Tutorial 11

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Topics Covered

This tutorial covers the following topics using Python and SymPy:

• Polar coordinates: converting a Cartesian integral and evaluating it
• Volume below a surface and between two cylinders
• Spherical coordinates: volume between a sphere and a cone
• Area of a polar region $r=f(\theta)$
• Cylindrical coordinates: curved wedges and setting up triple integrals
• Mass with a variable density

Introduction

The three coordinate systems and their volume elements:

System	Substitution	$dV$
Polar	$x=r\cos\theta,\ y=r\sin\theta$	$dA=r\,dr\,d\theta$
Cylindrical	$x=r\cos\theta,\ y=r\sin\theta,\ z=z$	$dV=r\,dz\,dr\,d\theta$
Spherical	$x=\rho\sin\phi\cos\theta,\ y=\rho\sin\phi\sin\theta,\ z=\rho\cos\phi$	$dV=\rho^{2}\sin\phi\,d\rho\,d\phi\,d\theta$

The factor $r$ (or $\rho^{2}\sin\phi$) is the Jacobian — forgetting it is the single most common mistake in this chapter.

import sympy as sy
from sympy.abc import x, y, z, t, r, theta, phi, rho, a, H, k
from sympy import (sqrt, sin, cos, tan, sec, pi, ln, exp, log, integrate, simplify,
                   diff, Rational, N, Matrix, Symbol)
sy.init_printing(use_latex=True)

Question 1

Convert each Cartesian integral to polar coordinates and evaluate.

a. $\displaystyle\int_{-1}^{1}\int_{0}^{\sqrt{1-x^{2}}}dy\,dx$

The region is the upper half of the unit disc: $0\le r\le1$, $0\le\theta\le\pi$.

I = integrate(r, (r, 0, 1), (theta, 0, pi))
display(I)                                       # pi/2

b. $\displaystyle\int_{0}^{2}\int_{0}^{x}y\,dy\,dx$

The region is the triangle $0\le y\le x$, $0\le x\le2$, i.e. $0\le\theta\le\pi/4$.

display(integrate(y, (y, 0, x), (x, 0, 2)))      # 4/3 (direct)
I = integrate(r*sin(theta)*r, (r, 0, 2/cos(theta)), (theta, 0, pi/4))
display(simplify(I))                             # 4/3 (polar)

c. $\displaystyle\int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\frac{2}{\left(1+x^{2}+y^{2}\right)^{2}}dy\,dx$

The region is the whole unit disc, and $1+x^{2}+y^{2}=1+r^{2}$.

I = integrate(2/(1 + r**2)**2 * r, (r, 0, 1), (theta, 0, 2*pi))
display(simplify(I))                             # pi

d. $\displaystyle\int_{0}^{\ln 2}\int_{0}^{\sqrt{(\ln 2)^{2}-y^{2}}}e^{\sqrt{x^{2}+y^{2}}}\,dx\,dy$

The region is the part of the disc of radius $\ln 2$ in the first quadrant, so $0\le r\le\ln 2$ and $0\le\theta\le\pi/2$, and $e^{\sqrt{r^{2}}}=e^{r}$.

I = integrate(exp(r)*r, (r, 0, ln(2)), (theta, 0, pi/2))
display(simplify(I))                             # pi*(log(2) - 1/2)
display(simplify(I - pi/2*(ln(4) - 1)))          # 0

$I=\frac{\pi}{2}\left(\ln 4-1\right)$

Question 2

$\displaystyle\iint\left(1-x^{2}-y^{2}\right)dA$ over the unit disc.

I = integrate((1 - r**2)*r, (r, 0, 1), (theta, 0, 2*pi))
display(simplify(I))                             # pi/2

Question 3

Volume below $z=\dfrac{y^{2}}{x^{2}+y^{2}}$, above the $xy$-plane, between $x^{2}+y^{2}=1$ and $x^{2}+y^{2}=2$.

In polar coordinates $z=\sin^{2}\theta$, so the volume is $\int_{0}^{2\pi}\int_{1}^{\sqrt{2}}\sin^{2}\theta\;r\,dr\,d\theta$.

I = integrate(sin(theta)**2*r, (r, 1, sqrt(2)), (theta, 0, 2*pi))
display(simplify(I))                             # pi/2

Question 4

Volume between the sphere $x^{2}+y^{2}+z^{2}=1$ and the cone $z=\sqrt{x^{2}+y^{2}}$.

In spherical coordinates the cone is $\phi=\pi/4$, so $0\le\theta\le2\pi$, $0\le\phi\le\pi/4$, $0\le\rho\le1$.

V = integrate(rho**2*sin(phi), (rho, 0, 1), (phi, 0, pi/4), (theta, 0, 2*pi))
display(simplify(V))                             # pi*(2 - sqrt(2))/3

$V=\frac{\pi}{3}\left(2-\sqrt{2}\right)$

Question 5

Area of the region bounded by $r=3\cos\theta$.

The curve is the circle $\left(x-\frac{3}{2}\right)^{2}+y^{2}=\frac{9}{4}$, traced for $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$; by symmetry we integrate from $0$ to $\frac{\pi}{2}$ and double.

