Chapter 15: Multiple Integrals

15.1 Double Integrals over Rectangles

Derivation

Consider a function $f$ of two variables bounded by a closed rectangle $R$ , i.e. bounded by $a\leq x\leq b$ and $c\leq y\leq d$ , for some $a,b,c,d\in \mathbb{R}$ . Then, define the 3D solid $S$ such that it lies between the rectangle $R$ lying in the $xy$ -plane and the surface with equation $z=f(x,y)$ , bounded by the above bounds. The below image may help you understand this:

We aim to find the volume of $S$ .

Consider dividing rectangle $R$ into subrectangles, analogous to how we might do a Riemann sum, but instead in 3 dimensions. For any subrectangle, we can approximate the volume of $S$ lying above the subrectangle by computing $f(x_{0},y_{0})\Delta A$ , where $x_{0},y_{0}$ represents an arbitrary point in the subrectangle and $\Delta A$ represents the area of the subrectangle.

Approximating the volume mathematically, we get

V\approx \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_{i},y_{j})\Delta A_{i,j}

Where $m,n$ denote the number of subdivisions in each dimension ( $x$ -axis and $y$ -axis).

To find the true volume, we of course just take the limit:

V=\lim_{ m,n \to \infty } \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_{i},y_{j})\Delta A_{i,j}

Double Integral

Mathematically, we often write instead that

Double Integral

V= \underset{ R }{\iint} f(x,y) \, dA = \lim_{ m,n \to \infty } \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_{i},y_{j})\Delta A_{i,j}

We don't have to use the limit definition to integrate, however. Instead, we can express a double integral, as an iterated integral:

\begin{align*} \int_{c}^{d} \int_{a}^{b} f(x,y) \, dx \, dy &= \int_{c}^{d} \left[ \int_{a}^{b} f(x,y) \, dx \right] \, dy \\ \int_{a}^{b} \int_{c}^{d} f(x,y) \,\;\mathrm{d} y \, dx &= \int_{a}^{b} \left[ \int_{c}^{d} f(x,y) \, dy \right] \, dx \end{align*}

Essentially, we can integrate independently of the other integrals. Note also that partial integration works analogously as to partial differentiation, i.e. you treat other variables as constant.

Additionally, note that many functions $f$ can be integrated in either order when finding the volume of the solid $S$ .

Fubini's Theorem

If $f$ is continuous over the rectangle $R=\{ (x,y) \mid a\leq x\leq b,c\leq y\leq d \}$

\underset{ R }{ \iint } f(x,y) \,\;\mathrm{d} A = \int_{c}^{d} \int_{a}^{b} f(x,y) \, \mathrm{d}x \, \mathrm{d}y = \int_{a}^{b} \int_{c}^{d} f(x,y) \, \mathrm{d}y \, \mathrm{d}x

Exception: Product of Univariate Functions

In the special case where $f(x,y)=g(x)h(y)$ , i.e. can be factored as a product of a function of only $x$ and a function of only $y$ , then we can write

\underset{ R }{ \iint } f(x,y) \, \mathrm{d}A =\underset{ R }{ \iint } g(x)h(y) \, \mathrm{d}A = \int_{a}^{b} g(x) \, \mathrm{d}x \int_{c}^{d} h(y) \, \mathrm{d}y

We can do this because $y$ is constant when integrating with respect to $x$ , and vice versa.

Average Value

This is calculated analogously to how it is in 1 dimension. Simply,

f_{\mathrm{avg}} = \frac{1}{A(R)}\underset{ R }{ \iint } f(x,y)\,\mathrm{d}A

Where $A(R)$ is the area of the rectangle $R$ . This should make sense intuitively.

15.2 Double Integrals over General Regions

We aim to integrate volumes of solids over arbitrary plane regions, not just rectangles. We will consider two types of regions, and their corresponding integrations.

