Chapter 16

Hard

Multiple Integrals

00 · Symbol Glossary

$\iint_D f(x,y)\,dA$double integral over a region D

The extension of Chapter 06's definite integral to two dimensions: sums up $f(x,y)$ times a tiny patch of area $dA$ , over every patch in a planar region $D$ . Where $\int_a^b f(x)\,dx$ computes area under a curve, $\iint_D f(x,y)\,dA$ computes volume under a surface.

$dA$d A — area element

An infinitesimal patch of area, $dA = dx\,dy$ in rectangular coordinates. Section 03 shows this changes form in polar coordinates.

$\int_a^b\!\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx$iterated integral

A double integral computed one variable at a time: the inner integral (with respect to $y$ ) is evaluated first, treating $x$ as fixed, producing a function of $x$ alone; the outer integral then integrates that result over $x$ .

$\iiint_E f(x,y,z)\,dV$triple integral over a solid E

The three-dimensional extension: sums $f(x,y,z)$ times a tiny patch of volume $dV$ over a solid region $E$ in space.

$(r,\theta,z)$cylindrical coordinates

A 3D coordinate system: $(r,\theta)$ locate a point's shadow in the $xy$ -plane exactly as in Chapter 13's polar coordinates, and $z$ is the ordinary height above that shadow.

$(\rho,\theta,\phi)$rho, theta, phi — spherical coordinates

A 3D coordinate system built around distance from the origin: $\rho$ is distance from the origin, $\theta$ is the same azimuthal angle as cylindrical coordinates, and $\phi$ is the angle measured down from the positive $z$ -axis — a new angle, distinct from $\theta$ .

01 · Double Integrals and Iterated Integrals

Chapter 06 built the definite integral by slicing a region under a curve into thin rectangles and summing. The same idea, one dimension up, slices a solid under a surface $z=f(x,y)$ into thin boxes.

Definition — Double Integral

For $f(x,y)\geq0$ on a region $D$ , $\displaystyle\iint_D f(x,y)\,dA$ is the volume of the solid between the surface $z=f(x,y)$ and the $xy$ -plane, over $D$ — the limit of a sum of box volumes $f(x_i^*,y_j^*)\,\Delta A$ , exactly mirroring Chapter 06, Section 01's Riemann sum, but with 2D patches $\Delta A$ instead of 1D strips $\Delta x$ .

Definition — Fubini's Theorem (Iterated Integrals)

If $D$ is the rectangle $a\leq x\leq b$ , $c\leq y\leq d$ , and $f$ is continuous on $D$ :

\iint_D f(x,y)\,dA = \int_a^b\int_c^d f(x,y)\,dy\,dx = \int_c^d\int_a^b f(x,y)\,dx\,dy

The double integral can be computed as two ordinary single-variable integrals (Chapter 06), one after another, in either order.

Step-by-step — Evaluate $\displaystyle\iint_D (x^2+y)\,dA$ over $D=[0,2]\times[0,3]$

Set up the iterated integral, integrating with respect to $y$ first (inner):

\displaystyle\int_0^2\left[\int_0^3 (x^2+y)\,dy\right]dx

Evaluate the inner integral, treating $x$ as a constant (matching Chapter 15's partial-derivative convention, run in reverse for integration):

\displaystyle\int_0^3(x^2+y)\,dy = \left[x^2y+\frac{y^2}{2}\right]_0^3 = \left(3x^2+\frac92\right)-0 = 3x^2+\frac92

Substitute the result into the outer integral:

\displaystyle\int_0^2\left(3x^2+\frac92\right)dx

Evaluate the outer integral (Chapter 06 power rule for integration):

\left[x^3+\frac92x\right]_0^2 = \left(8+9\right)-0 = 17

Conclude:

\displaystyle\iint_D(x^2+y)\,dA = 17

02 · Double Integrals over General Regions

Most regions of interest are not rectangles — they're bounded above and below by curves. The bounds of the inner integral are then allowed to depend on the outer variable.

Definition — Type I Region

A Type I region is $D = \{(x,y) \mid a\leq x\leq b,\ g_1(x)\leq y\leq g_2(x)\}$ — for each fixed $x$ , $y$ ranges between two curves.

