣ r≤2, 0≤θ≤2π, r2≤z≤6−r2 }E = \Set{(r\cos\theta,r\sin\theta,z)}{ r\le \sqrt{2},\ 0\le\theta\le 2\pi,\ r^2\le z\le \sqrt{6-r^2}}, and

  • dV\dee{V} is r dr dθ dzr\,\dee{r}\,\dee{\theta}\,\dee{z}

so

∭E(x2+y2) dV=∫02dr∫02πdθ∫r26−r2dz r r2=2π∫02dr r3(6−r2−r2)=2π∫02dr r r26−r2−2π∫02dr r5=2π∫64du−2 (6−u)u−2π236with u=6−r2, du=−2r dr=−π[6u3/23/2−u5/25/2]64−8π3=−π[4(8−66)−25(32−366)]−8π3=π[645−32−83+(24−725)6]=π[4856−32815]≈1.65π\begin{align*} \tripInt_E (x^2+y^2)\ \dee{V} &=\int_0^{\sqrt{2}}\dee{r} \int_0^{2\pi}\dee{\theta} \int_{r^2}^{\sqrt{6-r^2}} \dee{z}\ r\ r^2 \\ &=2\pi\int_0^{\sqrt{2}}\dee{r} \ r^3\big(\sqrt{6-r^2}-r^2\big) =2\pi\int_0^{\sqrt{2}}\dee{r}\,r\ r^2\sqrt{6-r^2} - 2\pi\int_0^{\sqrt{2}}\dee{r} \ r^5\\ &=2\pi\int_6^4\frac{\dee{u}}{-2} \ (6-u)\sqrt{u} -2\pi\frac{2^3}{6} \qquad\text{with } u = 6-r^2,\ \dee{u} = -2r\,\dee{r} \\ &=-\pi\left[6\frac{u^{3/2}}{3/2}-\frac{u^{5/2}}{5/2}\right]_6^4-\frac{8\pi}{3}\\ &=-\pi\left[4\big(8-6\sqrt{6}\big)-\frac{2}{5}\big(32-36\sqrt{6}\big)\right] -\frac{8\pi}{3}\\ &=\pi\left[\frac{64}{5}-32-\frac{8}{3} +\left(24-\frac{72}{5}\right)\sqrt{6}\right] \\ &=\pi\left[\frac{48}{5}\sqrt{6}-\frac{328}{15}\right] \approx 1.65\pi \end{align*}
Q10Stage 2Past exam · M200 2013D

Let a>0a > 0 be a fixed positive real number. Consider the solid inside both the cylinder x2+y2=axx^2 + y^2 = ax and the sphere x2+y2+z2=a2x^2 + y^2 + z^2 = a^2. Compute its volume.

You may use that ∫sin⁡3(θ)=112cos⁡(3θ)−34cos⁡(θ)+C\int \sin^3(\theta) =\frac{1}{12}\cos(3\theta) -\frac{3}{4}\cos(\theta) +C

Answer

4a33[π2−23]\frac{4a^3}{3}\left[\frac{\pi}{2} - \frac{2}{3} \right]

Full solution

We'll use cylindrical coordinates. In cylindrical coordinates

  • the sphere x2+y2+z2=a2x^2+y^2+z^2=a^2 becomes r2+z2=a2r^2+z^2=a^2 and

  • the circular cylinder x2+y2=axx^2+y^2=ax (or equivalently (x−a/2)2+y2=a2/4(x-a/2)^2+ y^2=a^2/4) becomes r2=arcos⁡θr^2=ar\cos\theta or r=acos⁡θr=a\cos\theta.

Here is a sketch of the top view of the solid.

Figure from prob_s3.6, line 693

Figure from prob_s3.6, line 693

The solid is

{ (rcos⁡θ , rsin⁡θ , z) ∣ −π ⁣/2≤θ≤π ⁣/2 , 0≤r≤acos⁡θ , −a2−r2≤z≤a2−r2 }\begin{equation*} \Set{(r\cos\theta\,,\,r\sin\theta\,,\,z)}{ -\nicefrac{\pi}{2}\le\theta\le\nicefrac{\pi}{2}\,,\, 0\le r\le a\cos\theta\,,\, -\sqrt{a^2-r^2}\le z\le \sqrt{a^2-r^2}} \end{equation*}

By symmetry, the volume of the specified solid is four times the volume of the solid