In order to evaluate $\iint_{D}f(x,y)\,\mathrm{d}A$ , we consider a rectangle $R=[a,b]\times[c,d]$ that contains the plane region $D$ , i.e. $D\subseteq R$ . We define a function $F$ with domain $R$ as

F(x,y)=\left\{ \begin{matrix} f(x,y) & \text{if } (x,y) \text{ is in } D \\ 0 & \text{if } (x,y) \text{ is in } R \text{ but not in } D \end{matrix} \right.

Then, if $F$ is integrable over $R$ , we have

\underset{ D }{ \iint } f(x,y)\,\mathrm{d}A = \underset{ R }{ \iint } F(x,y)\,\mathrm{d}A

Type I

A plane region $D$ is said to be of type I if it lies between the graphs of two continuous functions of $x$ , that is:

D=\{ (x,y)\mid a\leq x\leq b,g_{1}(x)\leq y\leq g_{2}(x) \}

Then, we may write

\begin{align*} \underset{ D }{ \iint } f(x,y)\,\mathrm{d}A &= \underset{ R }{ \iint } F(x,y)\,\mathrm{d}A = \int_{a}^{b} \int_{c}^{d} F(x,y) \, \mathrm{d}y \, \mathrm{d}x \\ \end{align*}

Note that $F(x,y)=0$ if $y<g_{1}(x)$ or $y>g_{2}(x)$ . Thus,

\int_{c}^{d} F(x,y) \, \mathrm{d}y = \int_{g_{1}(x)}^{g_{2}(x)} F(x,y) \, \mathrm{d}y = \int_{g_{1}(x)}^{g_{2}(x)} f(x,y) \, \mathrm{d}y

Therefore,

Type I Volume

If $f$ is continuous on a type I region $D$ such that $D=\{ (x,y) \mid a\leq x\leq b,g_{1}(x)\leq y\leq g_{2}(x) \}$ , then

\underset{ D }{ \iint } f(x,y)\,\mathrm{d}A = \int_{a}^{b} \int_{g_{1}(x)}^{g_{2}(x)} f(x,y) \, \mathrm{d}y \, \mathrm{d}x

Type II

A plane region $D$ is said to be type II if it is between the graphs of two continuous functions of $y$ , i.e. type I but $x$ is bounded instead of $y$ :

D=\{ (x,y)\mid c\leq y\leq y,h_{1}(y)\leq x\leq h_{2}(y) \}

Thus, by similar derivation as in [[#Type I]], we find

Type II Volume

\underset{ D }{ \iint } f(x,y)\,\mathrm{d}A=\int_{c}^{d} \int_{h_{1}(y)}^{h_{2}(y)} f(x,y) \, \mathrm{d}x \, \mathrm{d}y

Properties of Double Integrals

\begin{align*} \mathbf{(1)}\;& \underset{ D }{ \iint } [f(x,y)+g(x,y)]\,\mathrm{d}A = \underset{ D }{ \iint }f(x,y)\,\mathrm{d}A + \underset{ D }{ \iint } g(x,y)\,\mathrm{d}A \\ \mathbf{(2)}\;& \underset{ D }{ \iint } cf(x,y)\,\mathrm{d}A = c\underset{ D }{ \iint }f(x,y)\,\mathrm{d}A, \mathrm{\;for\;some\;constant\;}c \\ \mathbf{(3)}\;& f(x,y)\geq g(x,y),\forall(x,y)\in D \implies \underset{ D }{ \iint }f(x,y)\,\mathrm{d}A \geq \underset{ D }{ \iint }g(x,y)\,\mathrm{d}A \\ \mathbf{(4)}\;& D = D_{1}\cup D_{2},D_{1}\cap D_{2}=\emptyset \implies \underset{ D }{ \iint }f(x,y)\,\mathrm{d}A = \underset{ D_{1} }{ \iint }f(x,y)\,\mathrm{d}A+\underset{ D_{2} }{ \iint }f(x,y)\,\mathrm{d}A \\ \mathbf{(5)}\;& \underset{ D }{ \iint }1\,\mathrm{d}A=A(D) \\ \mathbf{(6)}\;& m\leq f(x,y)\leq M,\forall (x,y)\in D\implies mA(D)\leq \underset{ D }{ \iint }f(x,y)\;\mathrm{d}A\leq MA(D) \end{align*}

Where,

(4) $D_{1}\cap D_{2}$ can contain boundary points, but only that.
(5) $A(D)$ is the area of the region $D$ .

