MechForge · Calculus · Chapter 15: Multiple Integrals

SECTION 15.1

Double Integrals over Rectangles

Just as the area problem motivated the single definite integral, the problem of finding the volume under a surface motivates the double integral. We start over a closed rectangle and approximate the solid by a grid of rectangular columns.

Closed rectangle and the solid above it

A closed rectangle is the set \(R=[a,b]\times[c,d]=\{(x,y)\mid a\le x\le b,\ c\le y\le d\}\). For a function \(f(x,y)\ge 0\) defined on \(R\), the solid \(S\) of interest lies above \(R\) and below the graph \(z=f(x,y)\). The goal is to assign \(S\) a volume.

Subrectangles and the double Riemann sum

Dividing \([a,b]\) into \(m\) pieces and \([c,d]\) into \(n\) pieces partitions \(R\) into subrectangles \(R_{ij}\), each of area \(\Delta A=\Delta x\,\Delta y\). Choosing a sample point \((x_{ij}^*,y_{ij}^*)\) in each subrectangle and summing the column volumes gives the double Riemann sum, an approximation to the total volume.

Double Riemann Sum

\[ V \approx \sum_{i=1}^{m}\sum_{j=1}^{n} f(x_{ij}^*,y_{ij}^*)\,\Delta A \]

The double integral

The double integral of \(f\) over \(R\) is the limit of the Riemann sums as the grid is refined, provided the limit exists. Every continuous function is integrable, as is any bounded function whose discontinuities lie on a finite number of smooth curves.

Definition of the Double Integral

\[ \iint_{R} f(x,y)\,dA=\lim_{m,n\to\infty}\sum_{i=1}^{m}\sum_{j=1}^{n} f(x_{ij}^*,y_{ij}^*)\,\Delta A \]

Volume as a Double Integral

\[ V=\iint_{R} f(x,y)\,dA \qquad (f(x,y)\ge 0) \]

Figure 15.1.1 — Redrawn The solid \(S\) lies above rectangle \(R\) and below the surface \(z=f(x,y)\); its volume is the limit of the summed column volumes.

EXAMPLE 1 Estimating volume by a Riemann sum

Estimate the volume of the solid above the square \(R=[0,2]\times[0,2]\) and below the elliptic paraboloid \(z=16-x^2-2y^2\). Divide \(R\) into four equal squares and use the upper-right corner of each as the sample point.

Solution

Each square has area \(\Delta A=1\). Sample at the upper-right corners \((1,1),(1,2),(2,1),(2,2)\).

\(V\approx f(1,1)+f(1,2)+f(2,1)+f(2,2)\) \(=13+7+10+4\).

\(V\approx 34\). Refining the grid (16, 64, 256 squares) gives successively better estimates approaching the exact value \(48\).

EXAMPLE 2 A double integral as a known volume

For \(R=\{(x,y)\mid -1\le x\le 1,\ -2\le y\le 2\}\), evaluate \(\displaystyle\iint_{R}\sqrt{1-x^2}\,dA\).

Solution

Since \(z=\sqrt{1-x^2}\ge 0\) describes \(x^2+z^2=1,\ z\ge 0\), the integral is the volume under a half-cylinder of radius 1, length 4.

Volume = (area of semicircle) \(\times\) (length) \(=\tfrac{1}{2}\pi(1)^2\cdot 4\).

\(\displaystyle\iint_{R}\sqrt{1-x^2}\,dA=2\pi\).

The Midpoint Rule

The single-variable Midpoint Rule has a direct two-dimensional counterpart: evaluate \(f\) at the centre of each subrectangle.

Midpoint Rule for Double Integrals

\[ \iint_{R} f(x,y)\,dA \approx \sum_{i=1}^{m}\sum_{j=1}^{n} f(\bar{x}_i,\bar{y}_j)\,\Delta A \]

Average value of a function of two variables

The average value of \(f\) over \(R\) is the integral divided by the area of \(R\). Geometrically, a box of base \(R\) and height \(f_{\text{ave}}\) has the same volume as the solid under \(f\).

Average Value

\[ f_{\text{ave}}=\frac{1}{A(R)}\iint_{R} f(x,y)\,dA \]

Properties of Double Integrals (Linearity & Comparison)

\(\displaystyle\iint_{R}\big[f+g\big]\,dA=\iint_{R}f\,dA+\iint_{R}g\,dA\)

\(\displaystyle\iint_{R} c\,f\,dA=c\iint_{R} f\,dA\) for any constant \(c\)

If \(f(x,y)\ge g(x,y)\) on \(R\), then \(\displaystyle\iint_{R}f\,dA\ge\iint_{R}g\,dA\).

