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Multiple Integrals

3.7 Triple Integrals in Spherical Coordinates

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

1Stage 1Conceptual

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

Q1Stage 1

Use (ρ,θ,φ)(\rho,\theta,\varphi) to denote spherical coordinates.

  1. Draw φ=0\varphi=0.

  2. Draw φ=π ⁣/4\varphi=\nicefrac{\pi}{4}.

  3. Draw φ=π ⁣/2\varphi=\nicefrac{\pi}{2}.

  4. Draw φ=3π ⁣/4\varphi=\nicefrac{3\pi}{4}.

  5. Draw φ=π\varphi=\pi.

Answer

Figure from prob_s3.7, line 32

Figure from prob_s3.7, line 32

Figure from prob_s3.7, line 32

Figure from prob_s3.7, line 32

Figure from prob_s3.7, line 32

Figure from prob_s3.7, line 32

Figure from prob_s3.7, line 32

Figure from prob_s3.7, line 32

Figure from prob_s3.7, line 32

Figure from prob_s3.7, line 32

Full solution

Since the spherical coordinate φ(x,y,z)\varphi(x,y,z) of a point (x,y,z)(x,y,z) is the angle between the positive zz-axis and the radius vector from (0,0,0)(0,0,0) to (x,y,z)(x,y,z), the sets

{ (x,y,z) ∣ φ(x,y,z)=0 }=the positive z-axis{ (x,y,z) ∣ φ(x,y,z)=π ⁣/2 }=the xy-plane{ (x,y,z) ∣ φ(x,y,z)=π }=the negative z-axis\begin{align*} \Set{(x,y,z)}{\varphi(x,y,z)=0} &=\text{the positive }z\text{-axis} \\ \Set{(x,y,z)}{\varphi(x,y,z)=\nicefrac{\pi}{2}} &=\text{the }xy\text{-plane} \\ \Set{(x,y,z)}{\varphi(x,y,z)=\pi} &=\text{the negative }z\text{-axis} \end{align*}

Alternatively, tan⁡φ(x,y,z)=zx2+y2\tan\varphi(x,y,z)=\frac{z}{\sqrt{x^2+y^2}}, so that, for any 0<Φ<π0<\Phi<\pi,

{ (x,y,z) ∣ φ(x,y,z)=Φ }={ (x,y,z) ∣ z=tan⁡Φx2+y2 }=the cone that makes the angle Φ with the positive z-axis\begin{align*} \Set{(x,y,z)}{\varphi(x,y,z)=\Phi} &=\Set{(x,y,z)}{z=\tan\Phi\sqrt{x^2+y^2}} \\ &=\text{the cone that makes the angle }\Phi\text{ with the positive }z\text{-axis} \end{align*}

Figure from prob_s3.7, line 32

Figure from prob_s3.7, line 32

Figure from prob_s3.7, line 32

Figure from prob_s3.7, line 32

Figure from prob_s3.7, line 32

Figure from prob_s3.7, line 32

Figure from prob_s3.7, line 32

Figure from prob_s3.7, line 32

Figure from prob_s3.7, line 32

Figure from prob_s3.7, line 32

Q2Stage 1

Sketch the point with the specified spherical coordinates.

  1. ρ=0\rho=0, θ=0.1π\theta=0.1\pi, φ=0.7π\varphi=0.7\pi

  2. ρ=1\rho=1, θ=0.3π\theta=0.3\pi, φ=0\varphi=0

  3. ρ=1\rho=1, θ=0\theta=0, φ=π2\varphi=\frac{\pi}{2}

  4. ρ=1\rho=1, θ=π3\theta=\frac{\pi}{3}, φ=π2\varphi=\frac{\pi}{2}

  5. ρ=1\rho=1, θ=π2\theta=\frac{\pi}{2}, φ=π2\varphi=\frac{\pi}{2}

  6. ρ=1\rho=1, θ=π3\theta=\frac{\pi}{3}, φ=π6\varphi=\frac{\pi}{6}

Answer

Figure from prob_s3.7, line 103

Figure from prob_s3.7, line 103

Full solution

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

  • The spherical coordinate ρ\rho is the distance of the point from the origin (0,0,0)(0,0,0). In particular if ρ=0\rho=0, then the point is the origin (regardless of the values of θ\theta and φ\varphi). If ρ=1\rho=1 then the point lies on the sphere of radius 11 centred on the origin.

  • The spherical coordinate φ\varphi is the angle between the positive zz-axis and the radial line segment from the origin to (x,y,z)(x,y,z). In particular, all points with φ=0\varphi=0 lie on the positive zz–axis (regardless of the value of θ\theta). All points with φ=π2\varphi=\frac{\pi}{2} lie in the xyxy-plane.

Figure from prob_s3.7, line 103

Figure from prob_s3.7, line 103

Q3Stage 1

Convert from Cartesian to spherical coordinates.

  1. (−2,0,0)(-2,0,0)

  2. (0,3,0)(0,3,0)

  3. (0,0,−4)(0,0,-4)

  4. (−12,12,3)\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\sqrt{3}\right)

Answer

(a) ρ=2\rho=2, θ=π\theta=\pi, φ=π2\varphi=\frac{\pi}{2}

(b) ρ=3\rho=3, θ=π2\theta=\frac{\pi}{2}, φ=π2\varphi=\frac{\pi}{2}

(c) ρ=4\rho=4, θ=arbitrary\theta=\text{arbitrary}, φ=π\varphi=\pi

(d) ρ=2\rho=2, θ=3π4\theta=\frac{3\pi}{4}, φ=π6\varphi=\frac{\pi}{6}

Full solution

(a) The point (−2,0,0)(-2,0,0)

  • lies in the xyxy-plane (i.e. has z=ρcos⁡φ=0z=\rho\cos\varphi=0) and so has φ=π2\varphi=\frac{\pi}{2} and

  • lies on the negative xx-axis and so has θ=π\theta=\pi and

  • is a distance 22 from the origin and so has ρ=2\rho=2.

(b) The point (0,3,0)(0,3,0)

  • lies in the xyxy-plane (i.e. has z=ρcos⁡φ=0z=\rho\cos\varphi=0) and so has φ=π2\varphi=\frac{\pi}{2} and

  • lies on the positive yy-axis and so has θ=π2\theta=\frac{\pi}{2} and

  • is a distance 33 from the origin and so has ρ=3\rho=3.

(c) The point (0,0,−4)(0,0,-4)

  • lies on the negative zz-axis and so has φ=π\varphi=\pi and θ\theta arbitrary and

  • is a distance 44 from the origin and so has ρ=4\rho=4.

(d) The point (−12,12,3)\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\sqrt{3}\right)

  • has ρ=x2+y2+z2=(−12)2+(12)2+(3)2=4=2\rho=\sqrt{x^2+y^2+z^2} =\sqrt{\left(-\frac{1}{\sqrt{2}}\right)^2 +\left(\frac{1}{\sqrt{2}}\right)^2 +\left(\sqrt{3}\right)^2}=\sqrt{4}=2 and

  • has 3=z=ρcos⁡φ=2cos⁡φ\sqrt{3}=z=\rho\cos\varphi = 2\cos\varphi so that cos⁡φ=32\cos\varphi = \frac{\sqrt{3}}{2} and φ=π6\varphi=\frac{\pi}{6} and

  • has −12=x=ρsin⁡φcos⁡θ=2(12)cos⁡θ-\frac{1}{\sqrt{2}}=x=\rho\sin\varphi\cos\theta = 2\big(\frac{1}{2}\big)\cos\theta so that cos⁡θ=−12\cos\theta = -\frac{1}{\sqrt{2}}. As (−12,12)\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right) is in the second quadrant, we have π2≤θ≤π\frac{\pi}{2}\le\theta\le\pi and so θ=3π4\theta=\frac{3\pi}{4}.

Q4Stage 1

Convert from spherical to Cartesian coordinates.

  1. ρ=1\rho=1, θ=π3\theta=\frac{\pi}{3}, φ=π6\varphi=\frac{\pi}{6}

  2. ρ=2\rho=2, θ=π2\theta=\frac{\pi}{2}, φ=π2\varphi=\frac{\pi}{2}

Answer

(a) (14 , 34 , 32)\left(\frac{1}{4}\,,\,\frac{\sqrt{3}}{4}\,,\,\frac{\sqrt{3}}{2}\right) (b) (0,2,0)(0,2,0)

Full solution

(a) The Cartesian coordinates corresponding to ρ=1\rho=1, θ=π3\theta=\frac{\pi}{3}, φ=π6\varphi=\frac{\pi}{6} are

x=ρsin⁡φcos⁡θ=sin⁡π6cos⁡π3=(12)(12)=14y=ρsin⁡φsin⁡θ=sin⁡π6sin⁡π3=(12)(32)=34z=ρcos⁡φ=cos⁡π6=32\begin{align*} x&=\rho\sin\varphi\cos\theta =\sin\frac{\pi}{6}\cos\frac{\pi}{3} =\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) =\frac{1}{4} \\ y&=\rho\sin\varphi\sin\theta =\sin\frac{\pi}{6}\sin\frac{\pi}{3} =\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) =\frac{\sqrt{3}}{4} \\ z&=\rho\cos\varphi =\cos\frac{\pi}{6} =\frac{\sqrt{3}}{2} \end{align*}

(b) The Cartesian coordinates corresponding to ρ=2\rho=2, θ=π2\theta=\frac{\pi}{2}, φ=π2\varphi=\frac{\pi}{2} are

x=ρsin⁡φcos⁡θ=2sin⁡π2cos⁡π2=0y=ρsin⁡φsin⁡θ=2sin⁡π2sin⁡π2=2z=ρcos⁡φ=2cos⁡π2=0\begin{align*} x&=\rho\sin\varphi\cos\theta =2\sin\frac{\pi}{2}\cos\frac{\pi}{2} =0 \\ y&=\rho\sin\varphi\sin\theta =2\sin\frac{\pi}{2}\sin\frac{\pi}{2} =2 \\ z&=\rho\cos\varphi =2\cos\frac{\pi}{2} =0 \end{align*}

Alternatively, we could just observe that

  • as φ=π2\varphi=\frac{\pi}{2} the point lies in the xyxy-plane and so has z=0z=0 and

  • as ρ=2\rho=2, θ=π2\theta=\frac{\pi}{2} the point lies on the positive yy-axis and is a distance 22 from the origin and so is (0,2,0)(0,2,0).

Q5Stage 1

Rewrite the following equations in spherical coordinates.

  1. z2=3x2+3y2z^2=3x^2+3y^2

  2. x2+y2+(z−1)2=1x^2+y^2+(z-1)^2=1

  3. x2+y2=4x^2+y^2=4

Answer

(a) φ=π6 or 5π6\varphi=\frac{\pi}{6}\text{ or }\frac{5\pi}{6} (b) ρ=2cos⁡φ\rho=2\cos\varphi (c) ρsin⁡φ=2\rho\sin\varphi=2

Full solution

(a) In spherical coordinates

z2=3x2+3y2  ⟺  ρ2cos⁡2φ=3ρ2sin⁡2φcos⁡2θ+3ρ2sin⁡2φsin⁡2θ=3ρ2sin⁡2φ  ⟺  tan⁡2φ=13  ⟺  tan⁡φ=±13  ⟺  φ=π6 or 5π6\begin{align*} z^2=3x^2+3y^2 &\iff \rho^2\cos^2\varphi = 3\rho^2\sin^2\varphi\cos^2\theta +3\rho^2\sin^2\varphi\sin^2\theta =3\rho^2\sin^2\varphi \\ &\iff \tan^2\varphi=\frac{1}{3} \iff \tan\varphi=\pm\frac{1}{\sqrt{3}} \\ &\iff \varphi=\frac{\pi}{6}\text{ or }\frac{5\pi}{6} \end{align*}

The surface z2=3x2+3y2z^2=3x^2+3y^2 is a cone. The upper half of the cone, i.e. the part with z≥0z\ge 0, is φ=π6\varphi=\frac{\pi}{6}. The lower half of the cone, i.e. the part with z≤0z\le 0, is φ=π−π6=5π6\varphi=\pi-\frac{\pi}{6}=\frac{5\pi}{6}.

(b) In spherical coordinates

x2+y2+(z−1)2=1  ⟺  ρ2sin⁡2φcos⁡2θ+ρ2sin⁡2φsin⁡2θ+(ρcos⁡φ−1)2=1  ⟺  ρ2sin⁡2φ+ρ2cos⁡2φ−2ρcos⁡φ=0  ⟺  ρ2−2ρcos⁡φ=0  ⟺  ρ=2cos⁡φ\begin{align*} x^2+y^2+(z-1)^2=1 &\iff \rho^2\sin^2\varphi\cos^2\theta +\rho^2\sin^2\varphi\sin^2\theta +\big(\rho\cos\varphi-1\big)^2 =1 \\ &\iff \rho^2\sin^2\varphi +\rho^2\cos^2\varphi-2\rho\cos\varphi =0 \\ &\iff \rho^2-2\rho\cos\varphi =0 \\ &\iff \rho=2\cos\varphi \end{align*}

(c) In spherical coordinates

x2+y2=4  ⟺  ρ2sin⁡2φcos⁡2θ+ρ2sin⁡2φsin⁡2θ=4  ⟺  ρ2sin⁡2φ=4  ⟺  ρsin⁡φ=2\begin{align*} x^2+y^2=4 &\iff \rho^2\sin^2\varphi\cos^2\theta +\rho^2\sin^2\varphi\sin^2\theta =4 \\ &\iff \rho^2\sin^2\varphi =4 \\ &\iff \rho\sin\varphi=2 \end{align*}

since ρ≥0\rho\ge 0 and 0≤φ≤π0\le\varphi\le\pi so that sin⁡φ≥0\sin\varphi\ge 0.

Q6Stage 1Past exam · M200 2008A

Using spherical coordinates and integration, show that the volume of the sphere of radius 11 centred at the origin is 4π/34\pi/3.

Answer

See the solution.

Full solution

In spherical coordinates, the sphere in question is

B={ (ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ) ∣ 0≤ρ≤1 , 0≤φ≤π , 0≤θ≤2π }\begin{align*} B =\Set{(\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)}{0\le\rho\le 1\,,\,0\le\varphi\le\pi\,,\, 0\le\theta\le 2\pi} \end{align*}

As dV=ρ2sin⁡φ dρ dφ dθ\dee{V} =\rho^2\sin\varphi\ \dee{\rho}\,\dee{\varphi}\,\dee{\theta},

Volume(S)=∭BdV=∫02πdθ∫0πdφ∫01dρ ρ2sin⁡φ=[∫02πdθ][∫0πdφ sin⁡φ][∫01dρ ρ2]=2π[−cos⁡φ]0π[ρ33]01=(2π)(2)(13)=4π3\begin{align*} \text{Volume}(S) =\tripInt_B\dee{V} &=\int_0^{2\pi}\dee{\theta}\int_0^{\pi}\dee{\varphi}\int_0^1\dee{\rho}\ \rho^2\sin\varphi \\ &=\left[\int_0^{2\pi}\dee{\theta}\right] \left[\int_0^{\pi}\dee{\varphi}\ \sin\varphi\right] \left[\int_0^1\dee{\rho}\ \rho^2\right] \\ &= 2\pi\Big[-\cos\varphi\Big]_0^\pi \left[\frac{\rho^3}{3}\right]_0^1 =(2\pi)(2)\left(\frac{1}{3}\right) \\ &=\frac{4\pi}{3} \end{align*}

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q7Stage 2Past exam · M200 2006A

Consider the region EE in 33–dimensions specified by the spherical inequalities

1≤ρ≤1+cos⁡φ\begin{equation*} 1 \le \rho \le 1 + \cos \varphi \end{equation*}

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

  2. Find the volume of E.

Answer

(a)

Figure from prob_s3.7, line 418

Figure from prob_s3.7, line 418

(b) 11π6\frac{11\pi}{6}

Full solution

(a) First observe that both boundaries of EE, namely ρ=1\rho=1 and ρ=1+cos⁡φ\rho= 1 + \cos\varphi, are independent of the spherical coordinate θ\theta. So EE is invariant under rotations about the zz–axis. To sketch EE we

  • first sketch the part of the boundary of EE with θ=0\theta=0 (i.e. in the half of the xzxz–plane with x>0x>0), and then

  • rotate about the zz–axis.

The part of the boundary of EE with θ=0\theta=0 (i.e. in the half–plane y=0y=0, x≥0x\ge 0), consists of two curves.

  • ρ=1+cos⁡φ\rho=1+\cos\varphi, θ=0\theta=0:

    • When φ=0\varphi=0 (i.e. on the positive zz–axis), We have cos⁡φ=1\cos\varphi =1 and hence ρ=2\rho=2. So this curve starts at (0,0,2)(0,0,2).

    • As φ\varphi increases cos⁡φ\cos\varphi, and hence ρ\rho, decreases.

    • When φ\varphi is π2\frac{\pi}{2} (i.e. in the xyxy–plane), we have cos⁡φ=0\cos\varphi =0 and hence ρ=1\rho=1.

    • When π2<φ≤π\frac{\pi}{2}<\varphi\le \pi, we have cos⁡φ<0\cos\varphi<0 and hence ρ<1\rho<1. All points in EE are required to obey ρ≥1\rho\ge 1. So this part of the boundary stops at the point (1,0,0)(1,0,0) in the xyxy–plane.

    • The curve ρ=1+cos⁡φ\rho=1+\cos\varphi, θ=0\theta=0, 0≤φ≤π20\le\varphi\le\frac{\pi}{2} is sketched in the figure on the left below. It is the outer curve from (0,0,2)(0,0,2) to (1,0,0)(1,0,0).

  • ρ=1\rho=1, θ=0\theta=0:

    • The surface ρ=1\rho=1 is the sphere of radius 11 centred on the origin.

    • As we observed above, the conditions 1≤ρ≤1+cos⁡φ1\le\rho\le 1+\cos\varphi force 0≤φ≤π20\le\varphi\le\frac{\pi}{2}, i.e. z≥0z\ge 0.

    • The sphere ρ=1\rho=1 intersects the quarter plane y=0y=0, x≥0x\ge 0, z≥0z\ge 0, in the quarter circle centred on the origin that starts at (0,0,1)(0,0,1) on the zz–axis and ends at (1,0,0)(1,0,0) in the xyxy–plane.