{ (rcos⁡θ , rsin⁡θ , z) ∣ 0≤θ≤π ⁣/2 , 0≤r≤acos⁡θ , 0≤z≤a2−r2 }\begin{equation*} \Set{(r\cos\theta\,,\,r\sin\theta\,,\,z)}{ 0\le\theta\le\nicefrac{\pi}{2}\,,\, 0\le r\le a\cos\theta\,,\, 0\le z\le \sqrt{a^2-r^2}} \end{equation*}

Since dV=r dr dθ dz\dee{V} = r\,\dee{r}\,\dee{\theta}\,\dee{z}, the volume of the solid is

4∫0π/2dθ∫0acos⁡θdr∫0a2−r2dz r=4∫0π/2dθ∫0acos⁡θdr ra2−r2=−43∫0π/2dθ (a2−r2)3/2∣0acos⁡θ=43∫0π/2dθ [a3−(a2−a2cos⁡2θ)3/2]=4a33∫0π/2dθ [1−sin⁡3θ]=4a33[θ−112cos⁡(3θ)+34cos⁡θ]0π/2=4a33[π2+112−34]=4a33[π2−23]\begin{align*} 4\int_0^{\pi/2}\dee{\theta} \int_0^{a\cos\theta} \dee{r} \int_0^{\sqrt{a^2-r^2}}\dee{z}\ r &=4\int_0^{\pi/2}\dee{\theta} \int_0^{a\cos\theta} \dee{r}\ r\sqrt{a^2-r^2} \\ &=-\frac{4}{3}\int_0^{\pi/2}\dee{\theta}\ \big(a^2-r^2\big)^{3/2}\Big|_0^{a\cos\theta} \\ &=\frac{4}{3}\int_0^{\pi/2}\dee{\theta}\ \Big[a^3-\big(a^2-a^2\cos^2\theta\big)^{3/2}\Big] \\ &=\frac{4a^3}{3}\int_0^{\pi/2}\dee{\theta}\ \big[1-\sin^3\theta\big] \\ &=\frac{4a^3}{3}\left[\theta - \frac{1}{12}\cos(3\theta) + \frac{3}{4}\cos\theta \right]_0^{\pi/2} \\ &=\frac{4a^3}{3}\left[\frac{\pi}{2} + \frac{1}{12} - \frac{3}{4} \right] \\ &=\frac{4a^3}{3}\left[\frac{\pi}{2} - \frac{2}{3} \right] \end{align*}
Q11Stage 2Past exam · M200 2014A

Let EE be the solid lying above the surface z=y2z = y^2 and below the surface z=4−x2z = 4 - x^2. Evaluate

∭Ey2 dV\begin{equation*} \tripInt_E y^2\ \dee{V} \end{equation*}

You may use the half angle formulas:

sin⁡2θ=1−cos⁡(2θ)2,cos⁡2θ=1+cos⁡(2θ)2\begin{equation*} \sin^2\theta = \frac{1-\cos(2\theta)}{2},\qquad \cos^2\theta = \frac{1+\cos(2\theta)}{2} \end{equation*}
Answer

16π3\frac{16\pi}{3}

Full solution

Note that the surfaces meet when z=y2=4−x2z=y^2=4-x^2 and then (x,y)(x,y) runs over the circle x2+y2=4x^2+y^2=4. So the domain of integration is

E={ (x,y,z) ∣ x2+y2≤4, y2≤z≤4−x2 }\begin{equation*} E = \Set{(x,y,z)}{x^2+y^2\le 4,\ y^2\le z\le 4-x^2} \end{equation*}

Let's switch to cylindrical coordinates. Then

E={ (rcos⁡θ,rsin⁡θ,z) ∣ 0≤r≤2, 0≤θ≤2π, r2sin⁡2θ≤z≤4−r2cos⁡2θ }\begin{equation*} E = \Set{(r\cos\theta,r\sin\theta,z)}{0\le r\le 2,\ 0\le\theta\le 2\pi, \ r^2\sin^2\theta\le z\le 4-r^2\cos^2\theta} \end{equation*}