In practice you almost never compute a double integral from the Riemann-sum definition — that is conceptual scaffolding. The payoff comes in 15.2 with Fubini's Theorem, which turns a double integral into two ordinary single integrals.

SECTION 15.2

Iterated Integrals

Evaluating a double integral from first principles is hard. The practical method expresses it as an iterated integral — two single integrals done one after the other, working from the inside out.

Partial integration and the iterated integral

Holding \(x\) fixed and integrating with respect to \(y\) gives a function \(A(x)\). Integrating \(A(x)\) over \([a,b]\) produces the iterated integral. Inner limits may depend on the outer variable; outer limits must be constants.

Iterated Integrals (two orders)

\[ \int_a^b\!\!\int_c^d f(x,y)\,dy\,dx \qquad\text{and}\qquad \int_c^d\!\!\int_a^b f(x,y)\,dx\,dy \]

Fubini's Theorem

If \(f\) is continuous on the rectangle \(R=[a,b]\times[c,d]\), then the double integral equals either iterated integral:

\[ \iint_{R} f(x,y)\,dA=\int_a^b\!\!\int_c^d f(x,y)\,dy\,dx=\int_c^d\!\!\int_a^b f(x,y)\,dx\,dy. \]

More generally this holds when \(f\) is bounded on \(R\), discontinuous only on a finite number of smooth curves, and the iterated integrals exist.

Fubini is the multivariable analogue of "the order of operations doesn't change the answer." But while both orders give the same number, one order is often dramatically easier to evaluate — see Example 3 below. Always glance at both before committing.

EXAMPLE 1 Order doesn't change the value

Evaluate \(\displaystyle\int_0^3\!\!\int_1^2 x^2 y\,dy\,dx\) and the reversed-order integral.

Solution

Inner (in \(y\)): \(\int_1^2 x^2 y\,dy=x^2\big[\tfrac{y^2}{2}\big]_1^2=\tfrac{3}{2}x^2\).

Outer (in \(x\)): \(\int_0^3 \tfrac{3}{2}x^2\,dx=\big[\tfrac{1}{2}x^3\big]_0^3=\tfrac{27}{2}\).

Reversing the order yields the same value, \(\tfrac{27}{2}\).

EXAMPLE 3 Choosing the easier order

Evaluate \(\displaystyle\iint_{R} y\sin(xy)\,dA\) over \(R=[1,2]\times[0,\pi]\).

Solution

Integrate in \(x\) first: \(\int_1^2 y\sin(xy)\,dx=\big[-\cos(xy)\big]_{x=1}^{x=2}=\cos y-\cos 2y\).

Then in \(y\): \(\int_0^\pi(\cos y-\cos 2y)\,dy=\big[\sin y-\tfrac{1}{2}\sin 2y\big]_0^\pi=0\).

\(\displaystyle\iint_{R} y\sin(xy)\,dA=0\). Integrating in \(y\) first would require integration by parts twice — far harder.

Separable integrands

When the integrand factors as a product \(g(x)h(y)\) over a rectangle, the double integral splits into the product of two single integrals.

Product of Single Integrals (separable case)

\[ \iint_{R} g(x)\,h(y)\,dA=\left(\int_a^b g(x)\,dx\right)\!\left(\int_c^d h(y)\,dy\right),\quad R=[a,b]\times[c,d] \]

EXAMPLE 5 Using separability

Evaluate \(\displaystyle\iint_{R}\sin x\cos y\,dA\) over \(R=[0,\tfrac{\pi}{2}]\times[0,\tfrac{\pi}{2}]\).

Solution

The integrand factors, so split: \(\big(\int_0^{\pi/2}\sin x\,dx\big)\big(\int_0^{\pi/2}\cos y\,dy\big)\).

Each factor evaluates to \(1\).

\(\displaystyle\iint_{R}\sin x\cos y\,dA=1\cdot 1=1\).

SECTION 15.3

Double Integrals over General Regions

Real regions are rarely rectangles. We classify bounded regions by how their boundaries are described, then convert the double integral into an iterated integral with variable limits.