    • The curve ρ=1\rho=1, θ=0\theta=0, 0≤φ≤π20\le\varphi\le\frac{\pi}{2} is sketched in the figure on the left below. It is the inner curve from (0,0,1)(0,0,1) to (1,0,0)(1,0,0).

    To get EE, rotate the shaded region in the figure on the left below about the zz-axis. The part of EE in the first octant is sketched in the figure on the right below. The part of EE in the xzxz–plane (with x≥0x\ge 0) is lightly shaded and the part of EE in the yzyz–plane (with y≥0y\ge 0) is shaded a little more darkly.

    Figure from prob_s3.7, line 427

    Figure from prob_s3.7, line 427

    Figure from prob_s3.7, line 427

    Figure from prob_s3.7, line 427

(b) In EE

  • φ\varphi runs from 00 (i.e. the positive zz–axis) to π2\frac{\pi}{2} (i.e. the xyxy–plane).

  • For each φ\varphi in that range ρ\rho runs from 11 to 1+cos⁡φ1+\cos\varphi and θ\theta runs from 00 to 2π2\pi.

  • In spherical coordinates dV=ρ2 sin⁡φ dρ dθ dφ\dee{V} = \rho^2\,\sin\varphi\,\dee{\rho}\, \dee{\theta}\,\dee{\varphi}.

So

Volume(E)=∫0π/2dφ∫11+cos⁡φdρ∫02πdθ ρ2sin⁡φ=2π∫0π/2dφ sin⁡φ (1+cos⁡φ)3−133=−2π3∫21(u3−1) duwith u=1+cos⁡φ, du=−sin⁡φ dφ=−2π3[u44−u]21=−2π3[14−1−4+2]=11π6\begin{align*} \text{Volume}(E) &=\int_0^{\pi/2}\dee{\varphi} \int_1^{1+\cos\varphi}\dee{\rho} \int_0^{2\pi}\dee{\theta}\ \rho^2\sin\varphi \\ &=2\pi \int_0^{\pi/2}\dee{\varphi}\ \sin\varphi\ \frac{(1+\cos\varphi)^3-1^3}{3} \\ &=-\frac{2\pi}{3} \int_2^1 \big(u^3-1\big)\ \dee{u} \qquad\text{with }u=1+\cos\varphi,\ \dee{u}=-\sin\varphi\,\dee{\varphi} \\ &= -\frac{2\pi}{3}\left[\frac{u^4}{4}-u\right]_2^1 \\ &=-\frac{2\pi}{3} \left[\frac{1}{4}-1-4+2\right] \\ &=\frac{11 \pi}{6} \end{align*}
Q8Stage 2Past exam · M200 2006D

Use spherical coordinates to evaluate the integral

I=∭Dz dV\begin{equation*} I=\tripInt_D z\ \dee{V} \end{equation*}

where DD is the solid enclosed by the cone z=x2+y2z = \sqrt{x^2 + y^2} and the sphere x2+y2+z2=4x^2 + y^2 + z^2 = 4. That is, (x,y,z)(x,y,z) is in DD if and only if x2+y2≤z\sqrt{x^2 + y^2}\le z and x2+y2+z2≤4x^2 + y^2 + z^2 \le 4.

Answer

2π2\pi

Full solution

Recall that in spherical coordinates,

x=ρsin⁡φcos⁡θy=ρsin⁡φsin⁡θz=ρcos⁡φx2+y2=ρ2sin⁡2φ\begin{align*} x&=\rho\sin\varphi\cos\theta \\ y&=\rho\sin\varphi\sin\theta \\ z&=\rho\cos\varphi \\ x^2+y^2 &=\rho^2\sin^2\varphi \end{align*}

so that x2+y2+z2=4x^2+y^2+z^2= 4 becomes ρ=2\rho = 2, and x2+y2=z\sqrt{x^2+y^2}= z becomes

ρsin⁡φ=ρcos⁡φ  ⟺  tan⁡φ=1  ⟺  φ=π4\begin{align*} \rho\sin\varphi = \rho\cos\varphi \iff \tan\varphi= 1 \iff \varphi=\frac{\pi}{4} \end{align*}

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

Figure from prob_s3.7, line 547

Figure from prob_s3.7, line 547

Looking at the figure above, we see that, on DD

  • φ\varphi runs from 00 (the positive zz–axis) to π4\frac{\pi}{4} (on the cone), and

  • for each φ\varphi is that range, ρ\rho runs from 00 to 22 and θ\theta runs from 00 to 2π2\pi.

So

D={ (ρsin⁡φcos⁡θ,ρsin⁡φsin⁡θ,ρcos⁡φ) ∣ 0≤φ≤π ⁣/4, 0≤θ≤2π, ρ≤2 }\begin{align*} D&=\Set{(\rho\sin\varphi\cos\theta,\rho\sin\varphi\sin\theta,\rho\cos\varphi)} {0\le\varphi\le\nicefrac{\pi}{4},\ 0\le \theta\le 2\pi,\ \rho\le 2} \end{align*}

and, as dV=ρ2sin⁡φ dρ dθ dφ\dee{V} =\rho^2\sin\varphi\,\dee{\rho}\,\dee{\theta}\,\dee{\varphi},

I=∫0π/4dφ∫02πdθ∫02dρ ρ2sin⁡φ ρcos⁡φ⏞z=∫0π/4dφ∫02πdθ∫02dρ ρ3sin⁡φ cos⁡φ=2π 244∫0π/4dφ sin⁡φ cos⁡φ=2π 244[sin⁡2φ2]0π/4=2π\begin{align*} I&= \int_0^{\pi/4}\dee{\varphi}\int_0^{2\pi}\dee{\theta} \int_0^{2}\dee{\rho}\ \rho^2\sin\varphi\ \overbrace{\rho\cos\varphi}^{z} \\ &= \int_0^{\pi/4}\dee{\varphi}\int_0^{2\pi}\dee{\theta} \int_0^{2}\dee{\rho}\ \rho^3\sin\varphi\,\cos\varphi \\ &= 2\pi\ \frac{2^4}{4} \int_0^{\pi/4}\dee{\varphi}\ \sin\varphi\,\cos\varphi \\ &= 2\pi\ \frac{2^4}{4}\left[\frac{\sin^2\varphi}{2}\right]_0^{\pi/4} \\ &= 2\pi \end{align*}
Q9Stage 2

Use spherical coordinates to find

  1. The volume inside the cone z=x2+y2z=\sqrt{x^2+y^2} and inside the sphere x2+y2+z2=a2x^2+y^2+z^2=a^2.

  2. ∭Rx dV\tripInt_R x\, \dee{V} and ∭Rz dV\tripInt_R z\, \dee{V} over the part of the sphere of radius aa that lies in the first octant.

  3. The mass of a spherical planet of radius aa whose density at distance ρ\rho from the center is δ=A/(B+ρ2)\de=A/(B+\rho^2).

  4. The volume enclosed by ρ=a(1−cos⁡φ).\rho=a(1-\cos\varphi). Here ρ\rho and φ\varphi refer to the usual spherical coordinates.

Answer

(a) 2πa33(1−12)2\pi\frac{a^3}{3}\big(1-\frac{1}{\sqrt{2}}\big) (b) πa416\frac{\pi a^4}{16} (c) 4πA(a−Btan⁡−1aB)4\pi A\big(a-\sqrt{B}\tan^{-1}\frac{a}{\sqrt{B}}\big) (d) 83πa3\frac{8}{3}\pi a^3

Full solution

(a) Recall that in spherical coordinates,

x=ρsin⁡φcos⁡θy=ρsin⁡φsin⁡θz=ρcos⁡φx2+y2=ρ2sin⁡2φ\begin{align*} x&=\rho\sin\varphi\cos\theta \\ y&=\rho\sin\varphi\sin\theta \\ z&=\rho\cos\varphi \\ x^2+y^2 &=\rho^2\sin^2\varphi \end{align*}

so that x2+y2+z2=a2x^2+y^2+z^2= a^2 becomes ρ=a\rho = a, and x2+y2=z\sqrt{x^2+y^2}= z becomes

ρsin⁡φ=ρcos⁡φ  ⟺  tan⁡φ=1  ⟺  φ=π4\begin{align*} \rho\sin\varphi = \rho\cos\varphi \iff \tan\varphi= 1 \iff \varphi=\frac{\pi}{4} \end{align*}

Here is a sketch of the y=0y=0 cross–section of the specified region.

Figure from prob_s3.7, line 627

Figure from prob_s3.7, line 627

Looking at the figure above, we see that, on that region,

  • φ\varphi runs from 00 (the positive zz–axis) to π4\frac{\pi}{4} (on the cone), and

  • for each φ\varphi is that range, ρ\rho runs from 00 to aa and θ\theta runs from 00 to 2π2\pi.

so that

Volume=∫0adρ∫02πdθ∫0π4dφ ρ2sin⁡φ={∫0adρ ρ2}{∫02πdθ}{∫0π4dφ sin⁡φ}=a33 2π [−cos⁡φ]0π4=2πa33(1−12)\begin{align*} \text{Volume} &= \int_{0}^{a}\dee{\rho} \int_{0}^{2\pi} \dee{\theta} \int_0^{{\pi\over 4}}\dee{\varphi}\ \rho^2\sin\varphi = \Big\{\int_{0}^{a}\dee{\rho}\ \rho^2 \Big\}\Big\{\int_{0}^{2\pi} \dee{\theta}\Big\}\Big\{ \int_0^{{\pi\over 4}}\dee{\varphi}\ \sin\varphi\Big\} \\ &= \frac{a^3}{3}\ 2\pi\ \Big[-\cos\varphi\Big]_0^{{\pi\over 4}} =2\pi\frac{a^3}{3}\left(1-\frac{1}{\sqrt{2}}\right) \end{align*}

(b) The part of the sphere in question is

R={ (x,y,z) ∣ x2+y2+z2≤a2, x≥0, y≥0, z≥0 }={ (ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ) ∣ ρ≤a, 0≤φ≤π2, 0≤θ≤π2 }\begin{align*} R&=\Set{(x,y,z)}{x^2+y^2+z^2\le a^2,\ x\ge0,\ y\ge 0,\ z\ge 0} \\ &=\Set{(\rho\sin\varphi\cos\theta\,,\, \rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)}{\rho\le a,\ 0\le\varphi\le\tfrac{\pi}{2}, \ 0\le\theta\le\tfrac{\pi}{2}} \end{align*}

By symmetry, the two specified integrals are equal, and are

∫0adρ ρ2∫0π2dφ sin⁡φ∫0π2dθ ρcos⁡φ⏞z=a44π2∫0π2dφ sin⁡φcos⁡φ=πa48∫01dt t where t=sin⁡φ, dt=cos⁡φ dφ=πa416\begin{align*} \int_{0}^{a}\dee{\rho}\ \rho^2 \int_0^{{\pi\over 2}}\dee{\varphi}\ \sin\varphi \int_{0}^{{\pi\over 2}} \dee{\theta}\ \overbrace{\rho\cos\varphi}^{z} &=\frac{a^4}{4}\frac{\pi}{2} \int_0^{{\pi\over 2}}\dee{\varphi}\ \sin\varphi\cos\varphi \\ &=\frac{\pi a^4}{8} \int_0^{1}\dee{t}\ t\qquad\hbox{ where $t=\sin\varphi$, $\dee{t}=\cos\varphi\,\dee{\varphi}$} \\ &=\frac{\pi a^4}{16} \end{align*}

(c) The planet in question is

P={ (x,y,z) ∣ x2+y2+z2≤a2 }={ (ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ) ∣ ρ≤a, 0≤φ≤π, 0≤θ≤2π }\begin{align*} P&=\Set{(x,y,z)}{x^2+y^2+z^2\le a^2} \\ &=\Set{(\rho\sin\varphi\cos\theta\,,\, \rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)}{\rho\le a,\ 0\le\varphi\le\pi, \ 0\le\theta\le2\pi} \end{align*}

So the

mass=∫0adρ ρ2∫0πdφ sin⁡φ∫02πdθ AB+ρ2⏞density=2πA{∫0πdφ sin⁡φ}{∫0adρ ρ2B+ρ2}=4πA∫0adρ (1−BB+ρ2)=4πAa−4πAB∫0a/Bds 11+s2 where ρ=B s, dρ=B ds=4πA(a−Btan⁡−1aB)\begin{align*} \text{mass} &= \int_{0}^{a}\dee{\rho}\ \rho^2 \int_0^{\pi}\dee{\varphi}\ \sin\varphi \int_{0}^{2\pi} \dee{\theta}\ \overbrace{\frac{A}{B+\rho^2}}^{\rm density} =2\pi A\Big\{ \int_0^{\pi}\dee{\varphi}\ \sin\varphi\Big\} \Big\{\int_{0}^{a}\dee{\rho}\ \frac{\rho^2}{B+\rho^2} \Big\} \\ &=4\pi A\int_{0}^{a}\dee{\rho}\ \left(1-\frac{B}{B+\rho^2} \right) \\ &=4\pi Aa-4\pi A\sqrt{B}\int_{0}^{a/\sqrt{B}}\dee{s}\ \frac{1}{1+s^2} \qquad\hbox{ where $\rho=\sqrt{B}\,s$, $\dee{\rho}=\sqrt{B}\,\dee{s}$} \\ &=4\pi A\left(a-\sqrt{B}\tan^{-1}\frac{a}{\sqrt{B}}\right) \end{align*}

(d) Observe that

  • when φ=0\varphi=0 (i.e. on the positive zz–axis), cos⁡φ=1\cos\varphi=1 so that ρ=a(1−cos⁡φ)=0\rho=a(1-\cos\varphi)=0 and

  • as φ\varphi increases from 00 to π2\frac{\pi}{2}, cos⁡φ\cos\varphi decreases so that ρ=a(1−cos⁡φ)\rho=a(1-\cos\varphi) increases and

  • when φ=π2\varphi=\frac{\pi}{2} (i.e. on the xyxy–plane), cos⁡φ=0\cos\varphi=0 so that ρ=a(1−cos⁡φ)=a\rho=a(1-\cos\varphi)=a and

  • as φ\varphi increases from π2\frac{\pi}{2} to π\pi, cos⁡φ\cos\varphi continues to decrease so that ρ=a(1−cos⁡φ)\rho=a(1-\cos\varphi) increases still more and

  • when φ=π\varphi=\pi (i.e. on the negative zz–axis), cos⁡φ=−1\cos\varphi=-1 so that ρ=a(1−cos⁡φ)=2a\rho=a(1-\cos\varphi)=2a

So we have the following sketch of the intersection of the specified volume with the right half of the yzyz–plane.

Figure from prob_s3.7, line 627

Figure from prob_s3.7, line 627

The volume in question is invariant under rotations about the zz–axis so that

Volume=∫02πdθ∫0πdφ sin⁡φ∫0a(1−cos⁡φ)dρ ρ2=2πa33∫0πdφ sin⁡φ(1−cos⁡φ)3=2πa33∫02dt t3 where t=1−cos⁡φ, dt=sin⁡φ dφ=2π a33 244=83πa3\begin{align*} \text{Volume} &= \int_{0}^{2\pi} \dee{\theta}\int_0^{\pi}\dee{\varphi}\ \sin\varphi\int_{0}^{a(1-\cos\varphi)}\dee{\rho}\ \rho^2 \\ &=2\pi\frac{a^3}{3} \int_0^{\pi}\dee{\varphi}\ \sin\varphi(1-\cos\varphi)^3 \\ &=2\pi\frac{a^3}{3} \int_0^2\dee{t}\ t^3 \qquad\hbox{ where $t=1-\cos\varphi$, $\dee{t}=\sin\varphi\,\dee{\varphi}$} \\ &=2\pi\ \frac{a^3}{3}\ \frac{2^4}{4} =\frac{8}{3}\pi a^3 \end{align*}
Q10Stage 2Past exam · M200 2008A

Consider the hemispherical shell bounded by the spherical surfaces

x2+y2+z2=9andx2+y2+z2=4\begin{equation*} x^2 + y^2 + z^2 = 9\qquad\text{and}\qquad x^2 + y^2 + z^2 = 4 \end{equation*}

and above the plane z=0z = 0. Let the shell have constant density DD.

  1. Find the mass of the shell.

  2. Find the location of the center of mass of the shell.

Answer

(a) 383πD\frac{38}{3}\pi D

(b) xˉ=yˉ=0\bar x = \bar y=0 zˉ=195152≈1.28\bar z =\frac{195}{152} \approx 1.28

Full solution

Let's use HH to denote the hemispherical shell. On that shell, the spherical coordinate φ\varphi runs from 00 (on the zz–axis) to π/2\pi/2 (on the xyxy–plane, z=0z=0) and the spherical coordinate ρ\rho runs from 22, on x2+y2+z2=4x^2+y^2+z^2=4, to 33, on x2+y2+z2=9x^2+y^2+z^2=9. So, in spherical coordinates,

H={ (ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ) ∣ 2≤ρ≤3, 0≤φ≤π/2, 0≤θ≤2π }\begin{align*} H=\Set{ (\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\,\rho\cos\varphi)} {2\le \rho\le 3,\ 0\le\varphi\le\pi/2,\ 0\le\theta\le 2\pi} \end{align*}

(a) In spherical coordinates dV=ρ2sin⁡φ dρ dφ dθ\dee{V}=\rho^2\sin\varphi\ \dee{\rho}\,\dee{\varphi}\,\dee{\theta}, so that, as the density is the constant DD,

Mass(H)=∫23dρ∫02πdθ∫0π/2dφ D ρ2sin⁡φ=D [∫23dρ ρ2] [∫02πdθ] [∫0π/2dφ sin⁡φ]=D[333−233] [2π] [cos⁡0−cos⁡(π/2)]=383πD\begin{align*} \text{Mass}(H) &=\int_2^3\dee{\rho} \int_0^{2\pi}\dee{\theta}\int_0^{\pi/2}\dee{\varphi}\ D\ \rho^2\sin\varphi \\ &=D\ \left[\int_2^3\dee{\rho}\ \rho^2\right]\ \left[\int_0^{2\pi}\dee{\theta}\right]\ \left[\int_0^{\pi/2}\dee{\varphi}\ \sin\varphi\right] \\ &= D\left[\frac{3^3}{3}-\frac{2^3}{3}\right]\ \big[2\pi\big]\ \big[\cos 0 -\cos(\pi/2)\big] \\ &=\frac{38}{3}\pi D \end{align*}

We could have gotten the same result by expressing the mass as

  • one half, times

  • the density DD, times

  • the difference between the volume of a sphere of radius 33 and a sphere of radius 22.

