Module 40: Triple integrals and cylindrical and spherical coordinates

8 exercises, each with a complete solution. 1 figure.

On this page

A. Prerequisites and observable goals
B. The central idea: integrate contributions throughout a solid
C. Coordinate conventions that must be explicit
D. Why the spherical factor contains two powers and a sine
E. Build the solid before computing
F. Worked examples
G. Independent practice
H. Complete solutions: Module 40
I. Retention and readiness

A. Prerequisites and observable goals#

Before

Double integrals, spatial regions, polar coordinates, and density-weighted moments.

After

Describe a solid with nested bounds, choose cylindrical or spherical coordinates, and compute volume, mass, or moments with the correct volume element.

B. The central idea: integrate contributions throughout a solid#

A triple integral $\iiint_Ef\dd V$ adds tiny-volume contributions. Integrating $1$ gives volume; integrating a mass density gives mass. The solid and the integrand jointly determine the most useful coordinates. Cylindrical coordinates suit rotational symmetry around an axis; spherical coordinates suit distances and angles from a point.

C. Coordinate conventions that must be explicit#

Cartesian volume is $\dd V=\dd x\dd y\dd z$ in an appropriately nested order. Cylindrical coordinates use

x=r\cos\theta,\quad y=r\sin\theta,\quad z=z, \qquad \dd V=r\dd r\dd\theta\dd z.

The symbol $r$ is distance from the $z$ -axis, not distance from the origin.

Our spherical convention uses $\rho\ge0$ for distance from the origin, $\theta$ for azimuth in the $xy$ -plane, and $0\le\phi\le\pi$ for angle down from the positive $z$ -axis:

x=\rho\sin\phi\cos\theta,\quad y=\rho\sin\phi\sin\theta,\quad z=\rho\cos\phi,

\dd V=\rho^2\sin\phi\dd\rho\dd\phi\dd\theta.

Some texts interchange angle names; the formulas and geometric definitions, not the letters alone, determine correctness. The factors $r$ and $\rho^2\sin\phi$ account for volume scaling. Coordinate and integral conventions: [O3Triple].

For density $\delta\ge0$ , $m=\iiint_E\delta\dd V$ and $\bar z=(1/m)\iiint_E z\delta\dd V$ , with analogous $\bar x,\bar y$ . The moment of inertia about the $z$ -axis is $I_z=\iiint_E(x^2+y^2)\delta\dd V$ : distance from the axis is squared, not distance from the origin.

D. Why the spherical factor contains two powers and a sine#

A small spherical cell has approximate side lengths $\Delta\rho$ , $\rho\Delta\phi$ , and $\rho\sin\phi\Delta\theta$ . Their product gives $\rho^2\sin\phi\Delta\rho\Delta\phi\Delta\theta$ . Near the poles, a fixed azimuth change sweeps a smaller distance, explaining the sine. This geometric argument is supported by the Jacobian determinant in the next module.

A cone and a sphere simplify in these coordinates for different reasons. The cone $z=\sqrt{x^2+y^2}$ becomes $\rho\cos\phi=\rho\sin\phi$ , or $\phi=\pi/4$ away from the origin. The sphere $x^2+y^2+z^2=R^2$ becomes $\rho=R$ . Their intersection can therefore produce constant angular and radial bounds.

E. Build the solid before computing#

Describe its projections and bounding surfaces. Choose an outer variable range and an inner interval that depends only on variables not yet integrated. For cylindrical coordinates, decide whether $z$ or $r$ is best innermost. For spherical coordinates, draw the polar angle from the positive $z$ -axis and convert every inequality. Include the volume factor, then inspect whether symmetry makes odd moments zero. Never use symmetry across a plane that the solid or density does not actually possess.

F. Worked examples#

A ball volume. A radius- $R$ ball uses $0\le\rho\le R$ , $0\le\phi\le\pi$ , $0\le\theta\le2\pi$ . Thus

V=\int_0^{2\pi}\int_0^\pi\int_0^R\rho^2\sin\phi\dd\rho\dd\phi\dd\theta =(2\pi)(2)(R^3/3)=\frac{4\pi R^3}3.

Recovering a known volume is a useful check on the Jacobian and angular ranges.

A region under a paraboloid. For $0\le z\le4-x^2-y^2$ , the base disk has $r\le2$ . Cylindrical bounds are $0\le\theta\le2\pi$ , $0\le r\le2$ , $0\le z\le4-r^2$ . Volume is $\int_0^{2\pi}\int_0^2(4-r^2)r\dd r\dd\theta=8\pi$ . The upper height must be nonnegative, which determines the radial boundary.

Schematic three-dimensional projection of spherical coordinates with an exact point at

\rho=2

\phi=\pi/4

\theta=\pi/3

. The polar angle is measured from the positive

z

-axis. Projection lengths on the page are not the spatial coordinate lengths.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

M40.01FSolution

Evaluate $\iiint_{[0,1]^3}(x+y+z)\dd V$ .

M40.02CSolution

Find the volume of the tetrahedron $x,y,z\ge0$ , $x+y+z\le1$ using Cartesian iterated integration.

M40.03CSolution

Evaluate $\iiint_E z\dd V$ over the cylinder $x^2+y^2\le4$ , $0\le z\le3$ .