Type I region

A region is type I if it lies between two continuous curves of \(x\): \(D=\{(x,y)\mid a\le x\le b,\ g_1(x)\le y\le g_2(x)\}\). The inner integral runs in \(y\) between the lower and upper boundary curves.

Type II region

A region is type II if it lies between two continuous curves of \(y\): \(D=\{(x,y)\mid c\le y\le d,\ h_1(y)\le x\le h_2(y)\}\). The inner integral runs in \(x\) between the left and right boundary curves.

Integral over a Type I Region

\[ \iint_{D} f(x,y)\,dA=\int_a^b\!\!\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx \]

Integral over a Type II Region

\[ \iint_{D} f(x,y)\,dA=\int_c^d\!\!\int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy \]

Figure 15.3.1 — Redrawn A type I region. The inner integral sweeps a vertical arrow from the lower curve \(y=g_1(x)\) up to the upper curve \(y=g_2(x)\).

EXAMPLE 1 Integrating over a region between parabolas

Evaluate \(\displaystyle\iint_{D}(x+2y)\,dA\) where \(D\) is bounded by \(y=2x^2\) and \(y=1+x^2\).

Solution

The parabolas meet where \(2x^2=1+x^2\Rightarrow x=\pm 1\). Region is type I: \(-1\le x\le 1,\ 2x^2\le y\le 1+x^2\).

Inner integral in \(y\) gives \(\big[xy+y^2\big]_{2x^2}^{1+x^2}\), then integrate in \(x\).

\(\displaystyle\iint_{D}(x+2y)\,dA=\frac{32}{15}\).

EXAMPLE 5 Reversing the order to integrate at all

Evaluate \(\displaystyle\int_0^1\!\!\int_x^1 \sin(y^2)\,dy\,dx\).

Solution

\(\int \sin(y^2)\,dy\) has no elementary form, so reverse the order. The region \(\{0\le x\le 1,\ x\le y\le 1\}\) is also \(\{0\le y\le 1,\ 0\le x\le y\}\).

Now \(\int_0^1\!\!\int_0^y \sin(y^2)\,dx\,dy=\int_0^1 y\sin(y^2)\,dy\).

\(=\tfrac{1}{2}\big[-\cos(y^2)\big]_0^1=\tfrac{1}{2}(1-\cos 1)\).

Properties over General Regions

Additivity: if \(D=D_1\cup D_2\) with non-overlapping interiors, then \(\displaystyle\iint_{D}f\,dA=\iint_{D_1}f\,dA+\iint_{D_2}f\,dA\).

Area: \(\displaystyle\iint_{D} 1\,dA=A(D)\).

Bounds: if \(m\le f(x,y)\le M\) on \(D\), then \(m\,A(D)\le\displaystyle\iint_{D}f\,dA\le M\,A(D)\).

Example 5 is the single most exam-tested idea in this section: an integral that looks impossible becomes trivial once you redraw the region and swap the order. Whenever the inner antiderivative doesn't exist in closed form, your first move is "sketch the region and reverse."

SECTION 15.4

Double Integrals in Polar Coordinates

When a region is a disk, ring, or sector, rectangular limits become ugly. Polar coordinates describe such regions cleanly — at the cost of one extra factor of \(r\) in the integrand.

Polar–rectangular relations

A point has rectangular coordinates \((x,y)\) and polar coordinates \((r,\theta)\) related by \(x=r\cos\theta\), \(y=r\sin\theta\), and \(x^2+y^2=r^2\). A polar rectangle is \(\{(r,\theta)\mid a\le r\le b,\ \alpha\le\theta\le\beta\}\).

The area element in polar form

The infinitesimal polar rectangle has dimensions \(dr\) and \(r\,d\theta\), so its area is \(dA=r\,dr\,d\theta\). Forgetting the factor \(r\) is the most common error in polar integration.

Change to Polar Coordinates (polar rectangle)

\[ \iint_{R} f(x,y)\,dA=\int_\alpha^\beta\!\!\int_a^b f(r\cos\theta,\,r\sin\theta)\,r\,dr\,d\theta \]

Polar Region with Variable Radius

\[ \iint_{D} f\,dA=\int_\alpha^\beta\!\!\int_{h_1(\theta)}^{h_2(\theta)} f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta \]

Figure 15.4.1 — Redrawn The polar area element \(dA=r\,dr\,d\theta\): a thin patch with radial width \(dr\) and arc width \(r\,d\theta\).