That is

Mass(H)=12D[43π33−43π23]=383πD\begin{align*} \text{Mass}(H) =\frac{1}{2}D\left[\frac{4}{3}\pi 3^3-\frac{4}{3}\pi 2^3\right] =\frac{38}{3}\pi D \end{align*}

(b) By definition, the centre of mass is (xˉ,yˉ,zˉ)(\bar x,\bar y,\bar z) where xˉ\bar x, yˉ\bar y and zˉ\bar z are the weighted averages of xx, yy and zz, respectively, over HH. That is

xˉ=∭Hx D dV∭HD dVyˉ=∭Hy D dV∭HD dVzˉ=∭Hz D dV∭HD dV\begin{align*} \bar x =\frac{\tripInt_H x\,D\,\dee{V}}{\tripInt_H D\,\dee{V}}\qquad \bar y =\frac{\tripInt_H y\,D\,\dee{V}}{\tripInt_H D\,\dee{V}}\qquad \bar z =\frac{\tripInt_H z\,D\,\dee{V}}{\tripInt_H D\,\dee{V}} \end{align*}

As HH is invariant under reflection in the yzyz–plane (i.e. under x→−xx\rightarrow-x) we have xˉ=0\bar x=0. As HH is also invariant under reflection in the xzxz–plane (i.e. under y→−yy\rightarrow-y) we have yˉ=0\bar y=0. So we just have to find zˉ\bar z. We have already found the denominator in part (a), so we just have evaluate the numerator

∭Hz D dV=∫23dρ∫02πdθ∫0π/2dφ D ρ2sin⁡φ ρcos⁡φ⏞z=D [∫23dρ ρ3] [∫02πdθ] [∫0π/2dφ sin⁡φ cos⁡φ]=D[344−244] [2π] [12sin⁡2π2−12sin⁡20]=81−164πD=654πD\begin{align*} \tripInt_H z\,D\,\dee{V} &=\int_2^3\dee{\rho} \int_0^{2\pi}\dee{\theta}\int_0^{\pi/2}\dee{\varphi}\ D\ \rho^2\sin\varphi\ \overbrace{\rho\cos\varphi}^{z} \\ &=D\ \left[\int_2^3\dee{\rho}\ \rho^3\right]\ \left[\int_0^{2\pi}\dee{\theta}\right]\ \left[\int_0^{\pi/2}\dee{\varphi}\ \sin\varphi\ \cos\varphi\right] \\ &= D\left[\frac{3^4}{4}-\frac{2^4}{4}\right]\ \big[2\pi\big]\ \left[\frac{1}{2}\sin^2\frac{\pi}{2} -\frac{1}{2}\sin^2 0\right] \\ &=\frac{81-16}{4}\pi D =\frac{65}{4}\pi D \end{align*}

All together

xˉ=yˉ=0zˉ=654πD383πD=195152≈1.28\begin{align*} \bar x = \bar y=0\qquad \bar z =\frac{\frac{65}{4}\pi D}{\frac{38}{3}\pi D} =\frac{195}{152} \approx 1.28 \end{align*}
Q11Stage 2Past exam · M200 2008D

Let

I=∭Txz dV\begin{equation*} I = \tripInt_T xz\ \dee{V} \end{equation*}

where TT is the eighth of the sphere x2+y2+z2≤1x^2 + y^2 + z^2 \le 1 with x,y,z≥0x,y,z \ge 0.

  1. Sketch the volume TT.

  2. Express II as a triple integral in spherical coordinates.

  3. Evaluate II by any method.

Answer

(a)

Figure from prob_s3.7, line 884

Figure from prob_s3.7, line 884

(b) I=∫0π/2dφ∫0π/2dθ∫01dρ ρ4sin⁡2φcos⁡φ cos⁡θI=\int_0^{\pi/2}\dee{\varphi}\int_0^{\pi/2}\dee{\theta}\int_0^1\dee{\rho}\ \rho^4\sin^2\varphi\cos\varphi\ \cos\theta

(c) 115\frac{1}{15}

Full solution

(a) Here is a sketch

Figure from prob_s3.7, line 884

Figure from prob_s3.7, line 884

(b) On TT,

  • the spherical coordinate φ\varphi runs from 00 (the positive zz–xis) to π ⁣/2\nicefrac{\pi}{2} (the xyxy–plane), and

  • for each fixed φ\varphi in that range, θ\theta runs from 00 to π ⁣/2\nicefrac{\pi}{2}, and

  • for each fixed φ\varphi and θ\theta, the spherical coordinate ρ\rho runs from 00 to 11.

  • In spherical coordinates dV=ρ2 sin⁡φ dρ dθ dφ\dee{V}=\rho^2\,\sin\varphi\,\dee{\rho}\,\dee{\theta}\, \dee{\varphi} and

    xz=(ρsin⁡φcos⁡θ)(ρcos⁡φ)=ρ2sin⁡φ cos⁡φ cos⁡θ\begin{equation*} xz=\big(\rho\sin\varphi\cos\theta\big)\big(\rho\cos\varphi\big) =\rho^2 \sin\varphi\ \cos\varphi\ \cos\theta \end{equation*}

So

I=∫0π/2dφ∫0π/2dθ∫01dρ ρ4sin⁡2φcos⁡φ cos⁡θ\begin{align*} I=\int_0^{\pi/2}\dee{\varphi}\int_0^{\pi/2}\dee{\theta}\int_0^1\dee{\rho}\ \rho^4\sin^2\varphi\cos\varphi\ \cos\theta \end{align*}

(c) In spherical coordinates,

I=[∫0π/2dφ sin⁡2φcos⁡φ][∫0π/2dθ cos⁡θ][∫01dρ ρ4]=[sin⁡3φ3]0π/2[sin⁡θ]0π/2[ρ55]01=115\begin{align*} I&= \left[\int_0^{\pi/2}\dee{\varphi}\ \sin^2\varphi\cos\varphi\right] \left[\int_0^{\pi/2}\dee{\theta}\ \cos\theta\right] \left[\int_0^1\dee{\rho}\ \rho^4\right] \\ &=\left[\frac{\sin^3\varphi}{3}\right]_0^{\pi/2} \left[\sin\theta\right]_0^{\pi/2} \left[\frac{\rho^5}{5}\right]_0^1 \\ &=\frac{1}{15} \end{align*}
Q12Stage 2Past exam · M200 2010D

Evaluate W=∭Qxz dVW = \tripInt_Q xz\ \dee{V}, where QQ is an eighth of the sphere x2+y2+z2≤9x^2 + y^2 + z^2 \le 9 with xx, yy, z≥0z \ge 0.

Answer

815\frac{81}{5}

Full solution

We'll use spherical coordinates. On QQ,

  • the spherical coordinate φ\varphi runs from 00 (the positive zz–axis) to π2\frac{\pi}{2} (the xyxy–plane),

  • the spherical coordinate θ\theta runs from 00 (the half of the xzxz–plane with x≥0x\ge 0) to π2\frac{\pi}{2} (the half of the yzyz–plane with y≥0y\ge 0) and

  • the spherical coordinate ρ\rho runs from 00 to 33.

As dV=ρ2sin⁡φ dρ dθ dφ\dee{V}=\rho^2\sin\varphi\,\dee{\rho}\,\dee{\theta}\,\dee{\varphi},

W=∭Qxz dV=∫03dρ∫0π/2dθ∫0π/2dφ ρ2sin⁡φ ρsin⁡φcos⁡θ⏞xρcos⁡φ⏞z=∫03dρ∫0π/2dθ ρ4cos⁡θ[sin⁡3φ3]φ=0φ=π/2=13∫03dρ ρ4[sin⁡θ]0π/2=3515=815\begin{align*} W&= \tripInt_Q xz\ \dee{V} =\int_0^3 \dee{\rho} \int_0^{\pi/2}\dee{\theta}\int_0^{\pi/2}\dee{\varphi} \ \rho^2\sin\varphi\ \overbrace{\rho\sin\varphi\cos\theta}^{x} \overbrace{\rho\cos\varphi}^{z} \\ &=\int_0^3 \dee{\rho} \int_0^{\pi/2}\dee{\theta} \ \rho^4\cos\theta\left[\frac{\sin^3\varphi}{3}\right] _{\varphi=0}^{\varphi=\pi/2} \\ &=\frac{1}{3}\int_0^3 \dee{\rho}\ \rho^4\Big[\sin\theta\Big]_0^{\pi/2} \\ &=\frac{3^5}{15}=\frac{81}{5} \end{align*}
Q13Stage 2Past exam · M200 2012A

Evaluate ∭R3[1+(x2+y2+z2)3]−1 dV\tripInt_{\bbbr^3} {\big[1+{(x^2+y^2+z^2)}^3\big]}^{-1}\ \dee{V}.

Answer

2π23\frac{2\pi^2}{3}

Full solution

Let's use spherical coordinates. This is an improper integral. So, to be picky, we'll take the limit as R→∞R\rightarrow\infty of the integral over 0≤ρ≤R0\le\rho\le R.

∭R3[1+(x2+y2+z2)3]−1 dV=lim⁡R→∞∫0Rdρ∫02πdθ∫0πdφ ρ2sin⁡φ 11+ρ6=lim⁡R→∞∫0Rdρ∫02πdθ ρ21+ρ6[−cos⁡φ]0π=4πlim⁡R→∞∫0Rdρ ρ21+ρ6=4π3lim⁡R→∞∫0R3du 11+u2with u=ρ3, du=3ρ2 dρ=4π3lim⁡R→∞[arctan⁡u]0R3=2π23since lim⁡R→∞arctan⁡R3=π2\begin{align*} \tripInt_{\bbbr^3} {\big[1+{(x^2+y^2+z^2)}^3\big]}^{-1}\ \dee{V} &=\lim_{R\rightarrow\infty} \int_0^R \dee{\rho} \int_0^{2\pi}\dee{\theta} \int_0^\pi\dee{\varphi}\ \rho^2\sin\varphi\ \frac{1}{1+\rho^6} \\ &=\lim_{R\rightarrow\infty} \int_0^R \dee{\rho} \int_0^{2\pi}\dee{\theta} \ \frac{\rho^2}{1+\rho^6}\Big[-\cos\varphi\Big]_0^\pi \\ &=4\pi \lim_{R\rightarrow\infty} \int_0^R \dee{\rho} \ \frac{\rho^2}{1+\rho^6} \\ &=\frac{4\pi}{3} \lim_{R\rightarrow\infty} \int_0^{R^3} \dee{u} \ \frac{1}{1+u^2}\qquad \text{with }u=\rho^3,\ \dee{u}=3\rho^2\,\dee{\rho} \\ &=\frac{4\pi}{3}\lim_{R\rightarrow\infty}\Big[\arctan u\Big]_0^{R^3} \\ &=\frac{2\pi^2}{3}\qquad\text{since } \lim_{R\rightarrow\infty}\arctan R^3=\frac{\pi}{2} \end{align*}
Q14Stage 2Past exam · M200 2012D

Evaluate

∫−11∫−1−x21−x2∫1−1−x2−y21+1−x2−y2(x2+y2+z2)5/2 dz dy dx\begin{equation*} \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{1-\sqrt{1-x^2-y^2}}^{1+\sqrt{1-x^2-y^2}} (x^2+y^2+z^2)^{5/2} \ \dee{z}\,\dee{y}\,\dee{x} \end{equation*}

by changing to spherical coordinates.

Answer

64π9\frac{64\pi}{9}

Full solution

On the domain of integration

  • xx runs from −1-1 to 11.

  • For each fixed xx in that range, yy runs from −1−x2-\sqrt{1-x^2} to 1−x2\sqrt{1-x^2}. In inequalities, that is −1−x2≤y≤1−x2-\sqrt{1-x^2}\le y\le \sqrt{1-x^2}, which is equivalent to x2+y2≤1x^2+y^2\le 1.

  • For each fixed (x,y)(x,y) obeying x2+y2≤1x^2+y^2\le 1, zz runs from 1−1−x2−y21-\sqrt{1-x^2-y^2} to 1+1−x2−y21+\sqrt{1-x^2-y^2}. In inequalities, that is 1−1−x2−y2≤z≤1+1−x2−y21-\sqrt{1-x^2-y^2}\le z\le 1+\sqrt{1-x^2-y^2}, which is equivalent to x2+y2+(z−1)2≤1x^2+y^2+(z-1)^2\le 1.

So the domain of integration is

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

In spherical coordinates, the condition x2+y2+(z−1)2≤1x^2+y^2+(z-1)^2\le 1 is

(ρsin⁡φcos⁡θ)2+(ρsin⁡φsin⁡θ)2+(ρcos⁡φ−1)2≤1  ⟺  ρ2sin⁡2φ+(ρcos⁡φ−1)2≤1  ⟺  ρ2sin⁡2φ+ρ2cos⁡2φ−2ρcos⁡φ+1≤1  ⟺  ρ2≤2ρcos⁡φ  ⟺  ρ≤2cos⁡φ\begin{align*} &(\rho\sin\varphi\cos\theta)^2 +(\rho\sin\varphi\sin\theta)^2 +(\rho\cos\varphi-1)^2\le 1 \\ &\iff \rho^2\sin^2\varphi + (\rho\cos\varphi-1)^2\le 1 \\ &\iff \rho^2\sin^2\varphi + \rho^2\cos^2\varphi -2 \rho\cos\varphi +1 \le 1 \\ &\iff \rho^2\le 2\rho\cos\varphi \\ &\iff \rho\le 2\cos\varphi \end{align*}

Note that VV is contained in the upper half, z≥0z\ge 0, of R3\bbbr^3 and that the xyxy–plane in tangent to VV. So as (x,y,z)(x,y,z) runs over VV, the spherical coordinate φ\varphi runs from 00 (the positive zz–axis) to π ⁣/2\nicefrac{\pi}{2} (the xyxy–plane). Here is a sketch of the side view of VV.

Figure from prob_s3.7, line 1038

Figure from prob_s3.7, line 1038

As dV=ρ2sin⁡φ dρ dφ dθ\dee{V} = \rho^2\sin\varphi\ \dee{\rho}\,\dee{\varphi}\,\dee{\theta} and (x2+y2+z2)5/2=ρ5\big(x^2+y^2+z^2\big)^{5/2}=\rho^5, the integral is

∫−11∫−1−x21−x2∫1−1−x2−y21+1−x2−y2(x2+y2+z2)5/2 dz dy dx=∫02πdθ∫0π/2dφ∫02cos⁡φdρ ρ2sin⁡φ ρ5=∫02πdθ∫0π/2dφ 28cos⁡8φ8sin⁡φ=32∫02πdθ [−cos⁡9φ9]0π/2=329(2π)=64π9\begin{align*} \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{1-\sqrt{1-x^2-y^2}}^{1+\sqrt{1-x^2-y^2}} (x^2+y^2+z^2)^{5/2} \ \dee{z}\,\dee{y}\,\dee{x} &=\int_0^{2\pi}\dee{\theta}\int_0^{\pi/2}\dee{\varphi} \int_0^{2\cos\varphi}\dee{\rho}\ \rho^2\sin\varphi \ \rho^5 \\ &=\int_0^{2\pi}\dee{\theta}\int_0^{\pi/2}\dee{\varphi} \ \frac{2^8\cos^8\varphi}{8}\sin\varphi \\ &=32\int_0^{2\pi}\dee{\theta}\ \left[-\frac{\cos^9\varphi}{9}\right]_0^{\pi/2} \\ &=\frac{32}{9}(2\pi) =\frac{64\pi}{9} \end{align*}
Q15Stage 2

Evaluate the volume of a circular cylinder of radius aa and height hh by means of an integral in spherical coordinates.

Answer

πa2h\pi a^2h

Full solution

The top of the cylinder has equation z=hz=h, i.e. ρcos⁡φ=h\rho\cos\varphi=h. The side of the cylinder has equation x2+y2=a2x^2+y^2=a^2, i.e. ρsin⁡φ=a\rho\sin\varphi=a. The bottom of the cylinder has equation z=0z=0, i.e. φ=π2\varphi=\frac{\pi}{2}.