15.3 Double Integrals in Polar Coordinates

The proof of the formula for this will be left as an exercise to the reader since it is similar to the proof of double integrals over rectangles. Hint: consider dividing a polar region into polar subrectangles.

Double Integral over Polar Rectangle

If $f$ is continuous on a polar rectangle $R$ describe by $0\leq a\leq r\leq b$ , $\alpha\leq {0}\leq\beta$ , where $0\leq\beta-\alpha\leq{2}\pi$ , then

\underset{ R }{ \iint } f(x,y) \, \mathrm{d}A = \int_{\alpha}^{\beta} \int_{a}^{b} f(r\cos\theta,r\sin\theta) \cdot r \, \mathrm{d}r \, \mathrm{d}\theta

Note the additional factor of

r

on the RHS. Careful not to forget this!

Generalizing to any polar region, similarly to how it was done in 15.2, we get

Double Integral over General Polar Region

If $f$ is continuous on a polar region of the form

D=\{ (r,\theta) \mid\alpha\leq \theta\leq \beta \},h_{1}(\theta)\leq r\leq h_{2}(\theta)

Then

\underset{ D }{ \iint }f(x,y)\,\mathrm{d}A = \int_{\alpha}^{\beta} \int_{h_{1}(\theta)}^{h_{2}(\theta)} f(r\cos\theta,r\sin\theta)\cdot r \, \mathrm{d}r \, \mathrm{d}\theta

15.4 Applications of Double Integrals

Density and Mass

Let $\rho(x,y)$ denote the density function of a region $D$ . Thus,

\rho(x,y)=\lim \frac{\Delta m}{\Delta A}

Where $\Delta m,\Delta A$ are the mass and area of an infinitesimal rectangle containing $(x,y)$ . Thus, it naturally follows that

m=\underset{ D }{ \iint }\rho(x,y)\,\mathrm{d}A

Physicists consider other types of density that aren't mass as well. For instance, if the charge density if given by $\sigma(x,y)$ at $(x,y)$ in $D$ , then the total charge $Q$ is simply

Q=\underset{ D }{ \iint }\sigma(x,y)\,\mathrm{d}A

Moments and Centers of Mass

The moment of a particle is the product of its mass and its directed distance from the axis. Consider an infinitesimal rectangle $R_{xy}$ containing the point $(x,y)$ . Then, the mass of $R_{xy}\approx \rho(x,y)\Delta A$ , and the moment of $R_{xy}$ with respect to the $x$ -axis is therefore

M_{x}(R_{xy})=\rho(x,y)\Delta A\cdot y_{xy}

Taking the limit of the sum of these quantities gives us the moment of the entire lamina about the $x$ -axis:

M_{x}=\underset{ D }{ \iint }y\rho(x,y)\,\mathrm{d}A

And about the $y$ -axis,

M_{y}=\underset{ D }{ \iint }x\rho(x,y)\,\mathrm{d}A

Using these, we derive the coordinates $(\overline{x},\overline{y})$ of the center of mass of a lamina occupying the region $D$ :

\overline{x}=\frac{M_{y}}{m}=\frac{1}{m}\underset{ D }{ \iint }x\rho(x,y)\,\mathrm{d}A \qquad \qquad \overline{y}=\frac{M_{x}}{m}=\frac{1}{m}\underset{ D }{ \iint }y\rho(x,y)\,\mathrm{d}A