EXAMPLE 1 Integral over a half-ring

Evaluate \(\displaystyle\iint_{R}(3x+4y^2)\,dA\) where \(R\) is the upper half-plane region between \(x^2+y^2=1\) and \(x^2+y^2=4\).

Solution

In polar: \(1\le r\le 2,\ 0\le\theta\le\pi\). Substitute and include the factor \(r\): \(\int_0^\pi\!\int_1^2(3r\cos\theta+4r^2\sin^2\theta)\,r\,dr\,d\theta\).

Integrate in \(r\), then use \(\sin^2\theta=\tfrac{1}{2}(1-\cos 2\theta)\) in \(\theta\).

\(\displaystyle\iint_{R}(3x+4y^2)\,dA=\frac{15\pi}{2}\).

EXAMPLE 2 Volume under a paraboloid

Find the volume bounded by \(z=0\) and the paraboloid \(z=1-x^2-y^2\).

Solution

The solid sits over the disk \(x^2+y^2\le 1\), i.e. \(0\le r\le 1,\ 0\le\theta\le 2\pi\), with \(1-x^2-y^2=1-r^2\).

\(V=\int_0^{2\pi}\!\int_0^1(1-r^2)\,r\,dr\,d\theta=2\pi\big[\tfrac{r^2}{2}-\tfrac{r^4}{4}\big]_0^1\).

\(V=\dfrac{\pi}{2}\).

Trigger checklist for going polar: you see \(x^2+y^2\) in the integrand, or the region is a circle/sector/annulus, or rectangular limits involve \(\sqrt{a^2-x^2}\). Any one of these is your cue. And write the \(r\) in \(r\,dr\,d\theta\) before you substitute — that way you never forget it.

SECTION 15.5

Applications of Double Integrals

Double integrals compute physical quantities for laminae (thin plates) of variable density: mass, charge, centre of mass, moments of inertia, and — via joint density functions — probabilities.

Density and mass

For a lamina occupying region \(D\) with continuous density \(\rho(x,y)\) (mass per unit area), the total mass is the double integral of the density. Replacing density with charge density \(\sigma(x,y)\) gives total charge instead.

Mass and Total Charge

\[ m=\iint_{D}\rho(x,y)\,dA \qquad Q=\iint_{D}\sigma(x,y)\,dA \]

Moments and centre of mass

The moments about the axes weight each mass element by its distance from the axis. The centre of mass \((\bar x,\bar y)\) is the balance point — the lamina behaves as if its entire mass sat there.

Moments and Centre of Mass

\[ M_x=\iint_{D} y\,\rho\,dA,\quad M_y=\iint_{D} x\,\rho\,dA,\qquad \bar x=\frac{M_y}{m},\ \ \bar y=\frac{M_x}{m} \]

Moments of inertia (second moments)

The moment of inertia measures resistance to rotation, weighting each mass element by the square of its distance from the axis. The polar moment \(I_0=I_x+I_y\) is taken about the origin (the axis perpendicular to the plate).

Moments of Inertia

\[ I_x=\iint_{D} y^2\rho\,dA,\quad I_y=\iint_{D} x^2\rho\,dA,\quad I_0=\iint_{D}(x^2+y^2)\rho\,dA \]

Moment of inertia in rotation plays exactly the role mass plays in straight-line motion — it is what makes a flywheel hard to spin up or stop. You'll meet \(I\) constantly in dynamics and machine design; the double integral here is where it comes from. Note the squared distance: that is why mass far from the axis matters so much more.

EXAMPLE 2 Mass and centre of mass of a triangular lamina

Find the mass and centre of mass of the triangle with vertices \((0,0),(1,0),(0,2)\) and density \(\rho(x,y)=1+3x+y\).

Solution

Upper boundary \(y=2-2x\). Mass \(m=\int_0^1\!\int_0^{2-2x}(1+3x+y)\,dy\,dx=\tfrac{8}{3}\).

Compute \(M_y\) and \(M_x\) with the same limits, then divide by \(m\).

Centre of mass at \(\big(\tfrac{3}{8},\tfrac{11}{16}\big)\).

EXAMPLE 4 Moments of inertia of a uniform disk

Find \(I_x,I_y,I_0\) for a uniform disk of density \(\rho\), centre origin, radius \(a\).