Figure from prob_s3.7, line 1113

Figure from prob_s3.7, line 1113

Figure from prob_s3.7, line 1113

Figure from prob_s3.7, line 1113

Figure from prob_s3.7, line 1113

Figure from prob_s3.7, line 1113

For each fixed φ\varphi, θ\theta runs from 00 to 2π2\pi and ρ\rho runs from 00 to either hcos⁡φ\frac{h}{\cos\varphi} (at the top of the can, if φ<tan⁡−1ah\varphi<\tan^{-1}\frac{a}{h}) or asin⁡φ\frac{a}{\sin\varphi} (at the side of the can, if φ>tan⁡−1ah\varphi>\tan^{-1}\frac{a}{h}). So the

Volume=∫0tan⁡−1ahdφ∫02πdθ∫0h/cos⁡φdρ ρ2sin⁡φ+∫tan⁡−1ahπ2dφ∫02πdθ∫0a/sin⁡φdρ ρ2sin⁡φ=2π∫0tan⁡−1ahdφ h3sin⁡φ3cos⁡3φ+2π∫tan⁡−1ahπ2dφ a3sin⁡φ3sin⁡3φ=2π{∫0ahdt h33t−∫ha0ds a33} where t=tan⁡φ, dt=sec⁡2φ dφ, s=cot⁡φ, ds=−csc⁡2φ dφ=2π{h33 12(ah)2+a33 ha}=2π{ah26+a2h3}=πa2h\begin{align*} \text{Volume} &= \int_{0}^{\tan^{-1}{a\over h}} \dee{\varphi} \int_{0}^{2\pi} \dee{\theta} \int_0^{h/\cos\varphi}\dee{\rho}\ \rho^2\sin\varphi +\int_{\tan^{-1}{a\over h}}^{{\pi\over 2}} \dee{\varphi} \int_{0}^{2\pi} \dee{\theta} \int_0^{a/\sin\varphi}\dee{\rho}\ \rho^2\sin\varphi \\ &= 2\pi\int_{0}^{\tan^{-1}{a\over h}} \dee{\varphi} \ \frac{h^3\sin\varphi}{3\cos^3\varphi} +2\pi\int_{\tan^{-1}{a\over h}}^{{\pi\over 2}} \dee{\varphi} \ \frac{a^3\sin\varphi}{3\sin^3\varphi} \\ &=2\pi\Big\{\int_0^{{a\over h}}\dee{t}\ \frac{h^3}{3}t -\int^0_{{h\over a}}\dee{s}\ \frac{a^3}{3}\Big\} \\ &\hskip1in\hbox{ where $t=\tan\varphi$, $\dee{t}=\sec^2\varphi\,\dee{\varphi}$, $s=\cot\varphi$, $\dee{s}=-\csc^2\varphi\,\dee{\varphi}$} \\ &=2\pi\left\{\frac{h^3}{3}\,\frac{1}{2}\left(\frac{a}{h}\right)^2 +\frac{a^3}{3}\,\frac{h}{a}\right\} =2\pi\left\{\frac{ah^2}{6}+\frac{a^2h}{3}\right\} =\pi a^2h \end{align*}
Q16Stage 2Past exam · M200 2003A

Let BB denote the region inside the sphere x2+y2+z2=4x^2+y^2+z^2=4 and above the cone x2+y2=z2x^2+y^2=z^2. Compute the moment of inertia

∭Bz2 dV\begin{equation*} \tripInt_B z^2\,\dee{V} \end{equation*}
Answer

6415π(1−122)≈8.665\frac{64}{15}\pi\big(1-\frac{1}{2\sqrt{2}}\big)\approx 8.665

Full solution

In spherical coordinates,

x=ρsin⁡φcos⁡θy=ρsin⁡φsin⁡θz=ρcos⁡φ\begin{equation*} x=\rho\sin\varphi\cos\theta\qquad y=\rho\sin\varphi\sin\theta\qquad z=\rho\cos\varphi \end{equation*}

so that the sphere x2+y2+z2=4x^2+y^2+z^2=4 is ρ2=4\rho^2=4 or ρ=2\rho=2 and the cone x2+y2=z2x^2+y^2=z^2 is ρ2sin⁡2φ=ρ2cos⁡2φ\rho^2\sin^2\varphi=\rho^2\cos^2\varphi or tan⁡φ=±1\tan\varphi=\pm1 or φ=π4, 3π4\varphi=\frac{\pi}{4},\ \frac{3\pi}{4}. So

moment=∫02dρ∫0π/4dφ∫02πdθ ρ2sin⁡φ (ρcos⁡φ)2=2π∫02dρ ρ4∫0π/4dφ sin⁡φcos⁡2φ=2π[ρ55]02[−13cos⁡3φ]0π/4=6415π(1−122)≈8.665\begin{align*} \text{moment} &=\int_0^2\dee{\rho}\int_0^{\pi/4}\dee{\varphi}\int_0^{2\pi}\dee{\theta}\ \rho^2\sin\varphi \ (\rho\cos\varphi)^2 =2\pi\int_0^2 \dee{\rho}\ \rho^4\int_0^{\pi/4}\dee{\varphi}\ \sin\varphi\cos^2\varphi\\ &=2\pi\left[\frac{\rho^5}{5}\right]_0^2 \left[-\frac{1}{3}\cos^3\varphi\right]_0^{\pi/4} =\frac{64}{15}\pi\left(1-\frac{1}{2\sqrt{2}}\right)\approx 8.665 \end{align*}
Q17Stage 2Past exam · M200 2002A

  1. Evaluate ∭Ωz dV\dst\tripInt_\Om z\,\dee{V} where Ω\Om is the three dimensional region in the first octant x≥0x\ge 0, y≥0y\ge 0, z≥0z\ge 0, occupying the inside of the sphere x2+y2+z2=1x^2+y^2+z^2=1.

  2. Use the result in part (a) to quickly determine the centroid of a hemispherical ball given by z≥0z\ge 0, x2+y2+z2≤1x^2+y^2+z^2\le 1.

Answer

(a) π16\frac{\pi}{16} (b) The centroid is (xˉ,yˉ,zˉ)(\bar x,\bar y,\bar z) with xˉ=yˉ=0\bar x=\bar y=0 and zˉ=38\bar z=\frac{3}{8}.

Full solution

(a) In spherical coordinates,

x=ρsin⁡ϕcos⁡θy=ρsin⁡ϕsin⁡θz=ρcos⁡ϕ\begin{align*} x=\rho\sin\phi\cos\theta\qquad y=\rho\sin\phi\sin\theta\qquad z=\rho\cos\phi \end{align*}

so that

  • the sphere x2+y2+z2=1x^2+y^2+z^2=1 is ρ=1\rho=1,

  • the xyxy-plane, z=0z=0, is ϕ=π2\phi=\frac{\pi}{2},

  • the positive half of the xzxz-plane, y=0y=0, x>0x>0, is θ=0\theta=0 and

  • the positive half of the yzyz-plane, x=0x=0, y>0y>0, is θ=π2\theta=\frac{\pi}{2}.

So

∭Ωz dV=∫01dρ∫0π/2dϕ∫0π/2dθ ρ2sin⁡ϕ(ρcos⁡ϕ)⏞z=π2∫01dρ∫0π/2dϕ ρ3sin⁡ϕcos⁡ϕ=π2∫01dρ ρ3 12sin⁡2ϕ∣0π/2=π4∫01dρ ρ3=π16\begin{align*} \tripInt_\Om z\,\dee{V}&= \int_0^1 \dee{\rho}\int_0^{\pi/2}d\phi\int _0^{\pi/2}\dee{\theta}\ \rho^2\sin\phi\overbrace{(\rho\cos\phi)}^{z} \\ &=\frac{\pi}{2}\int_0^1 \dee{\rho}\int_0^{\pi/2}d\phi\ \rho^3\sin\phi \cos\phi \\ &=\frac{\pi}{2}\int_0^1 \dee{\rho}\ \rho^3\ \frac{1}{2}\sin^2\phi \Big|_0^{\pi/2} =\frac{\pi}{4}\int_0^1 \dee{\rho}\ \rho^3 =\frac{\pi}{16} \end{align*}

(b) The hemispherical ball given by z≥0z\ge 0, x2+y2+z2≤1x^2+y^2+z^2\le 1 (call it HH) has centroid (xˉ,yˉ,zˉ)(\bar x,\bar y,\bar z) with xˉ=yˉ=0\bar x=\bar y=0 (by symmetry) and

zˉ=∭Hz dV∭H dV=4∭Ωz dV12×43π=π42π3=38\begin{align*} \bar z=\frac{\tripInt_H z\,\dee{V}}{\tripInt_H\,\dee{V}} =\frac{4\tripInt_\Om z\,\dee{V}}{\half\times\frac{4}{3}\pi} =\frac{\frac{\pi}{4}}{\frac{2\pi}{3}} =\frac{3}{8} \end{align*}
Q18Stage 2Past exam · M200 2000D

Consider the top half of a ball of radius 2 centred at the origin. Suppose that the ball has variable density equal to 9z9z units of mass per unit volume.

  1. Set up a triple integral giving the mass of this half–ball.

  2. Find out what fraction of that mass lies inside the cone

    z=x2+y2\begin{equation*} z=\sqrt{x^2+y^2} \end{equation*}
Answer

(a) 9∫02dρ∫0π/2dϕ∫02πdθ ρ3 sin⁡ϕ cos⁡ϕ\dst 9\int_0^2d\rho\int_0^{\pi/2}d\phi\int_0^{2\pi}d\theta\ \rho^3\,\sin\phi\,\cos\phi (b) 12\frac{1}{2}

Full solution

(a) In spherical coordinates,

x=ρsin⁡ϕcos⁡θy=ρsin⁡ϕsin⁡θz=ρcos⁡ϕ\begin{equation*} x=\rho\sin\phi\cos\theta\qquad y=\rho\sin\phi\sin\theta\qquad z=\rho\cos\phi \end{equation*}

the sphere x2+y2+z2=4x^2+y^2+z^2=4 is ρ2=4\rho^2=4 or ρ=2\rho=2 and the xyxy-plane is ϕ=π2\phi=\frac{\pi}{2}. So

mass=∫02dρ∫0π/2dϕ∫02πdθ ρ2sin⁡ϕ (9ρcos⁡ϕ)⏞density\begin{equation*} \text{mass} =\int_0^2d\rho\int_0^{\pi/2}d\phi\int_0^{2\pi}d\theta\ \rho^2\sin\phi \ \overbrace{(9 \rho\cos\phi)}^{\rm density} \end{equation*}

(b) The mass of the half ball is

9∫02dρ∫0π/2dϕ∫02πdθ ρ3sin⁡ϕcos⁡ϕ=9[∫02dρ ρ3][∫0π/2dϕ sin⁡ϕcos⁡ϕ][∫02πdθ]\begin{equation*} 9 \int_0^2d\rho\int_0^{\pi/2}d\phi\int_0^{2\pi}d\theta\ \rho^3\sin\phi \cos\phi =9 \bigg[\int_0^2d\rho\ \rho^3\bigg] \bigg[\int_0^{\pi/2}d\phi\ \sin\phi \cos\phi\bigg] \bigg[\int_0^{2\pi}d\theta\bigg] \end{equation*}

In spherical coordinates, the cone x2+y2=z2x^2+y^2=z^2 is ρ2sin⁡2ϕ=ρ2cos⁡2ϕ\rho^2\sin^2\phi=\rho^2\cos^2\phi or tan⁡ϕ=±1\tan\phi=\pm1 or ϕ=π4, 3π4\phi=\frac{\pi}{4},\ \frac{3\pi}{4}. So the mass of the part that is inside the cone is

9∫02dρ∫0π/4dϕ∫02πdθ ρ3sin⁡ϕcos⁡ϕ=9[∫02dρ ρ3][∫0π/4dϕ sin⁡ϕcos⁡ϕ][∫02πdθ]\begin{equation*} 9 \int_0^2d\rho\int_0^{\pi/4}d\phi\int_0^{2\pi}d\theta\ \rho^3\sin\phi \cos\phi =9 \bigg[\int_0^2d\rho\ \rho^3\bigg] \bigg[\int_0^{\pi/4}d\phi\ \sin\phi \cos\phi\bigg] \bigg[\int_0^{2\pi}d\theta\bigg] \end{equation*}

The fraction inside the cone is

∫0π/4dϕ sin⁡ϕcos⁡ϕ∫0π/2dϕ sin⁡ϕcos⁡ϕ=12sin⁡2ϕ∣0π/412sin⁡2ϕ∣0π/2=12\begin{equation*} \frac{\int_0^{\pi/4}d\phi\ \sin\phi \cos\phi}{\int_0^{\pi/2}d\phi\ \sin\phi \cos\phi} =\frac{\half\sin^2\phi\big|_0^{\pi/4}}{\half\sin^2\phi\big|_0^{\pi/2}} =\frac{1}{2} \end{equation*}

3Stage 3Application

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

Q19Stage 3Past exam · M253 2010D

Find the limit or show that it does not exist

lim⁡(x,y,z)→(0,0,0)xy+yz2+xz2x2+y2+z4\begin{equation*} \lim_{(x,y,z)\to(0,0,0)}\frac{xy+yz^2+xz^2}{x^2+y^2+z^4} \end{equation*}
Hint

Switch to spherical coordinates.

Answer

It does not exist.

Full solution

In spherical coordinates, x=ρsin⁡φcos⁡θx=\rho\sin\varphi\cos\theta, y=ρsin⁡φsin⁡θy=\rho\sin\varphi\sin\theta, z=ρcos⁡φz=\rho\cos\varphi so that

ρ2sin⁡2φcos⁡θsin⁡θ+ρ3sin⁡φsin⁡θ cos⁡2φ+ρ3sin⁡φ cos⁡2φρ2sin⁡2φ+ρ4cos⁡4φ=sin⁡2φcos⁡θsin⁡θ+ρsin⁡φsin⁡θ cos⁡2φ+ρsin⁡φcos⁡θ cos⁡2φsin⁡2φ+ρ2cos⁡4φ\begin{align*} &\frac{\rho^2\sin^2\varphi\cos\theta\sin\theta +\rho^3\sin\varphi\sin\theta\,\cos^2\varphi +\rho^3\sin\varphi\,\cos^2\varphi} {\rho^2\sin^2\varphi+\rho^4\cos^4\varphi} \\ &\hskip1in =\frac{\sin^2\varphi\cos\theta\sin\theta +\rho\sin\varphi\sin\theta\,\cos^2\varphi +\rho\sin\varphi\cos\theta\,\cos^2\varphi} {\sin^2\varphi+\rho^2\cos^4\varphi} \end{align*}

As (x,y,z)→(0,0,0)(x,y,z)\to (0,0,0), the radius ρ→0\rho\to 0 and the second and third terms in the numerator and the second term in the denominator converge to 00. But that leaves

sin⁡2φcos⁡θsin⁡θsin⁡2φ=cos⁡θsin⁡θ\begin{equation*} \frac{\sin^2\varphi\cos\theta\sin\theta}{\sin^2\varphi} =\cos\theta\sin\theta \end{equation*}

which takes many different values. In particular, if we send (x,y,z)→(0,0,0)(x,y,z)\to (0,0,0) along either the xx– or yy–axis, that is with z=0z=0 and either x=0x=0 or y=0y=0, then

xy+yz2+xz2x2+y2+z4∣x=0 or y=0z=0=0\begin{equation*} \frac{xy+yz^2+xz^2}{x^2+y^2+z^4}\bigg|_{\Atop{x=0\text{ or }y=0}{z=0}}=0 \end{equation*}

converges to 00. But, if we send (x,y,z)→(0,0,0)(x,y,z)\to (0,0,0) along the line y=xy=x, z=0z=0

xy+yz2+xz2x2+y2+z4∣y=xz=0=x22x2=12\begin{equation*} \frac{xy+yz^2+xz^2}{x^2+y^2+z^4}\bigg|_{\Atop{y=x}{z=0}}=\frac{x^2}{2x^2} =\frac{1}{2} \end{equation*}

converges to 1/21/2. So xy+yz2+xz2x2+y2+z4\frac{xy+yz^2+xz^2}{x^2+y^2+z^4} does not approach a single value as (x,y,z)→(0,0,0)(x,y,z)\to(0,0,0) and the limit does not exist.

Q20Stage 3Past exam · M200 2007A

A certain solid VV is a right–circular cylinder. Its base is the disk of radius 22 centred at the origin in the xyxy–plane. It has height 22 and density x2+y2\sqrt{x^2 + y^2}.

A smaller solid UU is obtained by removing the inverted cone, whose base is the top surface of VV and whose vertex is the point (0,0,0)(0, 0, 0).

  1. Use cylindrical coordinates to set up an integral giving the mass of UU.

  2. Use spherical coordinates to set up an integral giving the mass of UU.

  3. Find that mass.

Answer

(a) Mass=∫02dz∫02πdθ∫z2dr r2\text{Mass} = \int_0^2\dee{z}\int_0^{2\pi}\dee{\theta}\int_z^2\dee{r}\ r^2

(b) Mass=∫π/4π/2dφ∫02πdθ∫02/sin⁡φdρ ρ3sin⁡2φ\text{Mass} = \int_{\pi/4}^{\pi/2}\dee{\varphi}\int_0^{2\pi}\dee{\theta} \int_0^{2/\sin\varphi}\dee{\rho}\ \rho^3\sin^2\varphi

(c) 8π8\pi

Full solution

The disk of radius 22 centred at the origin in the xyxy–plane is x2+y2≤4x^2+y^2\le 4. So

V={ (x,y,z) ∣ x2+y2≤4, 0≤z≤2 }\begin{align*} V&=\Set{(x,y,z)}{x^2+y^2\le 4,\ 0\le z\le 2} \end{align*}

The cone with vertex at the origin that contains the top edge, x+y2=4x+y^2=4, z=2z=2, of UU is x2+y2=z2x^2+y^2=z^2. So

U={ (x,y,z) ∣ x2+y2≤4, 0≤z≤2, x2+y2≥z2 }\begin{align*} U&=\Set{(x,y,z)}{x^2+y^2\le 4,\ 0\le z\le 2,\ x^2+y^2\ge z^2} \end{align*}

Here are sketches of the y=0y=0 cross–section of VV, on the left, and UU, on the right.

Figure from prob_s3.7, line 1405

Figure from prob_s3.7, line 1405

Figure from prob_s3.7, line 1405

Figure from prob_s3.7, line 1405

(a) In cylindrical coordinates, x2+y2≤4x^2+y^2\le 4 becomes r≤2r\le 2 and x2+y2≥z2x^2+y^2\ge z^2 is r≥∣z∣r\ge |z|, and the density is x2+y2=r\sqrt{x^2+y^2}=r. So

U={ (rcos⁡θ,rsin⁡θ,z) ∣ r≤2, 0≤z≤2, r≥z }\begin{align*} U&=\Set{(r\cos\theta,r\sin\theta,z)}{r\le 2,\ 0\le z\le 2,\ r\ge z} \end{align*}

Looking at the figure on the left below, we see that, on UU

  • zz runs from 00 to 22, and

  • for each zz is that range, rr runs from zz to 22 and θ\theta runs from 00 to 2π2\pi.

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

So

Mass=∫02dz∫02πdθ∫z2dr r r⏞density=∫02dz∫02πdθ∫z2dr r2\begin{align*} \text{Mass}&= \int_0^2\dee{z}\int_0^{2\pi}\dee{\theta}\int_z^2\dee{r}\ r\ \overbrace{r}^{\text{density}} = \int_0^2\dee{z}\int_0^{2\pi}\dee{\theta}\int_z^2\dee{r}\ r^2 \end{align*}

Figure from prob_s3.7, line 1405

Figure from prob_s3.7, line 1405

Figure from prob_s3.7, line 1405

Figure from prob_s3.7, line 1405

(b) Recall that in spherical coordinates,

x=ρsin⁡φcos⁡θy=ρsin⁡φsin⁡θz=ρcos⁡φx2+y2=ρ2sin⁡2φ\begin{align*} x&=\rho\sin\varphi\cos\theta \\ y&=\rho\sin\varphi\sin\theta \\ z&=\rho\cos\varphi \\ x^2+y^2 &=\rho^2\sin^2\varphi \end{align*}

so that x2+y2≤4x^2+y^2\le 4 becomes ρsin⁡φ≤2\rho\sin\varphi\le 2, and x2+y2≥z2x^2+y^2\ge z^2 becomes

ρsin⁡φ≥ρcos⁡φ  ⟺  tan⁡φ≥1  ⟺  φ≥π4\begin{align*} \rho\sin\varphi \ge \rho\cos\varphi \iff \tan\varphi\ge 1 \iff \varphi\ge\frac{\pi}{4} \end{align*}

and the density x2+y2=ρsin⁡φ\sqrt{x^2+y^2}=\rho\sin\varphi. So

U={ (ρsin⁡φcos⁡θ,ρsin⁡φsin⁡θ,ρcos⁡φ) ∣ π ⁣/4≤φ≤π ⁣/2, 0≤θ≤2π, ρsin⁡φ≤2 }\begin{align*} U&=\Set{(\rho\sin\varphi\cos\theta,\rho\sin\varphi\sin\theta,\rho\cos\varphi)} {\nicefrac{\pi}{4}\le\varphi\le\nicefrac{\pi}{2},\ 0\le \theta\le 2\pi,\ \rho\sin\varphi\le 2} \end{align*}

Looking at the figure on the right above, we see that, on UU

  • φ\varphi runs from π4\frac{\pi}{4} (on the cone) to π2\frac{\pi}{2} (on the xyxy–plane), and

  • for each φ\varphi is that range, ρ\rho runs from 00 to 2sin⁡φ\frac{2}{\sin\varphi} and θ\theta runs from 00 to 2π2\pi.