Where the mass $m$ is given by

m=\underset{ D }{ \iint }\rho(x,y)\,\mathrm{d}A

Moment of Inertia

The moment of inertia, or second moment, of a particle of mass $m$ about an axis is defined to be $mr^{2}$ , where $r$ is the distance from the particle to the axis. For a lamina with density function $\rho(x,y)$ and occupying a region $D$ , we can similarly derive equations for this.

\begin{align*} I_{x} &= \underset{ D }{ \iint }y^{2}\rho(x,y)\,\mathrm{d}A \\ I_{y} &= \underset{ D }{ \iint }x^{2}\rho(x,y)\,\mathrm{d}A \end{align*}

The moment of inertia about the origin, or the polar moment of inertia, is meanwhile

I_{0}=\underset{ D }{ \iint }(x^{2}+y^{2})\rho(x,y)\,\mathrm{d}A=I_{x}+I_{y}

Moreover, we may define the radius of gyration of a lamina about an axis is the number $R$ such that

mR^{2}=I

In particular, we may define

m \overline{\overline{y}}^{2}=I_{x} \qquad \qquad m \overline{\overline{x}}^{2}=I_{y}

Probability

The probability density function (PDF) $f$ of a continuous random variable $X$ has the property that $f(x)\geq 0,\forall x$ , and $\int_{-\infty}^{\infty} f(x) \, \mathrm{d}x=1$ . Additionally,

P(a\leq X\leq b)=\int_{a}^{b} f(x) \, \mathrm{d}x

Now, consider a pair of continuous random variables $X,Y$ . The joint density function of $X,Y$ is a function $f$ of two variables such that the probability that $(X,Y)$ lies in a region $D$ is

P((X,Y)\in D)=\underset{ D }{ \iint }f(x,y)\,\mathrm{d}A

Note also that

\underset{ \mathbb{R}^{2} }{ \iint }f(x,y)\,\mathrm{d}A=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y) \, \mathrm{d}x \, \mathrm{d}y = 1

Expected Values

If $X$ is a random variable with PDF $f$ , then its mean is

\mu=\int_{-\infty}^{\infty} xf(x) \, \mathrm{d}x

If $X,Y$ are random variables with joint density function $f$ , we define the $X$ -mean and $Y$ -mean, i.e. the expected values of $X,Y$ , to be

\mu_{1}=\underset{ \mathbb{R}^{2} }{ \iint }xf(x,y)\,\mathrm{d}A \qquad \qquad \mu_{2}=\underset{ \mathbb{R}^{2} }{ \iint }yf(x,y)\,\mathrm{d}A

15.5 Surface Area

Consider a point $P_{xy}=(x_{i},y_{j},f(x_{i},y_{j}))$ on some surface $S$ described by $z=f(x,y)$ . Now, consider some infinitesimal rectangle $R_{ij}$ lying in the $xy$ -plane that contains the projection of $P_{ij}$ down into the $xy$ -plane. Then, we may approximate the surface area of the curve lying above this rectangle $R_{ij}$ with the area of the parallelogram of the tangent plane $T_{ij}$ at point $P_{ij}$ bounded by the bounds of $R_{ij}$ (in $x$ and $y$ ). We will let one vertex of this parallelogram lie at point $P_{ij}$ (i.e. $P_{ij}$ lies directly above a vertex of $R_{ij}$ ). Let this area be denoted by $\Delta T_{ij}$ . Then, we may approximate the surface area of $S$ as