and, since dV=r dr dθ dz\dee{V} = r\,\dee{r}\,\dee{\theta}\,\dee{z},

∭Ey2 dV=∫02dr∫02πdθ∫r2sin⁡2θ4−r2cos⁡2θdz r r2sin⁡2θ⏞y2=∫02dr∫02πdθ r3sin⁡2θ[4−r2cos⁡2θ−r2sin⁡2θ][0]=∫02dr [4r3−r5]∫02πdθ 1−cos⁡(2θ)2[0]=12∫02dr [4r3−r5] [θ−sin⁡(2θ)2]02π[0]=π[r4−r66]02=16π3\begin{align*} \tripInt_E y^2\ \dee{V} &=\int_0^2\dee{r} \int_0^{2\pi} \dee{\theta} \int_{r^2\sin^2\theta}^{4-r^2\cos^2\theta}\dee{z}\ r\ \overbrace{r^2\sin^2\theta}^{y^2} \\ &=\int_0^2\dee{r} \int_0^{2\pi} \dee{\theta}\ r^3\sin^2\theta\big[4-r^2\cos^2\theta - r^2\sin^2\theta\big] [0]\\ &=\int_0^2\dee{r}\ \big[4r^3-r^5\big] \int_0^{2\pi} \dee{\theta}\ \frac{1-\cos(2\theta)}{2} [0]\\ &=\frac{1}{2}\int_0^2\dee{r}\ \big[4r^3-r^5\big]\ \left[\theta-\frac{\sin(2\theta)}{2}\right]_0^{2\pi} [0]\\ &=\pi\left[r^4-\frac{r^6}{6}\right]_0^2 \\ &=\frac{16\pi}{3} \end{align*}

For an efficient, sneaky, way to evaluate ∫02πsin⁡2θ dθ\int_0^{2\pi}\sin^2\theta\ \dee{\theta}, see Remark 3.3.5 in the CLP-3 text.

Q12Stage 2

The centre of mass (xˉ,yˉ,zˉ)(\bar x,\bar y, \bar z) of a body BB having density ρ(x,y,z)\rho(x,y,z) (units of mass per unit volume) at (x,y,z)(x,y,z) is defined to be

xˉ=1M∭Bxρ(x,y,z) dVyˉ=1M∭Byρ(x,y,z) dVzˉ=1M∭Bzρ(x,y,z) dV\begin{equation*} \bar x=\frac{1}{M}\tripInt_B x\rho(x,y,z)\ \dee{V}\quad \bar y=\frac{1}{M}\tripInt_B y\rho(x,y,z)\ \dee{V}\quad \bar z=\frac{1}{M}\tripInt_B z\rho(x,y,z)\ \dee{V} \end{equation*}

where

M=∭Bρ(x,y,z) dV\begin{equation*} M=\tripInt_B \rho(x,y,z)\ \dee{V} \end{equation*}

is the mass of the body. So, for example, xˉ\bar x is the weighted average of xx over the body. Find the centre of mass of the part of the solid ball x2+y2+z2≤a2x^2+y^2+z^2\le a^2 with x≥0x\ge 0, y≥0y\ge 0 and z≥0z\ge 0, assuming that the density ρ\rho is constant.

Hint

Use cylindrical coordinates.

Answer

xˉ=yˉ=zˉ=38a\bar x=\bar y=\bar z=\frac{3}{8}a

Full solution

By symmetry, xˉ=yˉ=zˉ\bar x=\bar y=\bar z, so it suffices to compute, for example, zˉ\bar z. The mass of the body is the density, ρ\rho, times its volume, which is one eighth of the volume of a sphere. So

M=ρ8 43πa3\begin{equation*} M=\frac{\rho}{8}\ \frac{4}{3}\pi a^3 \end{equation*}

In cylindrical coordinates, the equation of the spherical surface of the body is r2+z2=a2r^2+z^2=a^2. The part of the body at height zz above the xyxy–plane is one quarter of a disk of radius a2−z2\sqrt{a^2-z^2}. The numerator of zˉ\bar z is

∭Bzρ dV=ρ∫0adz∫0π/2dθ∫0a2−z2dr r z=ρ∫0adz∫0π/2dθ z r22∣0a2−z2=ρ2∫0adz∫0π/2dθ z(a2−z2)=π4ρ∫0adz z(a2−z2)=π4ρ[a2z22−z44]0a=π16ρa4\begin{alignat*}{3} \tripInt_B z\rho\ \dee{V} &=\rho\int_0^a \dee{z}\int_0^{\pi/2}d\theta\int_0^{\sqrt{a^2-z^2}}dr\ r\ z &&=\rho\int_0^a \dee{z}\int_0^{\pi/2}d\theta\ z\ \frac{r^2}{2} \bigg|_0^{\sqrt{a^2-z^2}} \\ &=\frac{\rho}{2}\int_0^a \dee{z}\int_0^{\pi/2}d\theta\ z(a^2-z^2) &&=\frac{\pi}{4}\rho\int_0^a \dee{z}\ z(a^2-z^2) \\ &=\frac{\pi}{4}\rho\left[a^2\frac{z^2}{2}-\frac{z^4}{4}\right]_0^a =\frac{\pi}{16}\rho a^4 \end{alignat*}

Dividing by M=π6ρa3M=\frac{\pi}{6}\rho a^3 gives xˉ=yˉ=zˉ=38a\bar x=\bar y=\bar z=\frac{3}{8}a.