\iint_D f(x,y)\,dA = \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx

Step-by-step — Evaluate $\displaystyle\iint_D 2xy\,dA$ where $D$ is bounded by $y=x^2$ and $y=x$, for $0\leq x\leq1$

Determine which curve is on top: for

0\leq x\leq1

x\geq x^2

(e.g. at

x=0.5

0.5>0.25

), so

y

ranges from

g_1(x)=x^2

(bottom) to

g_2(x)=x

(top).

Set up the iterated integral:

\displaystyle\int_0^1\int_{x^2}^{x} 2xy\,dy\,dx

Evaluate the inner integral with respect to $y$ , treating $x$ as constant:

\displaystyle\int_{x^2}^{x} 2xy\,dy = \left[xy^2\right]_{x^2}^{x} = x(x)^2 - x(x^2)^2 = x^3-x^5

Evaluate the outer integral:

\displaystyle\int_0^1(x^3-x^5)\,dx = \left[\frac{x^4}{4}-\frac{x^6}{6}\right]_0^1 = \left(\frac14-\frac16\right)-0 = \frac{3}{12}-\frac{2}{12}=\frac{1}{12}

03 · Double Integrals in Polar Coordinates

A region with circular symmetry is far easier to describe using Chapter 13's polar coordinates than rectangular ones — but the area element itself changes shape.

Definition — Area Element in Polar Coordinates

dA = r\,dr\,d\theta

Where the extra factor of r comes from

A small polar "rectangle" spanning $dr$ in radius and $d\theta$ in angle is not a true rectangle — its two arc-length sides have lengths $dr$ (radial) and $r\,d\theta$ (the arc length of a circle of radius $r$ subtending angle $d\theta$ , matching Chapter 13, Section 04's sector-area reasoning). Multiplying these two side lengths gives $dr\cdot r\,d\theta = r\,dr\,d\theta$ — the factor of $r$ is not optional, it's the actual width of the arc side of the patch.

Step-by-step — Find the volume under $z=x^2+y^2$ over the disk $x^2+y^2\leq4$

Convert the integrand to polar coordinates (Chapter 13):

x^2+y^2 = r^2

Convert the region: the disk

x^2+y^2\leq4

becomes

0\leq r\leq2

0\leq\theta\leq2\pi

(the full disk of radius

2

Set up the polar double integral, including the extra factor of $r$ from Section 03's area element:

\displaystyle\int_0^{2\pi}\int_0^2 r^2\cdot r\,dr\,d\theta = \int_0^{2\pi}\int_0^2 r^3\,dr\,d\theta

Evaluate the inner integral:

\displaystyle\int_0^2 r^3\,dr = \left[\frac{r^4}{4}\right]_0^2 = \frac{16}{4}-0=4

Evaluate the outer integral:

\displaystyle\int_0^{2\pi} 4\,d\theta = 4(2\pi)=8\pi

❌ Forgetting the extra factor of r when converting to polar

On the same problem, converting $dA$ to just $dr\,d\theta$ (dropping the $r$ ): $\displaystyle\int_0^{2\pi}\int_0^2 r^2\,dr\,d\theta = \int_0^{2\pi}\left[\frac{r^3}{3}\right]_0^2 d\theta = \int_0^{2\pi}\frac83\,d\theta = \frac{16\pi}{3}$ .

Why it breaks: as Section 03's note explains, the polar area element is $r\,dr\,d\theta$ , not $dr\,d\theta$ — a bare $dr\,d\theta$ patch does not correctly represent an area element at all except right at $r=1$ ; everywhere else, the true patch is wider or narrower than $dr\,d\theta$ by exactly the missing factor of $r$ .

Consequence: $\dfrac{16\pi}{3}\approx16.76 \neq 8\pi\approx25.13$ (the correct answer from step 5) — omitting the Jacobian-like scaling factor $r$ silently shrinks patches near the outer edge of the disk (where they should count for more, being wider arcs) relative to patches near the center, systematically undercounting the volume.