  • dV=ρ2sin⁡φ dρ dθ dφ\dee{V} =\rho^2\sin\varphi\,\dee{\rho}\,\dee{\theta}\,\dee{\varphi}

So

Mass=∫π/4π/2dφ∫02πdθ∫02/sin⁡φdρ ρ2sin⁡φ ρsin⁡φ⏞density=∫π/4π/2dφ∫02πdθ∫02/sin⁡φdρ ρ3sin⁡2φ\begin{align*} \text{Mass}&= \int_{\pi/4}^{\pi/2}\dee{\varphi}\int_0^{2\pi}\dee{\theta} \int_0^{2/\sin\varphi}\dee{\rho}\ \rho^2\sin\varphi\ \overbrace{\rho\sin\varphi}^{\text{density}} \\ &= \int_{\pi/4}^{\pi/2}\dee{\varphi}\int_0^{2\pi}\dee{\theta} \int_0^{2/\sin\varphi}\dee{\rho}\ \rho^3\sin^2\varphi \end{align*}

(c) We'll use the cylindrical form.

Mass=∫02dz∫02πdθ∫z2dr r2=2π∫02dz 8−z33=2π3 [16−244]=8π\begin{align*} \text{Mass} &= \int_0^2\dee{z}\int_0^{2\pi}\dee{\theta}\int_z^2\dee{r}\ r^2 \\ &=2\pi \int_0^2\dee{z}\ \frac{8-z^3}{3} \\ &=\frac{2\pi}{3}\ \left[16-\frac{2^4}{4}\right] \\ &=8\pi \end{align*}
Q21Stage 3Past exam · M200 2009A

A solid is bounded below by the cone z ⁣= ⁣x2 ⁣+ ⁣y2z\!=\!\sqrt{x^2\!+\!y^2} and above
by the sphere x2+y2+z2=2x^2+y^2+z^2 = 2. It has density δ(x,y,z)=x2+y2\de(x,y,z) = x^2 + y^2.

  1. Express the mass MM of the solid as a triple integral, with limits, in cylindrical coordinates.

  2. Same as (a) but in spherical coordinates.

  3. Evaluate MM.

Answer

(a) ∫01dr∫02πdθ∫r2−r2dz r3\int_0^1\dee{r}\int_0^{2\pi}\dee{\theta} \int_r^{\sqrt{2-r^2}}\dee{z}\ r^3

(b) ∫02dρ∫02πdθ∫0π/4dφ ρ4sin⁡3φ\int_0^{\sqrt{2}}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/4}\dee{\varphi}\ \rho^4\sin^3\varphi

(c) π[16215−43]≈0.5503\pi \left[\frac{16\sqrt{2}}{15} - \frac{4}{3}\right] \approx 0.5503

Full solution

(a) Call the solid VV. In cylindrical coordinates

  • x2+y2+z2≤2x^2+y^2+z^2\le 2 is r2+z2≤2r^2+z^2\le 2 and

  • x2+y2≤z\sqrt{x^2+y^2}\le z is r≤zr\le z and

  • the density δ=r2\de=r^2, and

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

Observe that r2+z2=2r^2+z^2= 2 and r=zr= z intersect when 2r2=22r^2=2 so that r=z=1r=z=1. Here is a sketch of the y=0y=0 cross–section of EE.

Figure from prob_s3.7, line 1543

Figure from prob_s3.7, line 1543

So

V={ (rcos⁡θ , rsin⁡θ , z) ∣ 0≤r≤1, 0≤θ≤2π, r≤z≤2−r2 }\begin{equation*} V =\Set{(r\cos\theta\,,\,r\sin\theta\,,\,z)}{0\le r\le 1,\ 0\le\theta\le2\pi,\ r\le z\le \sqrt{2-r^2}} \end{equation*}

and

M=∭Vρ(x,y,z) dV=∫01dr∫02πdθ∫r2−r2dz r(r2)⏞δ=∫01dr∫02πdθ∫r2−r2dz r3\begin{align*} M &=\tripInt_V \rho(x,y,z)\ \dee{V} =\int_0^1\dee{r}\int_0^{2\pi}\dee{\theta} \int_r^{\sqrt{2-r^2}}\dee{z}\ r\overbrace{(r^2)}^{\de} =\int_0^1\dee{r}\int_0^{2\pi}\dee{\theta} \int_r^{\sqrt{2-r^2}}\dee{z}\ r^3 \end{align*}

(b)

In spherical coordinates

  • x2+y2+z2≤2x^2+y^2+z^2\le 2 is ρ≤2\rho\le\sqrt{2}, and

  • x2+y2≤z\sqrt{x^2+y^2}\le z is ρsin⁡φ≤ρcos⁡φ\rho\sin\varphi\le \rho\cos\varphi, or tan⁡φ≤1\tan\varphi\le 1 or φ≤π4\varphi\le\frac{\pi}{4}, and

  • the density x2+y2=ρ2sin⁡2φx^2+y^2=\rho^2\sin^2\varphi, and

  • dV\dee{V} is ρ2sin⁡φ dρ dθ dφ\rho^2\sin\varphi\,\dee{\rho}\,\dee{\theta}\,\dee{\varphi}

So

V={ (ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ) ∣ 0≤ρ≤2, 0≤θ≤2π, 0≤φ≤π ⁣/4 }\begin{equation*} V =\Set{(\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)}{0\le\rho\le \sqrt{2},\ 0\le\theta\le2\pi,\ 0\le \varphi\le \nicefrac{\pi}{4}} \end{equation*}

and, since the integrand x2+y2=ρ2sin⁡2φx^2+y^2=\rho^2\sin^2\varphi,

M=∭V(x2+y2) dV=∫02dρ∫02πdθ∫0π/4dφ ρ2sin⁡φ ρ2sin⁡2φ=∫02dρ∫02πdθ∫0π/4dφ ρ4sin⁡3φ\begin{align*} M &=\tripInt_V \big(x^2+y^2\big)\ \dee{V} =\int_0^{\sqrt{2}}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/4}\dee{\varphi}\ \rho^2\sin\varphi\ \rho^2\sin^2\varphi \\ &=\int_0^{\sqrt{2}}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/4}\dee{\varphi}\ \rho^4\sin^3\varphi \end{align*}

(c) We'll use the spherical coordinate form.

M=∫02dρ∫02πdθ∫0π/4dφ ρ4sin⁡3φ=∫02dρ∫02πdθ∫0π/4dφ ρ4sin⁡φ[1−cos⁡2φ]=2π∫02dρ ρ4[−cos⁡φ+cos⁡3φ3]0π/4=2π[23−562]∫02dρ ρ4=2π425[23−562]=π[16215−43]≈0.5503\begin{align*} M &=\int_0^{\sqrt{2}}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/4}\dee{\varphi}\ \rho^4\sin^3\varphi \\ &=\int_0^{\sqrt{2}}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/4}\dee{\varphi}\ \rho^4\sin\varphi \big[1-\cos^2\varphi\big] \\ &=2\pi \int_0^{\sqrt{2}}\dee{\rho}\ \rho^4 \left[-\cos\varphi +\frac{\cos^3\varphi}{3}\right]_0^{\pi/4} =2\pi \left[\frac{2}{3} - \frac{5}{6\sqrt{2}}\right] \int_0^{\sqrt{2}}\dee{\rho}\ \rho^4 \\ &=2\pi \frac{4\sqrt{2}}{5}\left[\frac{2}{3} - \frac{5}{6\sqrt{2}}\right] =\pi \left[\frac{16\sqrt{2}}{15} - \frac{4}{3}\right] \approx 0.5503 \end{align*}
Q22Stage 3Past exam · M200 2009D

Let

I=∭Exz dV\begin{align*} I = \tripInt_E xz\ \dee{V} \end{align*}

where EE is the eighth of the sphere x2+y2+z2≤1x^2+y^2+z^2\le 1 with x,y,z≥0x,y,z\ge 0.

  1. Express II as a triple integral in spherical coordinates.

  2. Express II as a triple integral in cylindrical coordinates.

  3. Evaluate II by any method.

Answer

(a) ∫0π/2dφ∫0π/2dθ∫01dρ ρ4sin⁡2φcos⁡φ cos⁡θ\int_0^{\pi/2}\dee{\varphi}\int_0^{\pi/2}\dee{\theta}\int_0^1\dee{\rho}\ \rho^4\sin^2\varphi\cos\varphi\ \cos\theta (b) ∫01dz∫0π/2dθ∫01−z2dr r2 z cos⁡θ\int_0^1\dee{z}\int_0^{\pi/2}\dee{\theta}\int_0^{\sqrt{1-z^2}}\dee{r} \ r^2\,z\,\cos\theta

(c) 115\frac{1}{15}

Full solution

(a) On EE,

  • the spherical coordinate φ\varphi runs from 00 (the positive zz–xis) to π ⁣/2\nicefrac{\pi}{2} (the xyxy–plane), and

  • for each fixed φ\varphi in that range, θ\theta runs from 00 to π ⁣/2\nicefrac{\pi}{2}, and

  • for each fixed φ\varphi and θ\theta, the spherical coordinate ρ\rho runs from 00 to 11.

  • In spherical coordinates dV=ρ2 sin⁡φ dρ dθ dφ\dee{V}=\rho^2\,\sin\varphi\,\dee{\rho}\,\dee{\theta}\, \dee{\varphi} and

    xz=(ρsin⁡φcos⁡θ)(ρcos⁡φ)=ρ2sin⁡φ cos⁡φ cos⁡θ\begin{equation*} xz=\big(\rho\sin\varphi\cos\theta\big)\big(\rho\cos\varphi\big) =\rho^2 \sin\varphi\ \cos\varphi\ \cos\theta \end{equation*}

So

I=∫0π/2dφ∫0π/2dθ∫01dρ ρ4sin⁡2φcos⁡φ cos⁡θ\begin{align*} I=\int_0^{\pi/2}\dee{\varphi}\int_0^{\pi/2}\dee{\theta}\int_0^1\dee{\rho}\ \rho^4\sin^2\varphi\cos\varphi\ \cos\theta \end{align*}

(b) In cylindrical coordinates, the condition x2+y2+z2≤1x^2+y^2+z^2\le 1 becomes r2+z2≤1r^2+z^2\le 1. So, on EE

  • the cylindrical coordinate zz runs from 00 (in the xyxy–plane) to 11 (at (0,0,1)(0,0,1)) and

  • for each fixed zz in that range, θ\theta runs from 00 to π/2\pi/2 and

  • for each such fixed zz and θ\theta, the cylindrical coordinate rr runs from 00 to 1−z2\sqrt{1-z^2} (recall that r2+z2≤1r^2+z^2\le 1).

  • In cylindrical coordinates dV=r dr dθ dz\dee{V}=r\,\dee{r}\,\dee{\theta}\, \dee{z} and

    xz=(rcos⁡θ)(z)=r z cos⁡θ\begin{equation*} xz=\big(r\cos\theta\big)\big(z\big) =r\,z\, \cos\theta \end{equation*}

    So

    I=∫01dz∫0π/2dθ∫01−z2dr r2 z cos⁡θ\begin{align*} I=\int_0^1\dee{z}\int_0^{\pi/2}\dee{\theta}\int_0^{\sqrt{1-z^2}}\dee{r} \ r^2\,z\,\cos\theta \end{align*}

    (c) Both spherical and cylindrical integrals are straight forward to evaluate. Here are both. First, in spherical coordinates,

    I=[∫0π/2dφ sin⁡2φcos⁡φ][∫0π/2dθ cos⁡θ][∫01dρ ρ4]=[sin⁡3φ3]0π/2[sin⁡θ]0π/2[ρ55]01=115\begin{align*} I&= \left[\int_0^{\pi/2}\dee{\varphi}\ \sin^2\varphi\cos\varphi\right] \left[\int_0^{\pi/2}\dee{\theta}\ \cos\theta\right] \left[\int_0^1\dee{\rho}\ \rho^4\right] \\ &=\left[\frac{\sin^3\varphi}{3}\right]_0^{\pi/2} \left[\sin\theta\right]_0^{\pi/2} \left[\frac{\rho^5}{5}\right]_0^1 \\ &=\frac{1}{15} \end{align*}

    Now in cylindrical coordinates

    I=∫01dz∫0π/2dθ∫01−z2dr r2 z cos⁡θ=13∫01dz∫0π/2dθ z(1−z2)3/2 cos⁡θ=13∫01dz z(1−z2)3/2=13[−12 (1−z2)5/25/2]01=115\begin{align*} I&=\int_0^1\dee{z}\int_0^{\pi/2}\dee{\theta}\int_0^{\sqrt{1-z^2}}\dee{r} \ r^2\,z\,\cos\theta \\ &=\frac{1}{3}\int_0^1\dee{z}\int_0^{\pi/2}\dee{\theta}\ z{(1-z^2\big)}^{3/2}\, \cos\theta \\ &=\frac{1}{3}\int_0^1\dee{z}\ z{(1-z^2\big)}^{3/2} \\ &=\frac{1}{3}\left[-\frac{1}{2}\ \frac{(1-z^2)^{5/2}}{5/2}\right]_0^1\\ &=\frac{1}{15} \end{align*}
Q23Stage 3Past exam · M200 2010A

Let

I=∭T(x2+y2) dV\begin{equation*} I = \tripInt_T (x^2+y^2)\ \dee{V} \end{equation*}

where TT is the solid region bounded below by the cone z=3x2+3y2z =\sqrt{3x^2+3y^2} and above by the sphere x2+y2+z2=9x^2 + y^2 + z^2 = 9.

  1. Express II as a triple integral in spherical coordinates.

  2. Express II as a triple integral in cylindrical coordinates.

  3. Evaluate II by any method.

Answer

(a) ∫03dρ∫02πdθ∫0π/6dφ ρ4sin⁡3φ\int_0^{3}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/6}\dee{\varphi}\ \rho^4\sin^3\varphi

(b) ∫03/2dr∫02πdθ∫3 r9−r2dz  r3\int_0^{3/2}\dee{r}\int_0^{2\pi}\dee{\theta} \int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}\dee{z}\ \ r^3

(c) 81π[45−9320]≈5.2481\pi \left[\frac{4}{5} - \frac{9\sqrt{3}}{20}\right] \approx 5.24

Full solution

(a) Recall that in spherical coordinates

x=ρ sin⁡φ cos⁡θy=ρ sin⁡φ sin⁡θz=ρ cos⁡φ\begin{align*} x=\rho\,\sin\varphi\,\cos\theta\qquad y=\rho\,\sin\varphi\,\sin\theta\qquad z=\rho\,\cos\varphi \end{align*}

so that

  • x2+y2+z2≤9x^2+y^2+z^2\le 9 is ρ≤3\rho\le 3, and

  • 3x2+3y2≤z\sqrt{3x^2+3y^2}\le z is 3ρsin⁡φ≤ρcos⁡φ\sqrt{3}\rho\sin\varphi\le \rho\cos\varphi, or tan⁡φ≤13\tan\varphi\le \frac{1}{\sqrt{3}} or φ≤π6\varphi\le\frac{\pi}{6}, and

  • the integrand x2+y2=ρ2sin⁡2φx^2+y^2=\rho^2\sin^2\varphi, and

  • dV\dee{V} is ρ2sin⁡φ dρ dθ dφ\rho^2\sin\varphi\,\dee{\rho}\,\dee{\theta}\,\dee{\varphi}

So

T={ (ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ) ∣ 0≤ρ≤3, 0≤θ≤2π, 0≤φ≤π ⁣/6 }\begin{equation*} T =\Set{(\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)}{0\le\rho\le 3,\ 0\le\theta\le2\pi,\ 0\le \varphi\le \nicefrac{\pi}{6}} \end{equation*}

and,

I=∭T(x2+y2) dV=∫03dρ∫02πdθ∫0π/6dφ ρ2sin⁡φ ρ2sin⁡2φ⏞x2+y2=∫03dρ∫02πdθ∫0π/6dφ ρ4sin⁡3φ\begin{align*} I &=\tripInt_T \big(x^2+y^2\big)\ \dee{V} =\int_0^{3}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/6}\dee{\varphi}\ \rho^2\sin\varphi\ \overbrace{\rho^2\sin^2\varphi}^{x^2+y^2} \\ &=\int_0^{3}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/6}\dee{\varphi}\ \rho^4\sin^3\varphi \end{align*}

(b) In cylindrical coordinates

  • x2+y2+z2≤9x^2+y^2+z^2\le 9 is r2+z2≤9r^2+z^2\le 9 and

  • 3x2+3y2≤z\sqrt{3 x^2+3 y^2}\le z is 3 r≤z\sqrt{3}\,r\le z and

  • the integand x2+y2=r2x^2+y^2=r^2, and

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

Observe that r2+z2=9r^2+z^2= 9 and 3 r=z\sqrt{3}\,r= z intersect when r2+3r2=9r^2+3r^2=9 so that r=32r=\frac{3}{2} and z=332z=\frac{3\sqrt{3}}{2}. Here is a sketch of the y=0y=0 cross–section of TT.