A(S)=\lim_{ m,n \to \infty } \sum_{i=1}^{m} \sum_{j=1}^{n} \Delta T_{ij}

Now, we aim to find an expression for $\Delta T_{ij}$ . Since it is the area of some parallelogram in $T_{ij}$ , we can let $\mathbf{a,b}$ be vectors that lie along the sides of this parallelogram. Additionally, let $\Delta x$ and $\Delta y$ be the (infinitesimal) side lengths of the rectangle $R_{ij}$ . Then, we may define

\begin{align*} \mathbf{a}&=\Delta x\mathbf{i} + f_{x}(x_{i},y_{j})\Delta x\mathbf{k} \\ \mathbf{b}&=\Delta y\mathbf{j}+f_{y}(x_{i},y_{j})\Delta y\mathbf{k} \end{align*}

Note that we are enabled to do this because we can just let the sides of the rectangle $R_{ij}$ be parallel to the $x$ and $y$ axes, and therefore one side of the parallelogram will have no component in the $x$ direction, and similarly the other side will have no component in the $y$ direction.

Since the area of the parallelogram with respect to $\mathbf{a,b}$ is $\lvert \mathbf{a}\times \mathbf{b} \rvert$ , we may write that

\Delta T_{ij} = \lvert \mathbf{a}\times \mathbf{b} \rvert =\det \begin{bmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \Delta x & 0 & f_{x}(x_{i},y_{j})\Delta x \\ 0 & \Delta y & f_{y}(x_{i},y_{j})\Delta y \end{bmatrix}

Simplifying, we eventually get that

\Delta T_{ij} = \sqrt{ [f_{x}(x_{i},y_{j})]^{2}+[f_{y}(x_{i},y_{j})]^{2}+1 }\;\Delta A

Where $\Delta A=\Delta x\Delta y$ , i.e. the area of $R_{ij}$ . And thus,

Surface Area

The area of the surface $S$ with equation $z=f(x,y),(x,y)\in D$ , where $f_{x},f_{y}$ are continuous, is

\begin{align*} A(S)&=\underset{ D }{ \iint }\sqrt{ [f_{x}(x_{i},y_{j})]^{2}+[f_{y}(x_{i},y_{j})]^{2}+1 }\,\mathrm{d}A \\ &= \underset{ D }{ \iint }\sqrt{ 1+\left( \frac{ \partial z }{ \partial x } \right)^{2}+\left( \frac{ \partial z }{ \partial y } \right)^{2} } \,\mathrm{d}A \end{align*}

Note that the similarity between the surface area formula and the arc length formula may help with remembering this:

L=\int_{a}^{b} \sqrt{ 1+\left( \frac{\mathrm{d} y }{\mathrm{d} x } \right)^{2} } \, \mathrm{d}x

15.6 Triple Integrals

First, let's consider the triple integral of a function $f$ defined on a rectangular box

B=\{ (x,y,z)\mid a\leq x\leq b,c\leq x\leq d,e\leq z\leq f \}

Similar to previous sections, we consider dividing $B$ into infinitesimal sub-boxes and then finding the value of the triple integral by summing over the boxes' volumes.

\underset{ B }{ \iiint }f(x,y,z)\,\mathrm{d}V=\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

Just like double integrals, we can express this in the form of iterated integrals

Fubini's Theorem for Triple Integrals

If $f$ is continuous on the rectangular box $B=[a,b]\times[c,d]\times[e,f]$ , then

\underset{ R }{ \iiint }f(x,y,z)\,\mathrm{d}V = \int_{e}^{f} \int_{c}^{d} \int_{a}^{b} f(x,y,z) \, \mathrm{d}x \, \mathrm{d}y \, \mathrm{d}z

We may similarly extend this to a triple integral of a function $f$ over a general bounded region $E$ in 3d space (i.e. a solid), nearly identically to how it is done in 2d. The derivation is left as an exercise to the reader.

Triple Integrals over General Regions

For a function $f$ bounded by a region $E$ such that

E=\{ (x,y,z)\mid a\leq x\leq b,g_{1}(x)\leq y\leq g_{2}(x),u_{1}(x,y)\leq z\leq u_{2}(x,y) \}

\underset{ E }{ \iiint }f(x,y,z)\,\mathrm{d}V = \int_{a}^{b} \int_{g_{1}(x)}^{g_{2}(x)} \int_{h_{1}(x,y)}^{h_{2}(x,y)} f(x,y) \, \mathrm{d}z \, \mathrm{d}y \, \mathrm{d}x

This can similarly be extended to work for $y$ or $z$ as the outermost integral.