04 · Triple Integrals

Adding a third dimension extends the same idea once more: summing $f(x,y,z)$ over tiny volume patches inside a solid region.

Definition — Triple Integral

\iiint_E f(x,y,z)\,dV = \int_{a}^{b}\int_{h_1(x)}^{h_2(x)}\int_{g_1(x,y)}^{g_2(x,y)} f(x,y,z)\,dz\,dy\,dx

for an appropriately described solid $E$ ; in rectangular coordinates, $dV=dz\,dy\,dx$ (or any order of the three differentials).

Step-by-step — Evaluate $\displaystyle\iiint_E z\,dV$ over the box $E=[0,1]\times[0,2]\times[0,3]$

Set up the triple iterated integral, integrating $z$ first:

\displaystyle\int_0^1\int_0^2\int_0^3 z\,dz\,dy\,dx

Innermost integral, with respect to $z$ :

\displaystyle\int_0^3 z\,dz = \left[\frac{z^2}{2}\right]_0^3 = \frac92

Middle integral, a constant $\frac92$ integrated with respect to $y$ from $0$ to $2$ :

\displaystyle\int_0^2 \frac92\,dy = \frac92(2) = 9

Outer integral, a constant $9$ integrated with respect to $x$ from $0$ to $1$ :

\displaystyle\int_0^1 9\,dx = 9

05 · Cylindrical Coordinates — Real Depth

When a solid region has symmetry around a single axis (a cylinder, a cone, a paraboloid opening along the $z$ -axis), rectangular coordinates force awkward bounds. Cylindrical coordinates describe such solids far more naturally — they are simply Chapter 13's polar coordinates in the $xy$ -plane, with an ordinary $z$ -height stacked on top.

Definition — Cylindrical Coordinates

x = r\cos\theta \qquad y = r\sin\theta \qquad z = z

$r\geq0$ is distance from the $z$ -axis (not from the origin); $\theta$ is the same azimuthal angle as Chapter 13's polar coordinates; $z$ is unchanged from rectangular coordinates.

Definition — Volume Element in Cylindrical Coordinates

dV = r\,dr\,d\theta\,dz

Deriving the cylindrical volume element geometrically

A small cylindrical "box" spans $dr$ radially, $r\,d\theta$ along the arc (Section 03's polar arc-length reasoning, carried straight up through every height), and $dz$ vertically. These three edges are mutually perpendicular at each point, so the box's volume is simply the product of its three edge lengths: $dr\times r\,d\theta\times dz = r\,dr\,d\theta\,dz$ — the same factor of $r$ from polar area, now multiplied by the height $dz$ .

Step-by-step — Find the volume of the solid cone $z=\sqrt{x^2+y^2}$ under $z=3$ (inside the cone, below height 3)

Convert the cone's equation to cylindrical coordinates:

z=\sqrt{x^2+y^2} = \sqrt{r^2} = r

(since

r\geq0

). The solid is bounded below by

z=r

(the cone surface) and above by

z=3

Determine the range of $r$ : the cone meets the plane

z=3

where

r=3

, and the solid region extends from the central axis outward, so

0\leq r\leq3

Determine the range of $\theta$ : the solid is symmetric all the way around the axis, so

0\leq\theta\leq2\pi

Determine the range of $z$ for fixed $r,\theta$ : from the cone surface

z=r

up to the flat top

z=3

, so

r\leq z\leq3

Set up the triple integral for volume ( $f=1$ ), using the cylindrical volume element:

\displaystyle\int_0^{2\pi}\int_0^3\int_r^3 r\,dz\,dr\,d\theta

Innermost integral, with respect to $z$ (integrand $r$ is constant here):

\displaystyle\int_r^3 r\,dz = r(3-r) = 3r-r^2

Middle integral, with respect to $r$ :

\displaystyle\int_0^3(3r-r^2)\,dr = \left[\frac{3r^2}{2}-\frac{r^3}{3}\right]_0^3 = \left(\frac{27}{2}-9\right)-0 = \frac{27}{2}-\frac{18}{2}=\frac92

Outer integral, with respect to $\theta$ :

\displaystyle\int_0^{2\pi} \frac92\,d\theta = \frac92(2\pi) = 9\pi

Check against the elementary cone-volume formula: a cone of base radius

3

and height

3

has volume

\frac13\pi r^2 h = \frac13\pi(9)(3) = 9\pi

— matches exactly. ✓

06 · Spherical Coordinates — Real Depth

For solids with symmetry around a single point rather than an axis (a solid ball, a region bounded by a sphere), a third coordinate system built entirely around distance-from-origin is more natural still.