Q13Stage 2Past exam · M200 2003D

A sphere of radius 2m2{\rm m} centred on the origin has variable density 53(z2+1)\frac{5}{\sqrt{3}}(z^2+1)kg/m3{\rm m}^3. A hole of diameter 1m is drilled through the sphere along the zz–axis.

  1. Set up a triple integral in cylindrical coordinates giving the mass of the sphere after the hole has been drilled.

  2. Evaluate this integral.

Answer

(a) mass=∫1/22dr∫−4−r24−r2dz∫02πdθ 53(z2+1)r\dst\text{mass} = \int_{1/2}^2 \dee{r} \int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} \dee{z}\int_0^{2\pi}\dee{\theta}\ \frac{5}{\sqrt{3}}(z^2+1)r

(b) 525245π≈153.7kg\frac{525}{24}\sqrt{5}\pi\approx 153.7\text{kg}

Full solution

(a) In cylindrical coordinates the equation of a sphere of radius 2 centred on the origin is r2+z2=22r^2+z^2=2^2. Since dV=r dr dθ dz\dee{V}=r\, \dee{r}\, \dee{\theta}\, \dee{z} and dm=53(z2+1)r dr dθ dz\dee{m} =\frac{5}{\sqrt{3}}(z^2+1)r\, \dee{r}\, \dee{\theta}\, \dee{z} and the hole has radius 1/21/2, the integral is

mass=∫1/22dr∫−4−r24−r2dz∫02πdθ 53(z2+1)r\begin{align*} \text{mass} = \int_{1/2}^2 \dee{r} \int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} \dee{z}\int_0^{2\pi}\dee{\theta}\ \frac{5}{\sqrt{3}}(z^2+1)r \end{align*}

(b) By part (a)

mass=∫1/22dr∫−4−r24−r2dz∫02πdθ 53(z2+1)r=4π53∫1/22dr r∫04−r2dz (z2+1)=4π53∫1/22dr r[z33+z]04−r2=4π53∫1/22dr r[13(4−r2)3/2+(4−r2)1/2]\begin{align*} \text{mass} = \int_{1/2}^2 \dee{r} \int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} \dee{z}\int_0^{2\pi}\dee{\theta}\ \frac{5}{\sqrt{3}}(z^2+1)r &=4\pi\frac{5}{\sqrt{3}}\int_{1/2}^2\dee{r}\ r\int_0^{\sqrt{4-r^2}}\dee{z}\ (z^2+1) \\ &=4\pi\frac{5}{\sqrt{3}}\int_{1/2}^2\dee{r}\ r\left[\frac{z^3}{3}+z\right]_0^{\sqrt{4-r^2}} \\ &=4\pi\frac{5}{\sqrt{3}}\int_{1/2}^2\dee{r}\ r\left[\frac{1}{3}{(4-r^2)}^{3/2}+{(4-r^2)}^{1/2}\right] \end{align*}

Make the change of variables s=4−r2s=4-r^2, ds=−2r dr\dee{s}=-2r\,\dee{r}. This gives

mass=4π53∫15/40ds−2 [13s3/2+s1/2]=−2π53[215s5/2+23s3/2]15/40=2π53[215155/232+23153/28]=2π53[116+112]153/2=525245π≈153.7kg\begin{align*} \text{mass} &= 4\pi\frac{5}{\sqrt{3}}\int_{15/4}^0\frac{\dee{s}}{-2}\ \left[\frac{1}{3}s^{3/2}+s^{1/2}\right] = -2\pi\frac{5}{\sqrt{3}} \left[\frac{2}{15}s^{5/2}+\frac{2}{3}s^{3/2} \right]_{15/4}^0 \\ &= 2\pi\frac{5}{\sqrt{3}} \left[\frac{2}{15}\frac{15^{5/2}}{32} +\frac{2}{3}\frac{15^{3/2}}{8}\right] \\ &=2\pi\frac{5}{\sqrt{3}} \left[\frac{1}{16}+\frac{1}{12}\right]15^{3/2} =\frac{525}{24}\sqrt{5}\pi\approx 153.7\text{kg} \end{align*}
Q14Stage 2Past exam · M200 2001A

Consider the finite solid bounded by the three surfaces: z=e−x2−y2z=e^{-x^2-y^2}, z=0z=0 and x2+y2=4x^2+y^2=4.