Solution

By symmetry \(I_x=I_y\). Compute \(I_0=\rho\int_0^{2\pi}\!\int_0^a r^2\cdot r\,dr\,d\theta=\tfrac{\pi\rho a^4}{2}\).

Then \(I_x=I_y=\tfrac{1}{2}I_0=\tfrac{\pi\rho a^4}{4}\).

In terms of mass \(m=\rho\pi a^2\): \(I_0=\tfrac{1}{2}m a^2\).

Joint density and probability

For two continuous random variables \(X,Y\), a joint density \(f(x,y)\ge 0\) with total integral \(1\) gives the probability that \((X,Y)\) lands in a region \(D\) as the double integral of \(f\) over \(D\). If \(X,Y\) are independent, \(f(x,y)=f_1(x)f_2(y)\).

Probability from a Joint Density

\[ P\big((X,Y)\in D\big)=\iint_{D} f(x,y)\,dA,\qquad \iint_{\mathbb{R}^2} f(x,y)\,dA=1 \]

SECTION 15.6

Triple Integrals

The same construction extends to functions of three variables. Over a box, a triple Riemann sum defines the triple integral; over a general solid, we project onto a coordinate plane and integrate inward.

Definition of the Triple Integral

\[ \iiint_{B} f(x,y,z)\,dV=\lim_{l,m,n\to\infty}\sum_{i=1}^{l}\sum_{j=1}^{m}\sum_{k=1}^{n} f(x_{ijk}^*,y_{ijk}^*,z_{ijk}^*)\,\Delta V \]

Fubini's Theorem for Triple Integrals

If \(f\) is continuous on the box \(B=[a,b]\times[c,d]\times[r,s]\), then

\[ \iiint_{B} f\,dV=\int_r^s\!\!\int_c^d\!\!\int_a^b f(x,y,z)\,dx\,dy\,dz, \]

and any of the six orders of integration gives the same value.

Type 1 solid region

A solid is type 1 if it lies between two surfaces over a planar projection \(D\): \(E=\{(x,y,z)\mid (x,y)\in D,\ u_1(x,y)\le z\le u_2(x,y)\}\). Integrate in \(z\) first, then over \(D\) as a double integral.

Triple Integral over a Type 1 Solid

\[ \iiint_{E} f\,dV=\iint_{D}\left[\int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z)\,dz\right]dA \]

Type 2 and type 3 solids are defined analogously by projecting onto the \(yz\)- and \(xz\)-planes respectively, with the inner integral taken in \(x\) or in \(y\).

EXAMPLE 2 Integral over a tetrahedron

Evaluate \(\displaystyle\iiint_{E} z\,dV\) where \(E\) is the tetrahedron bounded by \(x=0,\ y=0,\ z=0,\ x+y+z=1\).

Solution

Lower surface \(z=0\), upper \(z=1-x-y\). Projection \(D\): \(0\le x\le 1,\ 0\le y\le 1-x\).

\(\int_0^1\!\int_0^{1-x}\!\int_0^{1-x-y} z\,dz\,dy\,dx=\tfrac{1}{2}\int_0^1\!\int_0^{1-x}(1-x-y)^2\,dy\,dx\).

\(\displaystyle\iiint_{E} z\,dV=\frac{1}{24}\).

Volume, mass, and centroids by triple integral

Setting \(f=1\) gives the volume of \(E\). With a density \(\rho(x,y,z)\), the triple integral gives mass; weighting by coordinate gives moments about the coordinate planes, and dividing by mass locates the centre of mass.

Volume and Mass

\[ V(E)=\iiint_{E} dV \qquad m=\iiint_{E}\rho(x,y,z)\,dV \]

The hardest part of any triple integral is never the calculus — it is setting up the limits. Remember the rule of thumb: the inner limits may contain at most two variables, the middle limits at most one, and the outer limits must be pure constants. Draw the solid and its projection every single time.

SECTION 15.7

Triple Integrals in Cylindrical Coordinates

Cylindrical coordinates are polar coordinates in the \(xy\)-plane with the rectangular \(z\) kept as is. They suit solids with an axis of symmetry — cylinders, cones, and paraboloids of revolution.

Cylindrical coordinates

A point is \((r,\theta,z)\), where \((r,\theta)\) are the polar coordinates of its projection on the \(xy\)-plane and \(z\) is the height. Conversions: \(x=r\cos\theta,\ y=r\sin\theta,\ z=z\), and \(r^2=x^2+y^2,\ \tan\theta=y/x\).