Definition — Spherical Coordinates

x = \rho\sin\phi\cos\theta \qquad y = \rho\sin\phi\sin\theta \qquad z = \rho\cos\phi

$\rho\geq0$ is distance from the origin (not from the $z$ -axis, unlike cylindrical $r$ ). $\theta$ is the same azimuthal angle as before. $\phi$ (with $0\leq\phi\leq\pi$ ) is the angle measured from the positive $z$ -axis down to the point — $\phi=0$ points straight up, $\phi=\pi/2$ lies in the $xy$ -plane, $\phi=\pi$ points straight down.

Why two different angles are needed, and why they mean different things

$\theta$ (shared with cylindrical and polar coordinates) sweeps a point around the $z$ -axis, like a line of longitude on a globe. $\phi$ sweeps a point from pole to pole, like a line of latitude, but measured from the top pole rather than from the equator. Confusing the two — or reusing $\theta$ 's role for $\phi$ — mislabels which direction a change in coordinate actually moves the point.

Definition — Volume Element in Spherical Coordinates

dV = \rho^2\sin\phi\,d\rho\,d\theta\,d\phi

Deriving the spherical volume element geometrically

A small spherical "box" has three edges: $d\rho$ radially outward; $\rho\,d\phi$ along a meridian (an arc of a great circle of radius $\rho$ , subtending angle $d\phi$ — the same arc-length idea as Chapter 13, Section 04); and $\rho\sin\phi\,d\theta$ around a parallel — a circle of latitude has radius $\rho\sin\phi$ (shrinking toward $0$ near the poles, where $\sin\phi\to0$ ), not the full $\rho$ , so its arc length for angle $d\theta$ is $\rho\sin\phi\,d\theta$ . Multiplying the three mutually perpendicular edge lengths: $d\rho\times\rho\,d\phi\times\rho\sin\phi\,d\theta = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ .

Step-by-step — Find the volume of a solid ball of radius 2 using spherical coordinates

Describe the ball in spherical coordinates: every point with

\rho\leq2

lies inside the ball, with

\theta

and

\phi

free to range over their entire natural spans:

0\leq\rho\leq2

0\leq\theta\leq2\pi

0\leq\phi\leq\pi

Set up the triple integral for volume ( $f=1$ ), using the spherical volume element:

\displaystyle\int_0^{2\pi}\int_0^{\pi}\int_0^2 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta

Innermost integral, with respect to $\rho$ :

\displaystyle\int_0^2 \rho^2\,d\rho = \left[\frac{\rho^3}{3}\right]_0^2 = \frac83

Middle integral, with respect to $\phi$ (the constant $\frac83$ carries through):

\displaystyle\int_0^\pi \frac83\sin\phi\,d\phi = \frac83\big[-\cos\phi\big]_0^\pi = \frac83\big(-\cos\pi-(-\cos0)\big) = \frac83(1+1) = \frac{16}{3}

Outer integral, with respect to $\theta$ :

\displaystyle\int_0^{2\pi}\frac{16}{3}\,d\theta = \frac{16}{3}(2\pi) = \frac{32\pi}{3}

Check against the elementary sphere-volume formula:

\frac43\pi r^3 = \frac43\pi(2)^3 = \frac43\pi(8)=\frac{32\pi}{3}

— matches exactly. ✓

❌ Forgetting the sin φ factor in the spherical volume element

On the same solid ball, a common mistake: writing $dV=\rho^2\,d\rho\,d\theta\,d\phi$ , omitting $\sin\phi$ .