Figure from prob_s3.7, line 1777

Figure from prob_s3.7, line 1777

So

T={ (rcos⁡θ , rsin⁡θ , z) ∣ 0≤r≤32, 0≤θ≤2π, 3 r≤z≤9−r2 }\begin{equation*} T =\Set{(r\cos\theta\,,\,r\sin\theta\,,\,z)}{0\le r\le \tfrac{3}{2},\ 0\le\theta\le2\pi,\ \sqrt{3}\,r\le z\le \sqrt{9-r^2}} \end{equation*}

and

I=∭V(x2+y2) dV=∫03/2dr∫02πdθ∫3 r9−r2dz r(r2)⏞x2+y2=∫03/2dr∫02πdθ∫3 r9−r2dz  r3\begin{align*} I &=\tripInt_V (x^2+y^2)\ \dee{V} =\int_0^{3/2}\dee{r}\int_0^{2\pi}\dee{\theta} \int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}\dee{z}\ r\overbrace{(r^2)}^{x^2+y^2} =\int_0^{3/2}\dee{r}\int_0^{2\pi}\dee{\theta} \int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}\dee{z}\ \ r^3 \end{align*}

(c) We'll use the spherical coordinate form.

I=∫03dρ∫02πdθ∫0π/6dφ ρ4sin⁡3φ=∫03dρ∫02πdθ∫0π/6dφ ρ4sin⁡φ[1−cos⁡2φ]=2π∫03dρ ρ4[−cos⁡φ+cos⁡3φ3]0π/6=2π[−32+38+1−13]∫03dρ ρ4=2π355[23−338]=81⏞34π[45−9320]≈5.24\begin{align*} I &=\int_0^{3}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/6}\dee{\varphi}\ \rho^4\sin^3\varphi \\ &=\int_0^{3}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/6}\dee{\varphi}\ \rho^4\sin\varphi \big[1-\cos^2\varphi\big] \\ &=2\pi \int_0^3\dee{\rho}\ \rho^4 \left[-\cos\varphi +\frac{\cos^3\varphi}{3}\right]_0^{\pi/6} =2\pi \left[-\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{8}+1-\frac{1}{3}\right] \int_0^3\dee{\rho}\ \rho^4 \\ &=2\pi \frac{3^5}{5}\left[\frac{2}{3} - \frac{3\sqrt{3}}{8}\right] =\overbrace{81}^{3^4}\pi \left[\frac{4}{5} - \frac{9\sqrt{3}}{20}\right] \approx 5.24 \end{align*}
Q24Stage 3Past exam · M200 2011A

Let EE be the “ice cream cone” x2+y2+z2≤1x^2 + y^2 + z^2 \le 1, x2+y2≤z2x^2 + y^2 \le z^2 , z≥0z \ge 0. Consider

J=∭Ex2+y2+z2 dV\begin{equation*} J =\tripInt_E \sqrt{x^2+y^2+z^2}\ \dee{V} \end{equation*}

  1. Write JJ as an iterated integral, with limits, in cylindrical coordinates.

  2. Write JJ as an iterated integral, with limits, in spherical coordinates.

  3. Evaluate JJ.

Answer

(a) ∫01/2dr∫02πdθ∫r1−r2dz rr2+z2\int_0^{1/\sqrt{2}}\dee{r}\int_0^{2\pi}\dee{\theta} \int_r^{\sqrt{1-r^2}}\dee{z}\ r\sqrt{r^2+z^2}

(b) ∫01dρ∫02πdθ∫0π/4dφ ρ3sin⁡φ\int_0^{1}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/4}\dee{\varphi}\ \rho^3\sin\varphi

(c) π2 [1−12]\frac{\pi}{2}\ \left[1-\frac{1}{\sqrt{2}}\right]

Full solution

(a) In cylindrical coordinates

  • x2+y2+z2≤1x^2+y^2+z^2\le 1 is r2+z2≤1r^2+z^2\le 1 and

  • x2+y2≤z2x^2+y^2\le z^2 is r2≤z2r^2\le z^2 and

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

Observe that r2+z2=1r^2+z^2= 1 and r2=z2r^2= z^2 intersect when r2=z2=12r^2=z^2=\frac{1}{2}. Here is a sketch of the y=0y=0 cross–section of EE.

Figure from prob_s3.7, line 1897

Figure from prob_s3.7, line 1897

So

E={ (rcos⁡θ , rsin⁡θ , z) ∣ 0≤r≤1 ⁣/2, 0≤θ≤2π, r≤z≤1−r2 }\begin{equation*} E =\Set{(r\cos\theta\,,\,r\sin\theta\,,\,z)}{0\le r\le\nicefrac{1}{\sqrt{2}},\ 0\le\theta\le2\pi,\ r\le z\le \sqrt{1-r^2}} \end{equation*}

and

J=∭Ex2+y2+z2 dV=∫01/2dr∫02πdθ∫r1−r2dz rr2+z2\begin{align*} J &=\tripInt_E \sqrt{x^2+y^2+z^2}\ \dee{V} =\int_0^{1/\sqrt{2}}\dee{r}\int_0^{2\pi}\dee{\theta} \int_r^{\sqrt{1-r^2}}\dee{z}\ r\sqrt{r^2+z^2} \end{align*}

(b) In spherical coordinates

  • x2+y2+z2≤1x^2+y^2+z^2\le 1 is ρ≤1\rho\le 1 and

  • x2+y2≤z2x^2+y^2\le z^2 is ρ2sin⁡2φ≤ρ2cos⁡2φ\rho^2\sin^2\varphi\le \rho^2\cos^2\varphi, or tan⁡φ≤1\tan\varphi\le 1 or φ≤π4\varphi\le\frac{\pi}{4}, and

  • dV\dee{V} is ρ2sin⁡φ dρ dθ dφ\rho^2\sin\varphi\,\dee{\rho}\,\dee{\theta}\,\dee{\varphi}

So

E={ (ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ) ∣ 0≤ρ≤1, 0≤θ≤2π, 0≤φ≤π ⁣/4 }\begin{equation*} E =\Set{(\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)}{0\le\rho\le 1,\ 0\le\theta\le2\pi,\ 0\le \varphi\le \nicefrac{\pi}{4}} \end{equation*}

and, since the integrand x2+y2+z2=ρ\sqrt{x^2+y^2+z^2}=\rho,

J=∭Ex2+y2+z2 dV=∫01dρ∫02πdθ∫0π/4dφ ρ2sin⁡φ ρ=∫01dρ∫02πdθ∫0π/4dφ ρ3sin⁡φ\begin{align*} J &=\tripInt_E \sqrt{x^2+y^2+z^2}\ \dee{V} =\int_0^{1}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/4}\dee{\varphi}\ \rho^2\sin\varphi\ \rho \\ &=\int_0^{1}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/4}\dee{\varphi}\ \rho^3\sin\varphi \end{align*}

(c) We'll use the spherical coordinate form to evaluate

J=∫01dρ∫02πdθ∫0π/4dφ ρ3sin⁡φ=2π∫01dρ ρ3[−cos⁡φ]0π/4=2π 14 [1−12]=π2 [1−12]\begin{align*} J &=\int_0^{1}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/4}\dee{\varphi}\ \rho^3\sin\varphi \\ &=2\pi \int_0^{1}\dee{\rho}\ \rho^3 \Big[-\cos\varphi\Big]_0^{\pi/4} =2\pi\ \frac{1}{4}\ \left[1-\frac{1}{\sqrt{2}}\right] \\ &=\frac{\pi}{2}\ \left[1-\frac{1}{\sqrt{2}}\right] \end{align*}
Q25Stage 3Past exam · M200 2011D

The body of a snowman is formed by the snowballs x2+y2+z2=12x^2 + y^2 + z^2 = 12 (this is its body) and x2+y2+(z−4)2=4x^2 + y^2 + (z - 4)^2 = 4 (this is its head).

  1. Find the volume of the snowman by subtracting the intersection of the two snow balls from the sum of the volumes of the snow balls. [Recall that the volume of a sphere of radius rr is 4π3r3\frac{4\pi}{3} r^3.]

  2. We can also calculate the volume of the snowman as a sum of the following triple integrals:

    1. ∫02π3∫02π∫02ρ2sin⁡φ dρ dθ dφ\begin{equation*} \int_0^{\frac{2\pi}{3}} \int_0^{2\pi} \int_0^2 \rho^2\sin{\varphi} \ \dee{\rho}\,\dee{\theta}\,\dee{\varphi} \end{equation*}
    2. ∫02π∫03∫3 r4−r3r dz dr dθ\begin{equation*} \int_0^{2\pi} \int_0^{\sqrt{3}} \int_{\sqrt{3}\,r}^{4-\frac{r}{\sqrt{3}}} r\ \dee{z}\,\dee{r}\,\dee{\theta} \end{equation*}
    3. ∫π6π∫02π∫023ρ2sin⁡(φ) dρ dθ dφ\begin{equation*} \int_{\frac{\pi}{6}}^\pi \int_0^{2\pi} \int_0^{2\sqrt{3}} \rho^2\sin(\varphi)\ \dee{\rho}\,\dee{\theta}\,\dee{\varphi} \end{equation*}

    Circle the right answer from the underlined choices and fill in the blanks in the following descriptions of the region of integration for each integral. [Note: We have translated the axes in order to write down some of the integrals above. The equations you specify should be those before the translation is performed.]

    1. The region of integration in (1) is a part of the snowman's

      body/head/body and head‾.\begin{align*}&\text{\underline{body/head/body and head}.}\end{align*}

      It is the solid enclosed by the

      sphere/cone‾ defined by the equation \begin{align*}&\text{\underline{sphere/cone} defined by the equation }\rule{1.2in}{0.15mm}\end{align*}

      and the

      sphere/cone‾ defined by the equation .\begin{align*}&\text{\underline{sphere/cone} defined by the equation }\rule{1.5in}{0.15mm}\text{.}\end{align*}

    2. The region of integration in (2) is a part of the snowman's

      body/head/body and head‾.\begin{align*}&\text{\underline{body/head/body and head}.}\end{align*}

      It is the solid enclosed by the

      sphere/cone‾ defined by the equation \begin{align*}&\text{\underline{sphere/cone} defined by the equation }\rule{1.2in}{0.15mm}\end{align*}

      and the

      sphere/cone‾ defined by the equation .\begin{align*}&\text{\underline{sphere/cone} defined by the equation }\rule{1.5in}{0.15mm}\text{.}\end{align*}

    3. The region of integration in (3) is a part of the snowman's

      body/head/body and head‾.\begin{align*}&\text{\underline{body/head/body and head}.}\end{align*}

      It is the solid enclosed by the

      sphere/cone‾ defined by the equation \begin{align*}&\text{\underline{sphere/cone} defined by the equation }\rule{1.2in}{0.15mm}\end{align*}

      and the

      sphere/cone‾ defined by the equation .\begin{align*}&\text{\underline{sphere/cone} defined by the equation }\rule{1.5in}{0.15mm}\text{.}\end{align*}
Answer

(a) 2π3[(12)3/2+54]\frac{2\pi}{3}\left[\big(12\big)^{3/2}+54\right]

(b) i. The top part is the part of the snowman's head that is inside the sphere
x2+y2+(z−4)2=4x^2+y^2+(z-4)^2 = 4 and above the cone z−4=−x2+y23z-4 = - \sqrt{\frac{x^2+y^2}{3}}.

ii. The middle part is the part of the snowman's head and body that is bounded on the top by the cone z−4=−x2+y23z-4 = - \sqrt{\frac{x^2+y^2}{3}} and is bounded on the bottom by the cone z=3(x2+y2)z = \sqrt{3(x^2+y^2)}.

iii. The bottom part is the part of the snowman's body that is inside the sphere
x2+y2+z2=12x^2+y^2+z^2 = 12 and is below the cone z=3(x2+y2)z = \sqrt{3(x^2+y^2)}.

Full solution

(a) As a check, the body of the snow man has radius 12=23≈3.46\sqrt{12} = 2\sqrt{3} \approx 3.46, which is between 22 (the low point of the head) and 44 (the center of the head). Here is a sketch of a side view of the snowman.

Figure from prob_s3.7, line 2073

Figure from prob_s3.7, line 2073

We want to determine the volume of the intersection of the body and the head, whose side view is the darker shaded region in the sketch.

  • The outer boundary of the body and the outer boundary of the head intersect when both x2+y2+z2=12x^2 + y^2 + z^2 = 12 and x2+y2+(z−4)2=4x^2 + y^2 + (z - 4)^2 = 4. Subtracting the second equation from the first gives

    z2−(z−4)2=12−4  ⟺  8z−16=8  ⟺  z=3\begin{equation*} z^2-(z-4)^2 =12 -4 \iff 8z-16 =8 \iff z=3 \end{equation*}

    Then substituting z=3z=3 into either equation gives x2+y2=3x^2+y^2=3. So the intersection of the outer boundaries of the head and body (i.e. the neck) is the circle x2+y2=3x^2+y^2=3, z=3z=3.

  • The top boundary of the intersection is part of the top half of the snowman's body and so has equation z=+12−x2−y2z=+\sqrt{12-x^2-y^2}.

  • The bottom boundary of the intersection is part of the bottom half of the snowman's head, and so has equation z=4−4−x2−y2z=4-\sqrt{4-x^2-y^2}

The intersection of the head and body is thus

V={ (x,y,z ∣ x2+y2≤3, 4−4−x2−y2≤z≤12−x2−y2 }\begin{align*} \cV = \Set{(x,y,z}{x^2+y^2\le 3,\ 4-\sqrt{4-x^2-y^2}\le z\le \sqrt{12-x^2-y^2}} \end{align*}

We'll compute the volume of V\cV using cylindrical coordinates

Volume(V)=∫03dr∫02πdθ∫4−4−r212−r2dz r=∫03dr 2π r[12−r2−4+4−r2]=2π[−13(12−r2)3/2−2r2−13(4−r2)3/2]03=2π[−13(9)3/2−2(3)−13(1)3/2+13(12)3/2+2(0)2+13(4)3/2][0]=2π[−9−6−13+13(12)3/2+83]=2π[13(12)3/2−383]\begin{align*} \text{Volume}(\cV) &= \int_0^{\sqrt{3}}\dee{r}\int_0^{2\pi}\dee{\theta} \int_{4-\sqrt{4-r^2}}^{\sqrt{12-r^2}}\dee{z}\ r \\ &=\int_0^{\sqrt{3}}\dee{r}\ 2\pi\ r\big[\sqrt{12-r^2}-4+\sqrt{4-r^2}\big] \\ &=2\pi\left[-\frac{1}{3}{\big(12-r^2\big)}^{3/2} -2r^2 -\frac{1}{3}{\big(4-r^2\big)}^{3/2}\right]_0^{\sqrt{3}} \\ &= 2\pi\left[-\frac{1}{3}{\big(9\big)}^{3/2} -2(3) -\frac{1}{3}{\big(1\big)}^{3/2} +\frac{1}{3}{\big(12\big)}^{3/2} +2(0)^2 +\frac{1}{3}{\big(4\big)}^{3/2} \right] [0]\\ &= 2\pi\left[-9 -6 -\frac{1}{3} +\frac{1}{3}{\big(12\big)}^{3/2} +\frac{8}{3} \right] \\ &=2\pi\left[\frac{1}{3}{\big(12\big)}^{3/2} -\frac{38}{3} \right] \end{align*}

So the volume of the snowman is

4π3(12)3/2+4π323−2π[13(12)3/2−383]=2π3[(12)3/2+54]\begin{align*} &\frac{4\pi}{3}\big(12\big)^{3/2} +\frac{4\pi}{3} 2^3 -2\pi\left[\frac{1}{3}{\big(12\big)}^{3/2} -\frac{38}{3} \right] \\ &\hskip0.5in=\frac{2\pi}{3}\left[\big(12\big)^{3/2}+54\right] \end{align*}

(b) The figure on the left below is another side view of the snowman. This time it is divided into a lighter gray top part, a darker gray middle part and a lighter gray bottom part. The figure on the right below is an enlarged view of the central part of the figure on the left.

Figure from prob_s3.7, line 2073

Figure from prob_s3.7, line 2073

Figure from prob_s3.7, line 2073

Figure from prob_s3.7, line 2073

i. The top part is the Pac–Man

Figure from prob_s3.7, line 2073

Figure from prob_s3.7, line 2073

part of the snowman's head. It is the part of the sphere

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

that is above the cone

z−4=−x2+y23\begin{equation*} z-4 = - \sqrt{\frac{x^2+y^2}{3}} \end{equation*}

(which contains the points (0,0,4)(0,0,4) and (3,0,3)(\sqrt{3},0,3)).

ii. The middle part is the diamond shaped

Figure from prob_s3.7, line 2073

Figure from prob_s3.7, line 2073

part of the snowman's head and body. It is bounded on the top by the cone

z−4=−x2+y23\begin{equation*} z-4 = - \sqrt{\frac{x^2+y^2}{3}} \end{equation*}

(which contains the points (0,0,4)(0,0,4) and (3,0,3)(\sqrt{3},0,3)) and is bounded on the bottom by the cone

z=3(x2+y2)\begin{equation*} z = \sqrt{3(x^2+y^2)} \end{equation*}

(which contains the points (0,0,0)(0,0,0) and (3,0,3)(\sqrt{3},0,3)).

iii. The bottom part is the Pac–Man

Figure from prob_s3.7, line 2073

Figure from prob_s3.7, line 2073

part of the snowman's body. It is the part of the sphere

x2+y2+z2≤12\begin{equation*} x^2+y^2+z^2\le 12 \end{equation*}

that is below the cone

z=3(x2+y2)\begin{equation*} z = \sqrt{3(x^2+y^2)} \end{equation*}

(which contains the points (0,0,0)(0,0,0) and (3,0,3)(\sqrt{3},0,3)).

Q26Stage 3Past exam · M200 2013D

  1. Find the volume of the solid inside the surface defined by the equation ρ=8sin⁡(φ)\rho = 8 \sin(\varphi) in spherical coordinates.

    You may use that

    ∫sin⁡4(φ)=132(12φ−8sin⁡(2φ)+sin⁡(4φ))+C\begin{equation*} \int \sin^4(\varphi) =\frac{1}{32}\big(12\varphi -8\sin(2\varphi) +\sin(4\varphi)\big) +C \end{equation*}

  2. Sketch this solid or describe what it looks like.

Hint

(b) it is a solid of revolution.

Answer

(a) 2(83) π3 12 π32=128π2\frac{2(8^3)\,\pi}{3}\ \frac{12\,\pi}{32}=128\pi^2

(b) The surface is a torus (a donut) but with the hole in the centre shrunk to a point. The figure below is a sketch of the part of the surface in the first octant.