Volume element

The cylindrical volume element is \(dV=r\,dz\,dr\,d\theta\) — the polar area element times the height \(dz\). As in polar integration, the factor \(r\) must not be dropped.

Triple Integration in Cylindrical Coordinates

\[ \iiint_{E} f\,dV=\int_\alpha^\beta\!\!\int_{h_1(\theta)}^{h_2(\theta)}\!\!\int_{u_1}^{u_2} f(r\cos\theta,r\sin\theta,z)\,r\,dz\,dr\,d\theta \]

EXAMPLE 3 Mass with density proportional to distance from the axis

A solid \(E\) lies within \(x^2+y^2=1\), below \(z=4\), above the paraboloid \(z=1-x^2-y^2\). Its density is proportional to distance from the \(z\)-axis. Find the mass.

Solution

In cylindrical: \(0\le\theta\le 2\pi,\ 0\le r\le 1,\ 1-r^2\le z\le 4\). Density \(=Kr\).

\(m=\int_0^{2\pi}\!\int_0^1\!\int_{1-r^2}^{4} (Kr)\,r\,dz\,dr\,d\theta=K\int_0^{2\pi}\!\int_0^1(3r^2+r^4)\,dr\,d\theta\).

\(m=\dfrac{12\pi K}{5}\).

EXAMPLE 4 Converting a hard rectangular integral

Evaluate \(\displaystyle\int_{-2}^{2}\!\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\!\int_{\sqrt{x^2+y^2}}^{2}(x^2+y^2)\,dz\,dy\,dx\).

Solution

The region is the disk \(x^2+y^2\le 4\) with \(\sqrt{x^2+y^2}\le z\le 2\) — a cone-capped solid. Cylindrical: \(0\le\theta\le 2\pi,\ 0\le r\le 2,\ r\le z\le 2\).

\(\int_0^{2\pi}\!\int_0^2\!\int_r^2 r^2\cdot r\,dz\,dr\,d\theta=2\pi\int_0^2 r^3(2-r)\,dr\).

\(=\dfrac{16\pi}{5}\).

Use cylindrical when the projection onto the \(xy\)-plane is a circle or sector, especially whenever \(x^2+y^2\) appears in the integrand or in a boundary surface. It is just "polar plus a \(z\)." Spherical (next section) is the better choice when the solid is bounded by spheres or cones about the origin.

SECTION 15.8

Triple Integrals in Spherical Coordinates

Spherical coordinates describe a point by its distance from the origin and two angles. They are unbeatable for regions bounded by spheres and cones centred at the origin.

Spherical coordinates

A point is \((\rho,\theta,\phi)\), where \(\rho=|OP|\ge 0\) is the distance from the origin, \(\theta\) is the same angle as in cylindrical coordinates, and \(\phi\in[0,\pi]\) is the angle from the positive \(z\)-axis to \(OP\).

Spherical–Rectangular Conversions

\[ x=\rho\sin\phi\cos\theta,\quad y=\rho\sin\phi\sin\theta,\quad z=\rho\cos\phi,\quad \rho^2=x^2+y^2+z^2 \]

Volume element

The spherical volume element is \(dV=\rho^2\sin\phi\,d\rho\,d\theta\,d\phi\). The \(\rho^2\sin\phi\) factor — derived from the dimensions of a spherical wedge — is essential.

Triple Integration in Spherical Coordinates

\[ \iiint_{E} f\,dV=\int_c^d\!\!\int_\alpha^\beta\!\!\int_a^b f(\rho\sin\phi\cos\theta,\,\rho\sin\phi\sin\theta,\,\rho\cos\phi)\,\rho^2\sin\phi\,d\rho\,d\theta\,d\phi \]

Figure 15.8.1 — Redrawn Spherical coordinates: \(\rho\) is the distance from the origin, \(\phi\) the angle down from the \(z\)-axis, and \(\theta\) the azimuthal angle.

EXAMPLE 3 Integrating over the unit ball

Evaluate \(\displaystyle\iiint_{B} e^{(x^2+y^2+z^2)^{3/2}}\,dV\) over the unit ball \(B:\ x^2+y^2+z^2\le 1\).

Solution

In spherical: \(0\le\rho\le 1,\ 0\le\theta\le 2\pi,\ 0\le\phi\le\pi\), and \(x^2+y^2+z^2=\rho^2\), so the exponent is \(\rho^3\).