  1. Set up (but do not evaluate) a triple integral in rectangular coordinates that describes the volume of the solid.

  2. Calculate the volume of the solid using any method.

Answer

(a) ∫−22dx∫−4−x24−x2dy∫0e−x2−y2dz\dst\int_{-2}^2 \dee{x} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\dee{y}\int_0^{e^{-x^2-y^2}}\dee{z} (b) π[1−e−4]≈3.084\pi\big[1-e^{-4}\big]\approx 3.084

Full solution

(a) The solid consists of all (x,y,z)(x,y,z) with

  • (x,y)(x,y) running over the disk x2+y2≤4x^2+y^2\le 4 and

  • for each fixed (x,y)(x,y) obeying x2+y2≤4x^2+y^2\le 4, zz running from 00 to e−x2−y2e^{-x^2-y^2}

On the disk x2+y2≤4x^2+y^2\le 4,

  • xx runs from −2-2 to 22 and

  • for each fixed xx obeying −2≤x≤2-2\le x\le 2, yy runs from −4−x2-\sqrt{4-x^2} to 4−x2\sqrt{4-x^2}

So

Volume=∫−22dx∫−4−x24−x2dy∫0e−x2−y2dz\begin{equation*} \text{Volume}=\int_{-2}^2 \dee{x} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\dee{y}\int_0^{e^{-x^2-y^2}}\dee{z} \end{equation*}

(b) Switching to cylindrical coordinates

Volume=∫02dr∫02πdθ∫0e−r2dz r=∫02dr∫02πdθ re−r2=∫02dr 2π re−r2=−πe−r2∣02=π[1−e−4]≈3.084\begin{align*} \text{Volume}&=\int_0^2 \dee{r}\int_0^{2\pi}\dee{\theta} \int_0^{e^{-r^2}}\dee{z}\ r =\int_0^2 \dee{r}\int_0^{2\pi}\dee{\theta} \ r e^{-r^2} =\int_0^2 \dee{r}\ 2\pi\ r e^{-r^2} \\ &=-\pi e^{-r^2}\Big|_0^2 =\pi\big[1-e^{-4}\big]\approx 3.084 \end{align*}
Q15Stage 2Past exam · M200 2000A

Find the volume of the solid which is inside x2+y2=4x^2 + y^2 = 4, above z=0z = 0 and below 2z=y2z = y.

Answer

83\frac{8}{3}

Full solution

The solid consists of the set of all points (x,y,z)(x,y,z) such that x2+y2≤4x^2+y^2\le 4 and 0≤z≤y20\le z\le\frac{y}{2}. In particular y≥0y\ge 0. When we look at the solid from above, we see all (x,y)(x,y) with x2+y2≤4x^2+y^2\le 4 and y≥0y\ge 0. This is sketched in the figure on the left below.

Figure from prob_s3.6, line 984

Figure from prob_s3.6, line 984

Figure from prob_s3.6, line 984

Figure from prob_s3.6, line 984

We'll use cylindrical coordinates. In the base region (the shaded region in the figure on the left above)

  • θ\theta runs from 00 to π\pi and

  • for each fixed θ\theta between 00 and π\pi, rr runs from 00 to 22.

  • For each fixed point (x,y)=(rcos⁡θ,rsin⁡θ)(x,y)=(r\cos\theta,r\sin\theta) in the base region, zz runs from 00) to y2=rsin⁡θ2\frac{y}{2}=\frac{r\sin\theta}{2}.

So the volume is

∫0πdθ∫02dr∫0rsin⁡θ/2dz r=∫0πdθ∫02dr 12r2sin⁡θ=∫0πdθ r36sin⁡θ ∣02=43∫0πdθ sin⁡θ=−43cos⁡θ ∣0π=83\begin{align*} \int_0^\pi \dee{\theta}\int_0^2 \dee{r}\int_0^{r\sin\theta/2} dz\ r &=\int_0^\pi \dee{\theta}\int_0^2 \dee{r}\ \frac{1}{2} r^2\sin\theta =\int_0^\pi \dee{\theta}\ \frac{r^3}{6}\sin\theta\,\bigg|_0^2 =\frac{4}{3}\int_0^\pi \dee{\theta}\ \sin\theta \\ &=-\frac{4}{3}\cos\theta\,\bigg|_0^\pi =\frac{8}{3} \end{align*}

3Stage 3Application

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

Q16Stage 3Past exam · M200 2014D

The density of hydrogen gas in a region of space is given by the formula

ρ(x,y,z)=z+2x21+x2+y2\begin{equation*} \rho(x,y,z) =\frac{z+2x^2}{1+x^2+y^2} \end{equation*}

  1. At (1,0,−1)(1,0,-1), in which direction is the density of hydrogen increasing most rapidly?

  2. You are in a spacecraft at the origin. Suppose the spacecraft flies in the direction of <0,0,1>\llt 0,0,1\rgt. It has a disc of radius 11, centred on the spacecraft and deployed perpendicular to the direction of travel, to catch hydrogen. How much hydrogen has been collected by the time that the spacecraft has traveled a distance 22?