Why it breaks: as the derivation above shows, the arc length around a parallel of latitude is $\rho\sin\phi\,d\theta$ , not $\rho\,d\theta$ — the radius of that circle shrinks toward the poles by exactly the factor $\sin\phi$ . Dropping $\sin\phi$ treats every latitude circle as if it had the same radius $\rho$ as the equator, which is only true at $\phi=\pi/2$ .

Consequence: integrating $\int_0^\pi 1\,d\phi=\pi$ in place of step 4's $\int_0^\pi\sin\phi\,d\phi=2$ changes the middle integral by a factor of $\frac{\pi}{2}$ , giving a "volume" of $\frac83\cdot\pi\cdot2\pi = \dfrac{16\pi^2}{3}\approx52.6$ — not even dimensionally sensible as a volume compared to the correct $\frac{32\pi}{3}\approx33.5$ , and it does not match the elementary sphere-volume formula at all.

07 · Choosing a Coordinate System

Definition — Coordinate System Selection Checklist

Rectangular boundaries, no circular or spherical symmetry? → rectangular coordinates ( $dV=dx\,dy\,dz$ ).
Symmetry around a single axis (a cylinder, a cone, a paraboloid opening along an axis)? → cylindrical coordinates ( $dV=r\,dr\,d\theta\,dz$ ).
Symmetry around a single point (a sphere, a ball, a region defined by distance from the origin)? → spherical coordinates ( $dV=\rho^2\sin\phi\,d\rho\,d\theta\,d\phi$ ).

08 · Exercises

EXERCISE 16.1

Integrate with respect to one variable first, treating the other as constant, then integrate the resulting function over the remaining variable.

$\displaystyle\iint_D (3x+2y)\,dA$ over $D=[0,1]\times[0,2]$ . Inner (over $y$ ): $\int_0^2(3x+2y)\,dy = \left[3xy+y^2\right]_0^2 = 6x+4$ . Outer: $\int_0^1(6x+4)\,dx = \left[3x^2+4x\right]_0^1 = 3+4=7$ .

Evaluate $\displaystyle\iint_D (3x+2y)\,dA$ over $D=[0,1]\times[0,2]$ .

EXERCISE 16.2

Determine which curve bounds the region from above and which from below over the given $x$ -range, then set up the iterated integral accordingly.

$D$ bounded by $y=x$ (top) and $y=0$ (bottom) for $0\leq x\leq2$ . $\displaystyle\int_0^2\int_0^x xy\,dy\,dx$ . Inner: $\int_0^x xy\,dy = \left[\frac{xy^2}{2}\right]_0^x = \frac{x^3}{2}$ . Outer: $\int_0^2\frac{x^3}{2}\,dx = \left[\frac{x^4}{8}\right]_0^2 = 2$ .

Evaluate $\displaystyle\iint_D xy\,dA$ where $D$ is bounded by $y=x$ and $y=0$ , for $0\leq x\leq2$ .

EXERCISE 16.3

Convert both the integrand and the region to polar coordinates, remembering the extra factor of $r$ in the area element.

Volume under $z=\sqrt{x^2+y^2}$ over the disk $x^2+y^2\leq9$ : in polar, integrand is $r$ , region is $0\leq r\leq3$ , $0\leq\theta\leq2\pi$ . $\displaystyle\int_0^{2\pi}\int_0^3 r\cdot r\,dr\,d\theta = \int_0^{2\pi}\int_0^3 r^2\,dr\,d\theta$ . Inner: $\left[\frac{r^3}{3}\right]_0^3=9$ . Outer: $\int_0^{2\pi}9\,d\theta=18\pi$ .

Find the volume under $z=\sqrt{x^2+y^2}$ over the disk $x^2+y^2\leq9$ , using polar coordinates.

EXERCISE 16.4

This solid has axis symmetry around the $z$ -axis — set up cylindrical bounds for $r$ , $\theta$ , and $z$ , then apply the cylindrical volume element.