Figure from prob_s3.7, line 2220

Figure from prob_s3.7, line 2220

Full solution

(a) Recall that, in spherical coordinates, φ\varphi runs from 00 (that's the positive zz–axis) to π\pi (that's the negative zz–axis), θ\theta runs from 00 to 2π2\pi (θ\theta is the regular polar or cylindrical coordinate) and dV=ρ2 sin⁡φ dρ dθ dφ\dee{V} = \rho^2\ \sin\varphi\ \dee{\rho}\,\dee{\theta}\,\dee{\varphi}. So

Volume=∫0πdφ∫02πdθ∫08sin⁡φdρ ρ2 sin⁡φ=∫0πdφ∫02πdθ (8sin⁡φ)33sin⁡φ=2(83) π3∫0πdφ sin⁡4φ=2(83) π3[132(12φ−8sin⁡(2φ)+sin⁡(4φ))]0π=2(83) π3 12 π32=128π2\begin{align*} \text{Volume} & = \int_0^\pi \dee{\varphi}\int_0^{2\pi}\dee{\theta} \int_0^{8\sin\varphi}\dee{\rho}\ \rho^2\ \sin\varphi \\ & = \int_0^\pi \dee{\varphi}\int_0^{2\pi}\dee{\theta}\ \frac{(8\sin\varphi)^3}{3} \sin\varphi \\ & = \frac{2(8^3)\,\pi}{3} \int_0^\pi \dee{\varphi}\ \sin^4\varphi \\ &= \frac{2(8^3)\,\pi}{3} \left[\frac{1}{32}\big(12\varphi -8\sin(2\varphi) +\sin(4\varphi)\big)\right]_0^\pi \\ &=\frac{2(8^3)\,\pi}{3}\ \frac{12\,\pi}{32} =128\pi^2 \end{align*}

(b) Fix any φ\varphi between 00 and π\pi. If ρ=8sin⁡φ\rho=8\sin\varphi, then as θ\theta runs from 00 to 2π2\pi,

(x,y,z)=(ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ)=(8sin⁡2φcos⁡θ , 8sin⁡2φsin⁡θ , 8sin⁡φcos⁡φ)=(Rcos⁡θ , Rsin⁡θ , Z)with R=8sin⁡2φ, Z=8sin⁡φcos⁡φ\begin{align*} (x,y,z) &= \big(\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi\big) \\ &= \big(8\sin^2\varphi\cos\theta\,,\,8\sin^2\varphi\sin\theta\,,\, 8\sin\varphi\cos\varphi\big) \\ &= \big(R\cos\theta\,,\,R\sin\theta\,,\,Z\big)\qquad \text{with }R=8\sin^2\varphi,\ Z= 8\sin\varphi\cos\varphi \end{align*}

sweeps out a circle of radius R=8sin⁡2φR=8\sin^2\varphi contained in the plane z=Z=8sin⁡φcos⁡φz=Z=8\sin\varphi\cos\varphi and centred on (0,0,Z=8sin⁡φcos⁡φ)\big(0,0,Z=8\sin\varphi\cos\varphi\big). So the surface is a bunch of circles stacked one on top of the other. It is a surface of revolution. We can sketch it by

  • first sketching the θ=0\theta=0 section of the surface (that's the part of the surface in the right half of the xzxz–plane)

  • and then rotate the result about the zz–axis.

The θ=0\theta=0 part of the surface is

{ (x,y,z) ∣ x=8sin⁡2φ, y=0, z=8sin⁡φcos⁡φ, 0≤φ≤π }={ (x,y,z) ∣ x=4−4cos⁡(2φ), y=0, z=4sin⁡(2φ), 0≤φ≤π }\begin{align*} &\Set{(x,y,z)}{x=8\sin^2\varphi,\ y=0,\ z=8\sin\varphi\cos\varphi,\ 0\le\varphi\le \pi} \\ &=\Set{(x,y,z)}{x=4-4\cos(2\varphi),\ y=0,\ z=4\sin(2\varphi),\ 0\le\varphi\le \pi} \end{align*}

It's a circle of radius 44, contained in the xzxz–plane (i.e. y=0y=0) and centred on (4,0,0)(4,0,0)! The figure on the left below is a sketch of the top half of the circle. When we rotate the circle about the zz–axis we get a torus (a donut) but with the hole in the centre shrunk to a point. The figure on the right below is a sketch of the part of the torus in the first octant.

Figure from prob_s3.7, line 2233

Figure from prob_s3.7, line 2233

Figure from prob_s3.7, line 2220

Figure from prob_s3.7, line 2220

Q27Stage 3Past exam · M200 2014A

Let EE be the solid

0≤z≤x2+y2,x2+y2≤1,\begin{equation*} 0 \le z \le \sqrt{x^2 + y^2},\qquad x^2 + y^2 \le 1, \end{equation*}

and consider the integral

I=∭Ezx2+y2+z2 dV.\begin{equation*} I = \tripInt_E z \sqrt{x^2 + y^2 + z^2}\ \dee{V}. \end{equation*}

  1. Write the integral II in cylindrical coordinates.

  2. Write the integral II in spherical coordinates.

  3. Evaluate the integral II using either form.

Answer

(a) I=∫01dr∫02πdθ∫0rdz r z r2+z2I = \int_0^1\dee{r} \int_0^{2\pi}\dee{\theta} \int_0^r\dee{z}\ r\ z\ \sqrt{r^2+z^2} (b) I=∫π/4π/2dφ∫02πdθ∫01/sin⁡φdρ ρ4sin⁡φ cos⁡φI = \int_{\pi/4}^{\pi/2}\dee{\varphi} \int_0^{2\pi}\dee{\theta} \int_0^{1/\sin\varphi}\dee{\rho}\ \rho^4\sin\varphi \ \cos\varphi

(c) 2(22−1)π15\frac{2(2\sqrt{2}-1)\pi}{15}

Full solution

(a) In cylindrical coordinates 0≤z≤x2+y20\le z\le \sqrt{x^2+y^2} becomes 0≤z≤r0\le z\le r, and x2+y2≤1x^2+y^2\le 1 becomes 0≤r≤10\le r\le 1. So

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

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

I=∭Ezx2+y2+z2 dV=∫01dr∫02πdθ∫0rdz r z r2+z2⏟x2+y2+z2\begin{align*} I &= \tripInt_E z \sqrt{x^2 + y^2 + z^2}\ \dee{V} = \int_0^1\dee{r} \int_0^{2\pi}\dee{\theta} \int_0^r\dee{z}\ r\ z\ \sqrt{\underbrace{r^2+z^2}_{x^2+y^2+z^2}} \end{align*}

(b) Here is a sketch of a constant θ\theta section of EE.

Figure from prob_s3.7, line 2334

Figure from prob_s3.7, line 2334

Recall that the spherical coordinate φ\varphi is the angle between the zz–axis and the radius vector. So, in spherical coordinates z=rz=r (which makes an angle π4\frac{\pi}{4} with the zz axis) becomes φ=π4\varphi=\frac{\pi}{4}, and the plane z=0z=0, i.e. the xyxy–plane, becomes φ=π2\varphi=\frac{\pi}{2}, and r=1r=1 becomes ρsin⁡φ=1\rho\sin\varphi = 1. So

E={(ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ) ∣ π4≤φ≤π2, 0≤θ≤2π, 0≤ρ≤1sin⁡φ }\begin{equation*} E = \left\{(\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)\ \left|\ \frac{\pi}{4}\le \varphi\le \frac{\pi}{2}, \ 0\le\theta\le 2\pi, \ 0\le \rho\le \frac{1}{\sin\varphi}\ \right.\right\} \end{equation*}

and, since dV=ρ2sin⁡φ dρ dθ dφ\dee{V} = \rho^2\sin\varphi\,\dee{\rho}\,\dee{\theta}\,\dee{\varphi},

I=∭Ezx2+y2+z2 dV=∫π/4π/2dφ∫02πdθ∫01/sin⁡φdρ ρ2sin⁡φ ρcos⁡φ⏞z ρ=∫π/4π/2dφ∫02πdθ∫01/sin⁡φdρ ρ4sin⁡φ cos⁡φ\begin{align*} I &= \tripInt_E z \sqrt{x^2 + y^2 + z^2}\ \dee{V} = \int_{\pi/4}^{\pi/2}\dee{\varphi} \int_0^{2\pi}\dee{\theta} \int_0^{1/\sin\varphi}\dee{\rho}\ \rho^2\sin\varphi \ \overbrace{\rho\cos\varphi}^{z}\ \rho \\ &= \int_{\pi/4}^{\pi/2}\dee{\varphi} \int_0^{2\pi}\dee{\theta} \int_0^{1/\sin\varphi}\dee{\rho}\ \rho^4\sin\varphi \ \cos\varphi \end{align*}

(c) We'll integrate using the spherical coordinate version.

I=∫π/4π/2dφ∫02πdθ∫01/sin⁡φdρ ρ4sin⁡φ cos⁡φ=∫π/4π/2dφ∫02πdθ 15sin⁡5φsin⁡φ cos⁡φ=2π5∫π/4π/2dφ cos⁡φsin⁡4φ=2π5∫1/21duu4with u=sin⁡φ, du=cos⁡φ dφ=2π5[u−3−3]1/21=2(22−1)π15\begin{align*} I &= \int_{\pi/4}^{\pi/2}\dee{\varphi} \int_0^{2\pi}\dee{\theta} \int_0^{1/\sin\varphi}\dee{\rho}\ \rho^4\sin\varphi \ \cos\varphi \\ &= \int_{\pi/4}^{\pi/2}\dee{\varphi} \int_0^{2\pi}\dee{\theta}\ \frac{1}{5\sin^5\varphi}\sin\varphi \ \cos\varphi \\ &= \frac{2\pi}{5}\int_{\pi/4}^{\pi/2}\dee{\varphi}\ \frac{\cos\varphi}{\sin^4\varphi} \\ &= \frac{2\pi}{5}\int_{1/\sqrt{2}}^{1} \frac{\dee{u}}{u^4}\qquad\text{with } u=\sin\varphi,\ \dee{u}=\cos\varphi\,\dee{\varphi} \\ &= \frac{2\pi}{5} \left[\frac{u^{-3}}{-3}\right]_{1/\sqrt{2}}^{1} \\ &= \frac{2(2\sqrt{2}-1)\pi}{15} \end{align*}
Q28Stage 3Past exam · M200 2014D

Consider the iterated integral

I=∫−a0∫−a2−x20∫0a2−x2−y2(x2+y2+z2)2014 dz dy dx\begin{equation*} I =\int_{-a}^0\int_{-\sqrt{a^2-x^2}}^0 \int_0^{\sqrt{a^2-x^2-y^2}} \big(x^2+y^2+z^2\big)^{2014}\ \dee{z}\,\dee{y}\,\dee{x} \end{equation*}

where aa is a positive constant.

  1. Write II as an iterated integral in cylindrical coordinates.

  2. Write II as an iterated integral in spherical coordinates.

  3. Evaluate I using whatever method you prefer.

Answer

(a) ∫0adz∫π3π/2dθ∫0a2−z2dr r(r2+z2)2014\int_0^a \dee{z}\int_{\pi}^{3\pi/2}\dee{\theta} \int_0^{\sqrt{a^2-z^2}}\dee{r}\ r\big(r^2+z^2\big)^{2014} (b) ∫0π/2dφ∫π3π/2dθ∫0adρ ρ4030sin⁡φ\int_0^{\pi/2} \dee{\varphi}\int_{\pi}^{3\pi/2}\dee{\theta} \int_0^a\dee{\rho}\ \rho^{4030}\sin\varphi

(c) a4031π8062\frac{a^{4031}\pi}{8062}

Full solution

The main step is to figure out what the domain of integration looks like.

  • The outside integral says that xx runs from −a-a to 00.

  • The middle integrals says that, for each xx in that range, yy runs from −a2−x2-\sqrt{a^2-x^2} to 00. We can rewrite y=−a2−x2y=-\sqrt{a^2-x^2} in the more familiar form x2+y2=a2x^2+y^2=a^2, y≤0y\le 0. So (x,y)(x,y) runs over the third quadrant part of the disk of radius aa, centred on the origin.

    Figure from prob_s3.7, line 2432

    Figure from prob_s3.7, line 2432

  • Finally, the inside integral says that, for each (x,y)(x,y) in the quarter disk, zz runs from 00 to a2−x2−y2\sqrt{a^2-x^2-y^2}. We can also rewrite z=a2−x2−y2z=\sqrt{a^2-x^2-y^2} in the more familiar form x2+y2+z2=a2x^2+y^2+z^2=a^2, z≥0z\ge 0.

So the domain of integration is the part of the interior of the sphere of radius aa, centred on the origin, that lies in the octant x≤0x\le 0, y≤0y\le 0, z≥0z\ge 0.

V={ (x,y,z) ∣ −a≤x≤0, −a2−x2≤y≤0, 0≤z≤a2−x2−y2 }={ (x,y,z) ∣ x2+y2+z2≤a2, x≤0, y≤0, z≥0 }\begin{align*} V&=\Set{(x,y,z)}{-a\le x\le 0,\ -\sqrt{a^2-x^2}\le y\le 0,\ 0\le z\le\sqrt{a^2-x^2-y^2} } \\ &=\Set{(x,y,z)}{x^2+y^2+z^2\le a^2,\ x\le 0,\ y\le 0,\ z\ge 0 } \end{align*}

Figure from prob_s3.7, line 2432

Figure from prob_s3.7, line 2432

(a) Note that, in VV, (x,y)(x,y) is restricted to the third quadrant, which in cylindrical coordinates is π≤θ≤3π2\pi\le\theta\le\frac{3\pi}{2}. So, in cylindrical coordinates,

V={(rcos⁡θ,rsin⁡θ,z) ∣ r2+z2≤a2, π≤θ≤3π2, z≥0}={(rcos⁡θ,rsin⁡θ,z) ∣ 0≤z≤a, π≤θ≤3π2, 0≤r≤a2−z2}\begin{align*} V&=\Big\{(r\cos\theta,r\sin\theta,z)\ \Big|\ r^2+z^2\le a^2,\ \pi\le\theta\le\frac{3\pi}{2},\ z\ge 0 \Big\} \\ &=\Big\{(r\cos\theta,r\sin\theta,z)\ \Big|\ 0\le z\le a,\ \pi\le\theta\le\frac{3\pi}{2},\ 0\le r\le\sqrt{a^2-z^2} \Big\} \end{align*}

and

I=∭V(x2+y2+z2)2014 dV=∭V(r2+z2)2014 r dr dθ dz=∫0adz∫π3π/2dθ∫0a2−z2dr r(r2+z2)2014\begin{align*} I&= \tripInt_V \big(x^2+y^2+z^2\big)^{2014}\ \dee{V} = \tripInt_V \big(r^2+z^2\big)^{2014}\ r\,\dee{r}\,\dee{\theta}\,\dee{z} \\ &=\int_0^a \dee{z}\int_{\pi}^{3\pi/2}\dee{\theta} \int_0^{\sqrt{a^2-z^2}}\dee{r}\ r\big(r^2+z^2\big)^{2014} \end{align*}

(b) The spherical coordinate φ\varphi runs from 00 (when the radius vector is along the positive zz–axis) to π ⁣/2\nicefrac{\pi}{2} (when the radius vector lies in the xyxy–plane) so that

I=∭V(x2+y2+z2)2014 dV=∭Vρ2×2014 ρ2sin⁡φ dρ dθ dφ=∫0π/2dφ∫π3π/2dθ∫0adρ ρ4030sin⁡φ\begin{align*} I&= \tripInt_V \big(x^2+y^2+z^2\big)^{2014}\ \dee{V} = \tripInt_V \rho^{2\times 2014}\ \rho^2\sin\varphi\ \dee{\rho}\,\dee{\theta}\,\dee{\varphi} \\ &=\int_0^{\pi/2} \dee{\varphi}\int_{\pi}^{3\pi/2}\dee{\theta} \int_0^a\dee{\rho}\ \rho^{4030}\sin\varphi \end{align*}

(c) Using the spherical coordinate version

I=∫0π/2dφ∫π3π/2dθ∫0adρ ρ4030sin⁡φ=a40314031∫0π/2dφ∫π3π/2dθ sin⁡φ=a4031π8062∫0π/2dφ sin⁡φ=a4031π8062\begin{align*} I&=\int_0^{\pi/2} \dee{\varphi}\int_{\pi}^{3\pi/2}\dee{\theta} \int_0^a\dee{\rho}\ \rho^{4030}\sin\varphi \\ &=\frac{a^{4031}}{4031} \int_0^{\pi/2} \dee{\varphi}\int_{\pi}^{3\pi/2}\dee{\theta} \ \sin\varphi \\ &=\frac{a^{4031}\pi}{8062} \int_0^{\pi/2} \dee{\varphi}\ \sin\varphi \\ &=\frac{a^{4031}\pi}{8062} \end{align*}
Q29Stage 3Past exam · M200 2015D

The solid EE is bounded below by the paraboloid z=x2+y2z = x^2 + y^2 and above by the cone z=x2+y2z=\sqrt{x^2+y^2}. Let

I=∭Ez(x2+y2+z2) dV\begin{equation*} I = \tripInt_E z\big(x^2+y^2+z^2\big)\ \dee{V} \end{equation*}

  1. Write II in terms of cylindrical coordinates. Do not evaluate.

  2. Write II in terms of spherical coordinates. Do not evaluate.

  3. Calculate II.

Answer

(a) ∫01dr r∫02πdθ∫r2rdz z(r2+z2)\int_0^1 \dee{r}\ r \int_0^{2\pi}\dee{\theta} \int_{r^2}^r\dee{z}\ z(r^2+z^2) (b) ∫π/4π/2dφ∫02πdθ∫0cos⁡φ/sin⁡2φdρ ρ5sin⁡φcos⁡φ\int_{\pi/4}^{\pi/2} \dee{\varphi} \int_0^{2\pi}\dee{\theta} \int_0^{\cos\varphi/\sin^2\varphi} \dee{\rho}\ \rho^5\sin\varphi \cos\varphi

(c) 3π40\frac{3\pi}{40}

Full solution

(a) In cylindrical coordinates, the paraboloid is z=r2z=r^2 and the cone is z=rz=r. The two meet when r2=rr^2=r. That is, when r=0r=0 and when r=1r=1. So, in cylindrical coordinates

I=∫01dr r∫02πdθ∫r2rdz z(r2+z2)\begin{align*} I = \int_0^1 \dee{r}\ r \int_0^{2\pi}\dee{\theta} \int_{r^2}^r\dee{z}\ z(r^2+z^2) \end{align*}

(b) In spherical coordinates, the paraboloid is

ρcos⁡φ=ρ2sin⁡2φorρ=cos⁡φsin⁡2φ\begin{equation*} \rho\cos\varphi=\rho^2\sin^2\varphi\qquad\text{or}\qquad \rho=\frac{\cos\varphi}{\sin^2\varphi} \end{equation*}

and the cone is

ρcos⁡φ=ρsin⁡φortan⁡φ=1orφ=π4\begin{equation*} \rho\cos\varphi=\rho\sin\varphi\quad\text{or}\quad \tan\varphi=1\quad\text{or}\quad \varphi=\frac{\pi}{4} \end{equation*}

The figure below shows a constant θ\theta cross–section of EE. Looking at that figure, we see that φ\varphi runs from π4\frac{\pi}{4} (i.e. the cone) to π2\frac{\pi}{2} (i.e. the xyxy–plane).