    You may use the fact that ∫02πcos⁡2θ dθ=π\int_0^{2\pi}\cos^2\theta\ \dee{\theta} =\pi.

Answer

(a) The unit vector in the direction of maximum rate of increse is 110(3,0,1)\frac{1}{\sqrt{10}}(3,0,1).

(b) 2π2\pi

Full solution

(a) The direction of maximum rate of increase is ∇ρ(1,0,−1)\vnabla\rho(1,0,-1). As

∂ρ∂x(x,y,z)=4x1+x2+y2−2x(z+2x2)(1+x2+y2)2∂ρ∂x(1,0,−1)=42−2(−1+2)(2)2=32∂ρ∂y(x,y,z)=−2y(z+2x2)(1+x2+y2)2∂ρ∂y(1,0,−1)=0∂ρ∂z(x,y,z)=11+x2+y2∂ρ∂z(1,0,−1)=12\begin{align*} \pdiff{\rho}{x}(x,y,z) &= \frac{4x}{1+x^2+y^2} -\frac{2x(z+2x^2)}{{(1+x^2+y^2)}^2} & \pdiff{\rho}{x}(1,0,-1) &= \frac{4}{2} -\frac{2(-1+2)}{{(2)}^2} =\frac{3}{2} \\ \pdiff{\rho}{y}(x,y,z) &= -\frac{2y(z+2x^2)}{{(1+x^2+y^2)}^2} & \pdiff{\rho}{y}(1,0,-1) &= 0\\ \pdiff{\rho}{z}(x,y,z) &= \frac{1}{1+x^2+y^2} & \pdiff{\rho}{z}(1,0,-1) &= \frac{1}{2} \end{align*}

So ∇ρ(1,0,−1)=12(3,0,1)\vnabla\rho(1,0,-1) = \frac{1}{2}(3,0,1). The unit vector in this direction is 110(3,0,1)\frac{1}{\sqrt{10}}(3,0,1).

(b) The region swept by the space craft is, in cylindrical coordinates,

V={ (rcos⁡θ , rsin⁡θ , z) ∣ 0≤θ≤2π, 0≤r≤1, 0≤z≤2 }\begin{equation*} V=\Set{(r\cos\theta\,,\,r\sin\theta\,,\,z)}{0\le\theta\le 2\pi,\ 0\le r\le 1,\ 0\le z\le 2} \end{equation*}

and the amount of hydrogen collected is

∭Vρ dV=∭Vz+2r2cos⁡2θ1+r2rdr dθ dz=∫02dz∫02πdθ∫01dr zr+(2cos⁡2θ)r31+r2=∫02dz∫02πdθ∫01dr [zr1+r2+2rcos⁡2θ−cos⁡2θ2r1+r2]since r31+r2=r+r3−r1+r2=r−r1+r2=∫02dz∫02πdθ [z2ln⁡(1+r2)+r2cos⁡2θ−ln⁡(1+r2)cos⁡2θ]01=∫02dz∫02πdθ [ln⁡22z+cos⁡2θ−ln⁡(2)cos⁡2θ]=∫02dz [(πln⁡2)z+π−πln⁡2]=2πln⁡2+2π−2πln⁡2=2π\begin{align*} \tripInt_V \rho\ \dee{V} &=\tripInt_V\frac{z+2r^2\cos^2\theta}{1+r^2}r\dee{r}\,\dee{\theta}\,\dee{z} \\ &=\int_0^2\dee{z} \int_0^{2\pi} \dee{\theta} \int_0^1\dee{r}\ \frac{zr+(2\cos^2\theta)r^3}{1+r^2} \\ &=\int_0^2\dee{z} \int_0^{2\pi} \dee{\theta} \int_0^1\dee{r}\ \left[ z \frac{r}{1+r^2} +2r\cos^2\theta -\cos^2\theta\frac{2r}{1+r^2}\right]\\ &\hskip1.5in\text{since }\frac{r^3}{1+r^2} = \frac{r+r^3 - r}{1+r^2} =r-\frac{r}{1+r^2} \\ &=\int_0^2\dee{z} \int_0^{2\pi} \dee{\theta}\ \left[\frac{z}{2}\ln(1+r^2)+r^2\cos^2\theta -\ln(1+r^2)\cos^2\theta\right]_0^1 \\ &=\int_0^2\dee{z} \int_0^{2\pi} \dee{\theta}\ \left[\frac{\ln 2}{2} z+ \cos^2\theta -\ln(2)\cos^2\theta\right] \\ &=\int_0^2\dee{z} \ \left[(\pi\ln 2) z+ \pi-\pi\ln 2\right] \\ &= 2\pi\ln 2 +2\pi -2\pi\ln 2 \\ &=2\pi \end{align*}
Q17Stage 3