Volume of the solid cylinder $x^2+y^2\leq4$ , $0\leq z\leq5$ : cylindrical bounds $0\leq r\leq2$ , $0\leq\theta\leq2\pi$ , $0\leq z\leq5$ . $\displaystyle\int_0^{2\pi}\int_0^2\int_0^5 r\,dz\,dr\,d\theta$ . Innermost: $\int_0^5 r\,dz = 5r$ . Middle: $\int_0^2 5r\,dr = \left[\frac{5r^2}{2}\right]_0^2=10$ . Outer: $\int_0^{2\pi}10\,d\theta=20\pi$ . Check: $\pi r^2h = \pi(4)(5)=20\pi$ ✓.

Find the volume of the solid cylinder $x^2+y^2\leq4$ , $0\leq z\leq5$ , using cylindrical coordinates, and check against the elementary cylinder-volume formula.

EXERCISE 16.5

This solid has point symmetry around the origin, but only occupies half the usual range of $\phi$ — determine the correct bound on $\phi$ for the upper half of the ball.

Volume of the upper half ( $z\geq0$ ) of a solid ball of radius $3$ : spherical bounds $0\leq\rho\leq3$ , $0\leq\theta\leq2\pi$ , and since only the upper half is wanted ( $z\geq0 \iff \cos\phi\geq0 \iff \phi\leq\pi/2$ ), $0\leq\phi\leq\pi/2$ . $\displaystyle\int_0^{2\pi}\int_0^{\pi/2}\int_0^3\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ . Innermost: $\int_0^3\rho^2\,d\rho=9$ . Middle: $\int_0^{\pi/2}9\sin\phi\,d\phi = 9\big[-\cos\phi\big]_0^{\pi/2}=9(0-(-1))=9$ . Outer: $\int_0^{2\pi}9\,d\theta=18\pi$ . Check: half of $\frac43\pi(27)=36\pi$ is $18\pi$ ✓.

Find the volume of the upper half ( $z\geq0$ ) of a solid ball of radius $3$ , using spherical coordinates.

EXERCISE 16.6

Identify the symmetry of the solid described (a single axis, or a single point) to choose between cylindrical and spherical coordinates before setting up any integral.

A commodity storage tank is modeled as the solid cone $z=2\sqrt{x^2+y^2}$ under the flat cap $z=6$ (axis symmetry around the $z$ -axis, so cylindrical coordinates apply, not spherical). Cone meets cap where $2r=6 \implies r=3$ . Bounds: $0\leq\theta\leq2\pi$ , $0\leq r\leq3$ , $2r\leq z\leq6$ . $\displaystyle\int_0^{2\pi}\int_0^3\int_{2r}^6 r\,dz\,dr\,d\theta$ . Innermost: $\int_{2r}^6 r\,dz = r(6-2r)=6r-2r^2$ . Middle: $\int_0^3(6r-2r^2)\,dr = \left[3r^2-\frac{2r^3}{3}\right]_0^3 = 27-18=9$ . Outer: $\int_0^{2\pi}9\,d\theta=18\pi$ cubic units of storage capacity.

A storage tank is shaped like the solid region bounded by the cone $z=2\sqrt{x^2+y^2}$ and the flat cap $z=6$ . Identify the appropriate coordinate system by its symmetry, then find its volume.

09 · Chapter Summary

Concept	Formula / Rule
Double integral	$\iint_D f\,dA$ ; volume under a surface
Fubini's theorem	Compute as two ordinary iterated single-variable integrals
Type I region	$a\leq x\leq b$ , $g_1(x)\leq y\leq g_2(x)$
Polar area element	$dA = r\,dr\,d\theta$
Triple integral	$\iiint_E f\,dV$
Cylindrical coordinates	$x=r\cos\theta$ , $y=r\sin\theta$ , $z=z$ ; $dV=r\,dr\,d\theta\,dz$ — use for axis symmetry
Spherical coordinates	$x=\rho\sin\phi\cos\theta$ , $y=\rho\sin\phi\sin\theta$ , $z=\rho\cos\phi$ ; $dV=\rho^2\sin\phi\,d\rho\,d\theta\,d\phi$ — use for point symmetry

Next: Chapter 17 — Vector Calculus: Gradient, Divergence, Curl, which combines the directed quantities of Chapter 14 with the partial derivatives of Chapter 15 to describe how fields spread out and rotate.