M40.04TSolution

Evaluate $\iiint_{x^2+y^2+z^2\le4}(x^2+y^2+z^2)\dd V$ in spherical coordinates.

M40.05TSolution

Find the centroid of the uniform upper half of the unit ball.

M40.06TSolution

Find the volume inside the radius-two sphere and above the cone $z=\sqrt{x^2+y^2}$ .

M40.07CSolution

For the solid $0\le z\le1$ , $0\le r\le z$ , $0\le\theta\le2\pi$ , density is $\delta=z$ . Find mass and $\bar z$ .

M40.08SSolution

Reverse the nested description $0\le z\le1$ , $0\le x\le z$ , $0\le y\le x$ to an order with $y$ outermost and $z$ innermost. Evaluate the volume in the new order.

Optional small hints

For M40.05, symmetry gives two coordinates, not the vertical one. For M40.06, above the cone means $0\le\phi\le\pi/4$ . For M40.08, rewrite the original inequalities as one chain $0\le y\le x\le z\le1$ .

H. Complete solutions: Module 40#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

M40.01Show solution

Separate the three symmetric contributions

Over the unit cube, $\iiint x\dd V=(\int_0^1x\dd x)(1)(1)=1/2$ . The $y$ and $z$ contributions are the same by symmetry, so the total is $\boxed{3/2}$ . This equals the cube volume times the affine function’s value at its center.

Back to M40.01

M40.02Show solution

The inner endpoint comes from the remaining plane height

Use $0\le x\le1$ , $0\le y\le1-x$ , and $0\le z\le1-x-y$ . Integrating in $z$ gives $1-x-y$ , then in $y$ gives $(1-x)^2/2$ . The outer integral is $\tfrac12\int_0^1(1-x)^2\dd x=\boxed{1/6}$ . All nested upper bounds are nonnegative on their stated outer ranges.

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M40.03Show solution

A constant horizontal cross-section separates the variables

Cylindrical coordinates give $I=\int_0^{2\pi}\int_0^2\int_0^3zr\dd z\dd r\dd\theta=(2\pi)(2)(9/2)=\boxed{18\pi}$ . The cylinder’s volume is $12\pi$ and its mean height is $3/2$ , giving the same first volume moment.

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M40.04Show solution

The radial integrand multiplies the radial volume factor

The integrand becomes $\rho^2$ and the volume element contributes another $\rho^2\sin\phi$ . Therefore

I=\int_0^{2\pi}\int_0^\pi\int_0^2\rho^4\sin\phi\dd\rho\dd\phi\dd\theta =(2\pi)(2)(32/5)=\boxed{128\pi/5}.

This is not a surface integral over the sphere: every radius from zero to two is included.

Back to M40.04

M40.05Show solution

Use symmetry for two moments and integrate the third

Symmetry gives $\bar x=\bar y=0$ . The hemisphere volume is $2\pi/3$ . Its $z$ moment in spherical coordinates is

M=\int_0^{2\pi}\int_0^{\pi/2}\int_0^1 (\rho\cos\phi)\rho^2\sin\phi\dd\rho\dd\phi\dd\theta =(2\pi)(1/2)(1/4)=\pi/4.

Thus $\boxed{\bar z=(\pi/4)/(2\pi/3)=3/8}$ . The centroid lies above the flat base and below the sphere’s top, as required.

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M40.06Show solution

Translate the cone into a polar-angle bound

Inside the sphere means $0\le\rho\le2$ . Above the cone means $z\ge r$ , hence $\cos\phi\ge\sin\phi$ in the upper half-space, giving $0\le\phi\le\pi/4$ . The azimuth covers $[0,2\pi]$ . Therefore $V=(2\pi)(8/3)\int_0^{\pi/4}\sin\phi\dd\phi =\boxed{\frac{16\pi}{3}(1-\frac{\sqrt2}{2})}$ . The origin’s angle ambiguity has no effect on the volume integral.

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M40.07Show solution

A growing cross-section and growing density both weight the top

At height $z$ , the cross-section has area $\pi z^2$ . Mass is $m=\int_0^1z(\pi z^2)\dd z=\boxed{\pi/4}$ . The $z$ moment adds another factor $z$ , giving $\pi\int_0^1z^4\dd z=\pi/5$ . Thus $\boxed{\bar z=4/5}$ . The centroid’s high position reflects both the widening solid and the increasing density.

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M40.08Show solution

Keep the same ordered inequalities

The original solid is $0\le y\le x\le z\le1$ . With $y$ outermost, use $0\le y\le1$ , $y\le x\le1$ , and $x\le z\le1$ . Thus

V=\int_0^1\int_y^1\int_x^1 1\dd z\dd x\dd y =\int_0^1\frac{(1-y)^2}{2}\dd y=\boxed{1/6}.

The order change is a new description of one solid, not a permutation of the old limits attached to new variable names.

Back to M40.08

I. Retention and readiness#

Retrieve M39.08 and M19.07. Explain whether $r$ means axial distance or $\rho$ means distance from the origin. Advance when your bounds reconstruct the intended cone, sphere, or cylinder and your units distinguish volume from mass or moment.