A torus of mass MM is generated by rotating a circle of radius aa about an axis in its plane at distance bb from the centre (b>a)(b>a). The torus has constant density. Find the moment of inertia about the axis of rotation. By definition the moment of intertia is ∭r2dm\tripInt r^2 \dee{m} where dm\dee{m} is the mass of an infinitesmal piece of the solid and rr is its distance from the axis.

Answer

M(34a2+b2)M\big(\frac{3}{4}a^2+b^2\big)

Full solution

We may choose our coordinate axes so that the torus is constructed by rotating the circle (x−b)2+z2=a2(x-b)^2+z^2=a^2 (viewed as lying in the xzxz–plane) about the zz–axis. On this circle, xx runs from b−ab-a to b+ab+a.

Figure from prob_s3.6, line 1122

Figure from prob_s3.6, line 1122

In cylindrical coordinates, the torus has equation (r−b)2+z2=a2(r-b)^2+z^2=a^2. (Recall that the cylindrical coordinate rr of a point is its distance from the zz–axis.) On this torus,

  • rr runs from b−ab-a to b+ab+a.

  • For each fixed rr, zz runs from −a2−(r−b)2-\sqrt{a^2-(r-b)^2} to a2−(r−b)2\sqrt{a^2-(r-b)^2}.

As the torus is symmetric about the xyxy–plane, its volume is twice that of the volume of the part with z≥0z\ge 0.

Volume=2∫02πdθ∫b−ab+adr r∫0a2−(r−b)2dz=2∫02πdθ∫b−ab+adr ra2−(r−b)2=4π∫−aads (s+b)a2−s2 where s=r−b\begin{align*} \text{Volume} &= 2\int_{0}^{2\pi} d\theta\int_{b-a}^{b+a}\dee{r}\ r\int_{0}^{\sqrt{a^2-(r-b)^2}}\dee{z} \\ &= 2\int_{0}^{2\pi} d\theta\int_{b-a}^{b+a}\dee{r}\ r\sqrt{a^2-(r-b)^2} \\ &= 4\pi \int_{-a}^{a}\dee{s}\ (s+b)\sqrt{a^2-s^2} \qquad\text{ where }s=r-b \end{align*}

As sa2−s2s\sqrt{a^2-s^2} is odd under s→−ss\rightarrow -s, ∫−aads sa2−s2=0\int_{-a}^{a}\dee{s}\ s\sqrt{a^2-s^2}=0. Also, ∫−aads a2−s2\int_{-a}^{a}\dee{s}\ \sqrt{a^2-s^2} is precisely the area of the top half of a circle of radius aa. So

Volume =4bπ∫−aads a2−s2=2π2a2b\text{Volume }= 4b\pi \int_{-a}^{a}\dee{s}\ \sqrt{a^2-s^2} =2\pi^2a^2b

So the mass density of the torus is M2π2a2b\frac{M}{2\pi^2a^2b} and dm=M2π2a2b dV=M2π2a2b r dr dθ dz\dee{m} = \frac{M}{2\pi^2a^2b}\,\dee{V} =\frac{M}{2\pi^2a^2b}\,r\,\dee{r}\,d\theta\,\dee{z} and

moment of inertia=2∫02πdθ∫b−ab+adr r∫0a2−(r−b)2dz M2π2a2br2=Mπ2a2b∫02πdθ∫b−ab+adr r3a2−(r−b)2=2Mπa2b∫−aads (s+b)3a2−s2 where s=r−b=2Mπa2b∫−aads (s3+3s2b+3sb+b3)a2−s2\begin{align*} \text{moment of inertia} &= 2\int_{0}^{2\pi} d\theta\int_{b-a}^{b+a}\dee{r}\ r\int_{0}^{\sqrt{a^2-(r-b)^2}}\dee{z}\ \frac{M}{2\pi^2a^2b} r^2 \\ &= \frac{M}{\pi^2a^2b}\int_{0}^{2\pi} d\theta\int_{b-a}^{b+a}\dee{r}\ r^3\sqrt{a^2-(r-b)^2} \\ &= \frac{2M}{\pi a^2b} \int_{-a}^{a}\dee{s}\ (s+b)^3\sqrt{a^2-s^2} \qquad\text{ where }s=r-b \\ &= \frac{2M}{\pi a^2b} \int_{-a}^{a}\dee{s}\ (s^3+3s^2b+3sb+b^3)\sqrt{a^2-s^2} \end{align*}