\(\int_0^\pi\!\int_0^{2\pi}\!\int_0^1 e^{\rho^3}\rho^2\sin\phi\,d\rho\,d\theta\,d\phi =\big(\int_0^\pi\sin\phi\,d\phi\big)\big(\int_0^{2\pi}d\theta\big)\big(\int_0^1\rho^2 e^{\rho^3}d\rho\big)\).

\(=2\cdot 2\pi\cdot\tfrac{1}{3}(e-1)=\dfrac{4\pi}{3}(e-1)\).

EXAMPLE 4 Volume above a cone, below a sphere

Find the volume above the cone \(z=\sqrt{x^2+y^2}\) and below the sphere \(x^2+y^2+z^2=z\).

Solution

Sphere becomes \(\rho=\cos\phi\); the cone becomes \(\phi=\pi/4\). So \(0\le\theta\le 2\pi,\ 0\le\phi\le\pi/4,\ 0\le\rho\le\cos\phi\).

\(V=\int_0^{2\pi}\!\int_0^{\pi/4}\!\int_0^{\cos\phi}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta =\tfrac{2\pi}{3}\int_0^{\pi/4}\sin\phi\cos^3\phi\,d\phi\).

\(V=\dfrac{\pi}{8}\).

Heads-up on notation: physics texts routinely swap the meanings of \(\theta\) and \(\phi\) and write \(r\) for \(\rho\). When you cross between a math course and a physics or engineering course, check the convention first — mixing them up is a classic source of sign and limit errors.

SECTION 15.9

Change of Variables in Multiple Integrals

Polar, cylindrical, and spherical coordinates are all special cases of one general idea: a substitution in a multiple integral introduces a scaling factor — the absolute value of the Jacobian.

Transformation and Jacobian

A \(C^1\) transformation \(T(u,v)=(x,y)\) with \(x=g(u,v),\ y=h(u,v)\) maps a region \(S\) in the \(uv\)-plane to a region \(R\) in the \(xy\)-plane. The Jacobian measures how \(T\) scales area locally.

The Jacobian

\[ \frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix}\dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\[6pt]\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u} \]

Change of Variables in a Double Integral

If \(T\) is a one-to-one \(C^1\) transformation with nonzero Jacobian mapping \(S\) onto \(R\), and \(f\) is continuous, then

\[ \iint_{R} f(x,y)\,dA=\iint_{S} f\big(x(u,v),y(u,v)\big)\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv. \]

This is the two-variable analogue of the substitution rule, with \(|\,\partial(x,y)/\partial(u,v)\,|\) replacing the derivative factor.

The same idea extends to triple integrals with a \(3\times 3\) Jacobian \(\partial(x,y,z)/\partial(u,v,w)\). For example, the spherical-coordinate Jacobian works out to \(\rho^2\sin\phi\), recovering the volume element of §15.8.

EXAMPLE 3 A substitution that simplifies the integrand

Evaluate \(\displaystyle\iint_{R} e^{(x+y)/(x-y)}\,dA\) over the trapezoid with vertices \((1,0),(2,0),(0,2),(0,1)\).

Solution

Let \(u=x+y,\ v=x-y\); then \(x=\tfrac{1}{2}(u+v),\ y=\tfrac{1}{2}(u-v)\) and the Jacobian is \(-\tfrac{1}{2}\), so \(\big|\,\partial(x,y)/\partial(u,v)\,\big|=\tfrac{1}{2}\).

The region maps to \(S:\ 1\le v\le 2,\ -v\le u\le v\). Integral becomes \(\tfrac{1}{2}\int_1^2\!\int_{-v}^{v} e^{u/v}\,du\,dv\).

\(=\tfrac{1}{2}\int_1^2(e-e^{-1})v\,dv=\dfrac{3}{4}\big(e-e^{-1}\big)\).

Two cues tell you to invent a change of variables: a nasty integrand (like the \((x+y)/(x-y)\) above — define \(u,v\) to be those expressions) or a nasty region (define \(u,v\) so the boundaries become constant). Then never forget the Jacobian factor — it is the whole point of the method, and the \(r\), \(r\) again, and \(\rho^2\sin\phi\) you have been using all chapter are just its special cases.

End of Chapter 15 · Exercises and review problems are intentionally omitted from this reference.