A = 2*integrate(r, (r, 0, 3*cos(theta)), (theta, 0, pi/2))
display(simplify(A))                             # 9*pi/4

note

A naive $0\le\theta\le\pi$ setup does not describe the region correctly, because $r=3\cos\theta<0$ for $\frac{\pi}{2}<\theta\le\pi$ while $r$ must be non-negative. It is still worth knowing that this setup does not give $0$: the inner integral is a square, $\int_{0}^{3\cos\theta}r\,dr=\frac{9}{2}\cos^{2}\theta$, so a negative upper limit cannot make it vanish. Do not rely on a sign argument here — split the region and double it.

Question 6

$\displaystyle\iiint y\,dV$ over the solid bounded by $z=4-x^{2}-y^{2}$ in the first octant.

# in cylindrical coordinates y = r*sin(theta) and dV = r dz dr dtheta
V = integrate(r*sin(theta)*r, (z, 0, 4 - r**2), (r, 0, 2), (theta, 0, pi/2))
display(simplify(V))                             # 64/15

Note the integrand in cylindrical coordinates: $y\,dV=y\cdot r\,dz\,dr\,d\theta=r\sin\theta\cdot r\,dz\,dr\,d\theta$.

Question 7

Cylindrical coordinates: volume of the wedge cut from the cylinder $(x-2)^{2}+y^{2}=4$ by $z=0$ and $z=-y$.

The cylinder becomes $r=4\cos\theta$ and $z\le-y$ forces $y\le0$, so $\frac{3\pi}{2}\le\theta\le2\pi$.

V = integrate(r, (z, 0, -r*sin(theta)), (r, 0, 4*cos(theta)), (theta, 3*pi/2, 2*pi))
display(simplify(V))                             # 16/3

Question 8

Set up (do not evaluate) the triple integral for the region $E$ inside $r=2\sin\theta$, above $z=0$ and below the sphere of radius $4$.

The bounds are $0\le\theta\le\pi$, $0\le r\le2\sin\theta$ and $0\le z\le\sqrt{16-r^{2}}$, so

$$\iiint_{E}f(r,\theta,z)\,dV=\int_{0}^{\pi}\int_{0}^{2\sin\theta}\int_{0}^{\sqrt{16-r^{2}}} f(r,\theta,z)\,r\,dz\,dr\,d\theta$$

ztop = sqrt(16 - r**2)
display(ztop)                                    # sqrt(16 - r**2) : the upper z-limit
display(sy.Integral(sy.Function('f')(r, theta, z)*r, (z, 0, ztop), (r, 0, 2*sin(theta)), (theta, 0, pi)))

Question 9

Volume of the solid bounded by $z=2$ and $z=\sqrt{x^{2}+y^{2}}$.

V = integrate(r, (z, r, 2), (r, 0, 2), (theta, 0, 2*pi))
display(simplify(V))                             # 8*pi/3

Question 10

Spherical coordinates: volume outside $\rho=2\cos\phi$ and inside $\rho=2$ with $\phi\in[0,\pi/2]$.

V = integrate(rho**2*sin(phi), (rho, 2*cos(phi), 2), (phi, 0, pi/2), (theta, 0, 2*pi))
display(simplify(V))                             # 4*pi

Question 11

Mass of the solid bounded by $z=2$ and $z=\sqrt{x^{2}+y^{2}}$ when the density is proportional to the square of the distance from the origin, $\delta=k\rho^{2}$.

M = integrate(k*rho**2 * rho**2*sin(phi), (rho, 0, 2/cos(phi)), (phi, 0, pi/4), (theta, 0, 2*pi))
display(simplify(M))                             # 48*k*pi/5

$M=\frac{48k\pi}{5}$

Question 12

Mass of $T$, bounded by $y=x^{2}+z^{2}$ and $y=4$, with density $\rho(x,y,z)=y$.

Using cylindrical coordinates with the axis along $y$, we have $x=r\cos\theta$, $z=r\sin\theta$ and $dV=r\,dy\,dr\,d\theta$ with $r^{2}\le y\le4$.

M = integrate(y*r, (y, r**2, 4), (r, 0, 2), (theta, 0, 2*pi))
display(simplify(M))                             # 64*pi/3

$M=\frac{64\pi}{3}$

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Extra Learning Resources

Key Concepts to Master

1. Always include the Jacobian: $r$ for polar/cylindrical, $\rho^{2}\sin\phi$ for spherical
2. Recognising a disc: $x^{2}+y^{2}\le a^{2}$ becomes $0\le r\le a$, $0\le\theta\le2\pi$
3. Circles through the origin: $r=2a\cos\theta$ (centre on the $x$-axis) and $r=2a\sin\theta$ (centre on the $y$-axis)
4. Cone $\to$ spherical: $z=\sqrt{x^{2}+y^{2}}$ is $\phi=\pi/4$
5. Plane $z=c$ in spherical: $\rho=c\sec\phi$

SymPy Resources

• SymPy integrate — nested integrals with variable limits
• SymPy simplify — clean up trig-heavy results
• SymPy Integral — display an integral without evaluating it (Q8)

Common Pitfalls

• Forgetting the Jacobian $r$ or $\rho^{2}\sin\phi$ — the most common error in this chapter
• $r$ must be non-negative; a region such as $r=3\cos\theta$ only exists where $\cos\theta\ge0$
• In spherical coordinates the polar angle is $\phi$ (from the $z$-axis) and the azimuthal angle is $\theta$ — don't swap them
• When the symmetry axis is not $z$ (Q12), relabel the cylindrical coordinates accordingly