Again, by oddness, the s3s^3 and 3sb3sb integrals are zero. For the others, substitute in s=asin⁡ts=a\sin t, ds=acos⁡t\dee{s}=a\cos t.

moment=2Mπa2b∫−π2π2(acos⁡t dt) (3a2bsin⁡2t+b3)acos⁡t=2Mπ∫−π2π2dt (3a2sin⁡2t+b2)cos⁡2t=4Mπ∫0π2dt (3a2cos⁡2t−3a2cos⁡4t+b2cos⁡2t)since sin⁡2t=1−cos⁡2t\begin{align*} \text{moment} &= \frac{2M}{\pi a^2b} \int_{-{\pi\over 2}}^{{\pi\over 2}}(a\cos t\, \dee{t}) \ (3a^2b\sin^2 t+b^3)a\cos t = \frac{2M}{\pi } \int_{-{\pi\over 2}}^{{\pi\over 2}}\dee{t}\ (3a^2\sin^2 t+b^2)\cos^2 t \\ &= \frac{4M}{\pi} \int_0^{{\pi\over 2}}\dee{t}\ (3a^2\cos^2 t-3a^2\cos^4 t+b^2\cos^2 t) \qquad\text{since }\sin^2t=1-\cos^2t \end{align*}

To integrate (For a general discussion of trigonometric integrals see §1.8 in the CLP-2 text. In particular the integral
∫cos⁡4x dx\int \cos^4 x\ \dee{x} is evaluated in Example 1.8.8 in the CLP-2 text. For an efficient, sneaky, way to evaluate ∫0π2cos⁡2t dt\int_0^{{\pi\over 2}} \cos^2 t\ \dee{t} see Remark 3.3.5 in the CLP-3 text.) cos⁡2t\cos^2t and cos⁡4t\cos^4t, we use the double angle formulae sin⁡2x=1−cos⁡(2x)2\sin^2 x= \frac{1-\cos(2x)}{2} and cos⁡2x=1+cos⁡(2x)2\cos^2 x= \frac{1+\cos(2x)}{2} to write

cos⁡2t=1+cos⁡(2t)2\begin{equation*} \cos^2 t= \frac{1+\cos(2t)}{2} \end{equation*}

and

cos⁡4t=[1+cos⁡(2t)2]2=14+12cos⁡(2t)+14cos⁡2(2t)=14+12cos⁡(2t)+18(1+cos⁡(4t))=38+12cos⁡(2t)+18cos⁡(4t)\begin{align*} \cos^4 t &= \left[ \frac{1+\cos(2t)}{2} \right]^2 \\ &= \frac{1}{4} + \frac{1}{2} \cos(2t) + \frac{1}{4}\cos^2(2t)\\ &= \frac{1}{4} + \frac{1}{2} \cos(2t) + \frac{1}{8}\left(1 + \cos(4t)\right)\\ &= \frac{3}{8} + \frac{1}{2} \cos(2t) + \frac{1}{8}\cos(4t) \end{align*}

So

moment=4Mπ[3a2(t2+sin⁡(2t)4)−3a2(3t8+14sin⁡(2t)+132sin⁡(4t))+b2(t2+sin⁡(2t)4)]0π2=4Mπ[3a2π4−3a23π16+b2π4]=M(34a2+b2)\begin{align*} \text{moment} &= \frac{4M}{\pi} \left[3a^2\left(\frac{t}{2}+\frac{\sin(2t)}{4}\right) -3a^2\left(\frac{3t}{8} + \frac{1}{4} \sin(2t) + \frac{1}{32}\sin(4t)\right) +b^2\left(\frac{t}{2}+\frac{\sin(2t)}{4}\right)\right]_0^{{\pi\over 2}} \\ &= \frac{4M}{\pi} \Big[3a^2\frac{\pi}{4}-3a^2\frac{3\pi}{16}+b^2\frac{\pi}{4}\Big] =M\left(\frac{3}{4}a^2+b^2\right) \end{align*}

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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.