Figure from prob_s3.7, line 2540

Figure from prob_s3.7, line 2540

So, is spherical coordinates,

I=∫π/4π/2dφ∫02πdθ∫0cos⁡φ/sin⁡2φdρ ρ2sin⁡φ ρcos⁡φ⏞zρ2⏞x2+y2+z2=∫π/4π/2dφ∫02πdθ∫0cos⁡φ/sin⁡2φdρ ρ5sin⁡φcos⁡φ\begin{align*} I &= \int_{\pi/4}^{\pi/2} \dee{\varphi} \int_0^{2\pi}\dee{\theta} \int_0^{\cos\varphi/\sin^2\varphi} \dee{\rho}\ \rho^2\sin\varphi \ \overbrace{\rho\cos\varphi}^{z} \overbrace{\rho^2}^{x^2+y^2+z^2} \\ &= \int_{\pi/4}^{\pi/2} \dee{\varphi} \int_0^{2\pi}\dee{\theta} \int_0^{\cos\varphi/\sin^2\varphi} \dee{\rho}\ \rho^5\sin\varphi \cos\varphi \end{align*}

(c) The cylindrical coordinates integral looks easier.

I=∫01dr r∫02πdθ∫r2rdz z(r2+z2)=∫01dr r∫02πdθ [r2z22+z44]r2r=2π∫01dr r[r2r22+r44−r2r42−r84]=2π[112+124−116−140]=π2[13+16−14−110]=3π40\begin{align*} I &= \int_0^1 \dee{r}\ r \int_0^{2\pi}\dee{\theta} \int_{r^2}^r\dee{z}\ z(r^2+z^2) \\ &= \int_0^1 \dee{r}\ r \int_0^{2\pi}\dee{\theta}\ \left[r^2\frac{z^2}{2}+\frac{z^4}{4}\right]_{r^2}^r \\ &= 2\pi \int_0^1 \dee{r}\ r \left[r^2\frac{r^2}{2}+\frac{r^4}{4} -r^2\frac{r^4}{2}-\frac{r^8}{4}\right] \\ &=2\pi \left[\frac{1}{12}+\frac{1}{24} -\frac{1}{16}-\frac{1}{40}\right] =\frac{\pi}{2} \left[\frac{1}{3}+\frac{1}{6} -\frac{1}{4}-\frac{1}{10}\right] \\ &=\frac{3\pi}{40} \end{align*}
Q30Stage 3Past exam · M200 2016D

Let SS be the region on the first octant (so that x,y,z≥0x,y,z\ge 0) which lies above the cone z=x2+y2z=\sqrt{x^2+y^2} and below the sphere (z−1)2+x2+y2=1(z-1)^2 +x^2+y^2=1. Let VV be its volume.

  1. Express VV as a triple integral in cylindrical coordinates.

  2. Express VV as an triple integral in spherical coordinates.

  3. Calculate VV using either of the integrals above.

Answer

(a) V=∫01dr∫0π/2dθ∫r1+1−r2dz rV = \int_0^1\dee{r} \int_0^{\pi/2}\dee{\theta}\int_r^{1+\sqrt{1-r^2}}\dee{z} \ r (b) V=∫0π/4dφ∫0π/2dθ∫02cos⁡φdρ ρ2sin⁡φV = \int_0^{\pi/4}\dee{\varphi} \int_0^{\pi/2}\dee{\theta} \int_0^{2\cos\varphi}\dee{\rho}\ \rho^2\sin\varphi

(c) π4\frac{\pi}{4}

Full solution

Note that both the sphere x2+y2+(z−1)2=1x^2+y^2+(z-1)^2=1 and the cone z=x2+y2z=\sqrt{x^2+y^2} are invariant under rotations around the zz–axis. The sphere x2+y2+(z−1)2=1x^2 + y^2 + (z-1)^2 = 1 and the cone z=x2+y2z = \sqrt{x^2 + y^2} intersect when z=x2+y2z=\sqrt{x^2+y^2}, so that x2+y2=z2x^2+y^2=z^2, and

x2+y2+(z−1)2=z2+(z−1)2=1  ⟺  2z2−2z=0  ⟺  2z(z−1)=0  ⟺  z=0,1\begin{align*} x^2 + y^2 + (z-1)^2 = z^2+(z-1)^2=1 &\iff 2z^2-2z=0 \iff 2z(z-1)=0 \\ &\iff z=0,1 \end{align*}

So the surfaces intersect on the circle z=1z=1, x2+y2=1x^2+y^2=1 and

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

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

Figure from prob_s3.7, line 2628

Figure from prob_s3.7, line 2628

(a) In cylindrical coordinates

  • the condition x,y≥0x,y\ge 0 is 0≤θ≤π ⁣/20\le\theta\le\nicefrac{\pi}{2},

  • the condition x2+y2≤1x^2+y^2\le 1 is r≤1r\le 1, and

  • the conditions x2+y2≤z≤1+1−x2−y2\sqrt{x^2+y^2}\le z\le 1+\sqrt{1-x^2-y^2} are r≤z≤1+1−r2r\le z\le 1+\sqrt{1-r^2}, and

  • dV=r dr dθ dz\dee{V} = r\,\dee{r}\,\dee{\theta}\,\dee{z}.

So

V=∭SdV=∫01dr∫0π/2dθ∫r1+1−r2dz r\begin{align*} V = \tripInt_S\dee{V} = \int_0^1\dee{r} \int_0^{\pi/2}\dee{\theta}\int_r^{1+\sqrt{1-r^2}}\dee{z} \ r \end{align*}

(b) In spherical coordinates,

  • the cone z=x2+y2z=\sqrt{x^2+y^2} becomes

    ρ cos⁡φ=ρ2sin⁡2φ cos⁡2θ+ρ2sin⁡2φ sin⁡2θ=ρsin⁡φ  ⟺  tan⁡φ=1  ⟺  φ=π4\begin{align*} \rho\,\cos\varphi =\sqrt{\rho^2\sin^2\varphi\,\cos^2\theta +\rho^2\sin^2\varphi\,\sin^2\theta} =\rho\sin\varphi \iff \tan\varphi=1 \iff \varphi=\frac{\pi}{4} \end{align*}

  • so that, on SS, the spherical coordinate φ\varphi runs from φ=0\varphi=0 (the positive zz –axis) to φ=π ⁣/4\varphi=\nicefrac{\pi}{4} (the cone z=x2+y2z=\sqrt{x^2+y^2}), which keeps us above the cone,

  • the condition x,y≥0x,y\ge 0 is 0≤θ≤π ⁣/20\le\theta\le\nicefrac{\pi}{2},

  • the condition x2+y2+(z−1)2≤1x^2+y^2+(z-1)^2\le 1, (which keeps us inside the sphere), becomes

    ρ2sin⁡2φ cos⁡2θ+ρ2sin⁡2φ sin⁡2θ+(ρcos⁡φ−1)2≤1  ⟺  ρ2 sin⁡2φ+ρ2cos⁡2φ−2ρcos⁡φ+1≤1  ⟺  ρ2−2ρcos⁡φ≤0  ⟺  ρ≤2cos⁡φ\begin{align*} &\rho^2\sin^2\varphi\,\cos^2\theta + \rho^2\sin^2\varphi\,\sin^2\theta +\big(\rho\cos\varphi-1\big)^2\le 1 \\ &\iff \rho^2\,\sin^2\varphi +\rho^2\cos^2\varphi -2\rho\cos\varphi +1 \le 1 \\ &\iff \rho^2-2\rho\cos\varphi\le 0 \\ &\iff \rho\le 2 \cos\varphi \end{align*}

  • and dV=ρ2sin⁡φ dρ dθ dφ\dee{V} = \rho^2\sin\varphi\,\dee{\rho}\,\dee{\theta}\,\dee{\varphi}.

So

V=∭SdV=∫0π/4dφ∫0π/2dθ∫02cos⁡φdρ ρ2sin⁡φ\begin{align*} V = \tripInt_S\dee{V} = \int_0^{\pi/4}\dee{\varphi} \int_0^{\pi/2}\dee{\theta} \int_0^{2\cos\varphi}\dee{\rho}\ \rho^2\sin\varphi \end{align*}

(c) We'll evaluate VV using the spherical coordinate integral of part (b).

V=∫0π/4dφ∫0π/2dθ∫02cos⁡φdρ ρ2sin⁡φ=83∫0π/4dφ∫0π/2dθ cos⁡3φ sin⁡φ=83 π2 [−cos⁡4φ4]0π/4=π3[1−1(2)4]=π4\begin{align*} V &= \int_0^{\pi/4}\dee{\varphi} \int_0^{\pi/2}\dee{\theta} \int_0^{2\cos\varphi}\dee{\rho}\ \rho^2\sin\varphi \\ &=\frac{8}{3} \int_0^{\pi/4}\dee{\varphi} \int_0^{\pi/2}\dee{\theta}\ \cos^3\varphi\ \sin\varphi \\ &=\frac{8}{3}\ \frac{\pi}{2}\ \left[-\frac{\cos^4\varphi}{4}\right]_0^{\pi/4} \\ &=\frac{\pi}{3} \left[1-\frac{1}{(\sqrt{2})^4}\right] \\ &=\frac{\pi}{4} \end{align*}
Q31Stage 3Past exam · M200 2004A

A solid is bounded below by the cone z=3x2+3y2z=\sqrt{3x^2+3y^2} and above by the sphere x2+y2+z2=9x^2+y^2+z^2=9. It has density δ(x,y,z)=x2+y2\de(x,y,z)=x^2+y^2.

  1. Express the mass mm of the solid as a triple integral in cylindrical coordinates.

  2. Express the mass mm of the solid as a triple integral in spherical coordinates.

  3. Evaluate mm.

Answer

(a) ∫02πdθ∫03/2dr r∫3 r9−r2dz r2\int_0^{2\pi}\dee{\theta}\int_0^{3/2}\dee{r}\ r\int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}dz\ r^2

(b) ∫02πdθ∫0π/6dφ∫03dρ (ρ2sin⁡φ)(ρ2sin⁡2φ)\int_0^{2\pi}\dee{\theta}\int_0^{\pi/6}\dee{\varphi}\int_0^3\dee{\rho}\ \big(\rho^2\sin\varphi\big) \big(\rho^2\sin^2\varphi\big)

(c) 2π355[23−338]2\pi \frac{3^5}{5} \big[\frac{2}{3}-\frac{3\sqrt{3}}{8}\big]

Full solution

(a) In cylindrical coordinates, the density of is δ=x2+y2=r2\de=x^2+y^2=r^2, the bottom of the solid is at z=3x2+3y2=3 rz=\sqrt{3x^2+3y^2}=\sqrt{3}\,r and the top of the solid is at z=9−x2−y2=9−r2z=\sqrt{9-x^2-y^2}=\sqrt{9-r^2}. The top and bottom meet when

3 r=9−r2  ⟺  3r2=9−r2  ⟺  4r2=9  ⟺  r=32\begin{equation*} \sqrt{3}\,r=\sqrt{9-r^2}\iff 3r^2=9-r^2\iff 4r^2=9\iff r=\frac{3}{2} \end{equation*}

The mass is

m=∫02πdθ∫03/2dr r∫3 r9−r2dz r2⏞δ\begin{align*} m=\int_0^{2\pi}\dee{\theta}\int_0^{3/2}\dee{r}\ r\int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}dz\ \overbrace{r^2}^{\de} \end{align*}

Figure from prob_s3.7, line 2748

Figure from prob_s3.7, line 2748

(b) In spherical coordinates, the density of is δ=x2+y2=ρ2sin⁡2φ\de=x^2+y^2=\rho^2\sin^2\varphi, the bottom of the solid is at

z=3 r  ⟺  ρcos⁡φ=3 ρsin⁡φ  ⟺  tan⁡φ=13  ⟺  φ=π6\begin{equation*} z=\sqrt{3}\,r\iff \rho\cos\varphi=\sqrt{3}\,\rho\sin\varphi\iff \tan\varphi=\frac{1}{\sqrt{3}}\iff\varphi=\frac{\pi}{6} \end{equation*}

and the top of the solid is at x2+y2+z2=ρ2=9x^2+y^2+z^2=\rho^2=9. The mass is

m=∫02πdθ∫0π/6dφ∫03dρ (ρ2sin⁡φ)(ρ2sin⁡2φ)⏞δ\begin{equation*} m=\int_0^{2\pi}\dee{\theta}\int_0^{\pi/6}\dee{\varphi}\int_0^3\dee{\rho}\ \big(\rho^2\sin\varphi\big) \overbrace{\big(\rho^2\sin^2\varphi\big)}^{\de} \end{equation*}

(c) Solution 1: Making the change of variables s=cos⁡φs=\cos\varphi, ds=−sin⁡φ dφ\dee{s}=-\sin\varphi\ \dee{\varphi}, in the integral of part (b),

m=∫02πdθ∫0π/6dφ∫03dρ ρ4sin⁡φ(1−cos⁡2φ)=355∫02πdθ∫0π/6dφ sin⁡φ(1−cos⁡2φ)=−355∫02πdθ∫13/2ds (1−s2)=−355∫02πdθ [s−s33]13/2=2π355[1−13−32+38]=2π355[23−338]\begin{align*} m&=\int_0^{2\pi}\dee{\theta}\int_0^{\pi/6}\dee{\varphi} \int_0^3\dee{\rho}\ \rho^4\sin\varphi \big(1-\cos^2\varphi\big) \\ &=\frac{3^5}{5}\int_0^{2\pi}\dee{\theta}\int_0^{\pi/6}\dee{\varphi}\ \sin\varphi \big(1-\cos^2\varphi\big) \\ &=-\frac{3^5}{5}\int_0^{2\pi}\dee{\theta}\int_1^{\sqrt{3}/2}\dee{s}\ (1-s^2) \\ &=-\frac{3^5}{5}\int_0^{2\pi}\dee{\theta}\ \left[s-\frac{s^3}{3}\right]_1^{\sqrt{3}/2} \\ &=2\pi \frac{3^5}{5} \left[1-\frac{1}{3}-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{8}\right] \\ &=2\pi \frac{3^5}{5} \left[\frac{2}{3}-\frac{3\sqrt{3}}{8}\right] \end{align*}

(c) Solution 2:
As an alternate solution, we can also evaluate the integral of part (a).

m=∫02πdθ∫03/2dr r∫3 r9−r2dz r2=∫02πdθ∫03/2dr r3(9−r2−3 r)=2π∫03/2dr r3(9−r2−3 r)\begin{align*} m&=\int_0^{2\pi}\dee{\theta}\int_0^{3/2}\dee{r}\ r\int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}dz\ r^2\cr &=\int_0^{2\pi}\dee{\theta}\int_0^{3/2}\dee{r}\ r^3\big(\sqrt{9-r^2}-\sqrt{3}\,r\big)\cr &=2\pi\int_0^{3/2}\dee{r}\ r^3\big(\sqrt{9-r^2}-\sqrt{3}\,r\big) \end{align*}

The second term

−2π∫03/2dr 3 r4=−2π3 r55∣03/2=−2π3 355×25\begin{equation*} -2\pi\int_0^{3/2}\dee{r}\ \sqrt{3}\,r^4=-2\pi\sqrt{3}\,\frac{r^5}{5}\bigg|_0^{3/2} =-2\pi\sqrt{3}\,\frac{3^5}{5\times 2^5} \end{equation*}

For the first term, we substitute s=9−r2s=9-r^2, ds=−2r dr\dee{s}=-2r\,\dee{r}.

2π∫03/2dr r39−r2=2π∫927/4ds−2⏞r dr(9−s)⏞r2s=−π[6s3/2−25s5/2]927/4=−π[3534−2×34−372453+2355]\begin{align*} 2\pi\int_0^{3/2}\dee{r}\ r^3\sqrt{9-r^2} &=2\pi\int_9^{27/4}\overbrace{\frac{\dee{s}}{-2}}^{r\,\dee{r}} \overbrace{(9-s)}^{r^2} \sqrt{s} =-\pi \left[6s^{3/2}-\frac{2}{5}s^{5/2}\right]_9^{27/4}\cr &=-\pi \left[\frac{3^5\sqrt{3}}{4}-2\times 3^4 -\frac{3^7}{2^4 5}\sqrt{3}+2\frac{3^5}{5}\right] \end{align*}

Adding the two terms together,

m=−2π355332−2π355538+2π35553+2π3559332−2π355=2π355[(53−1)−3(132+58−932)]=2π355[23−338]\begin{align*} m&=-2\pi\frac{3^5}{5} \frac{\sqrt{3}}{32} -2\pi\frac{3^5}{5} \frac{5\sqrt{3}}{8} +2\pi\frac{3^5}{5} \frac{5}{3} +2\pi\frac{3^5}{5} \frac{9\sqrt{3}}{32} -2\pi\frac{3^5}{5} \\ &=2\pi\frac{3^5}{5}\left[\left(\frac{5}{3}-1\right) -\sqrt{3}\left(\frac{1}{32}+\frac{5}{8}-\frac{9}{32}\right)\right] \\ &=2\pi \frac{3^5}{5} \left[\frac{2}{3}-\frac{3\sqrt{3}}{8}\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.