Note that the problem may sometimes be made easier by first focusing on reducing the problem to 2d, typically by projecting the solid to the $xy$ , $yz$ , or $xz$ plane, depending on the solid type.

For instance, a type 2 region $E$ (as it is termed in the textbook) is defined by

E=\{ (x,y,z)\mid(y,z)\in D,g_{1}(y,z)\leq x\leq g_{2}(y,z) \}

Where $D$ is the projection of $E$ onto the $yz$ plane. Then,

\underset{ E }{ \iiint }f(x,y,z)\,\mathrm{d}V=\underset{ D }{ \iint }\left[ \int_{g_{1}(x,y)}^{g_{2}(x,y)} f(x,y,z) \, \mathrm{d}x \right] \,\mathrm{d}A

This can be extended to work for projections onto the $xy$ and $xz$ planes (type 1 and 3, respectively).

We can modify the applications discussed in [[#15.4 Applications of Double Integrals|15.4]] for triple integrals.

\begin{align*} &&& m=\underset{ E }{ \iiint }\rho(x,y,z)\,\mathrm{d}V && \\ & M_{yz} = \underset{ E }{ \iiint } x\rho(x,y,z)\,\mathrm{d}V && M_{xz} = \underset{ E }{ \iiint } y\rho(x,y,z)\,\mathrm{d}V && M_{xy} = \underset{ E }{ \iiint } z\rho(x,y,z)\,\mathrm{d}V \\ & \overline{x} = \frac{M_{yz}}{m} && \overline{y} = \frac{M_{xz}}{m} && \overline{z} = \frac{M_{xy}}{m} \\ & I_{x} = \underset{ E }{ \iiint } (y^{2}+z^{2})\rho(x,y,z)\,\mathrm{d}V && I_{y} = \underset{ E }{ \iiint } (x^{2}+z^{2})\rho(x,y,z)\,\mathrm{d}V && I_{z} = \underset{ E }{ \iiint } (x^{2}+y^{2})\rho(x,y,z)\,\mathrm{d}V \\ &&& Q = \underset{ E }{ \iiint }\sigma(x,y,z)\,\mathrm{d}V && \\ &&& P((X,Y,Z)\in E)=\underset{ E }{ \iiint }f(x,y,z)\,\mathrm{d}V \end{align*}

15.7 Triple Integrals in Cylindrical Coordinates

First, some conversions:

\begin{align*} &=r\cos \theta & y &=r\sin\theta & z&=z \\ r^{2}&=x^{2}+y^{2} & \tan\theta&=\frac{y}{x} & z&=z \end{align*}

Now, consider $E$ , a type 1 region whose projection $D$ onto the $xy$ -plane is described in polar coordinates. Suppose $f$ is continuous and

E=\{ (x,y,z)\mid(x,y)\in D,u_{1}(x,y)\leq z\leq u_{2}(x,y) \}

Where $D$ is

D=\{ (r,\theta)\mid\alpha\leq \theta\leq \beta,h_{1}(\theta)\leq r\leq h_{2}(\theta) \}

We know that

\underset{ E }{ \iiint }f(x,y,z)\,\mathrm{d}V = \underset{ D }{ \iint }\left[ \int_{u_{1}(x,y)}^{u_{2}(x,y)} f(x,y,z) \, \mathrm{d}z \right]\,\mathrm{d}A

Thus, we can write the following formula:

Triple Integration in Cylindrical Coordinates

\underset{ E }{ \iiint }f(x,y,z)\,\mathrm{d}V = \int_{\alpha}^{\beta} \int_{h_{1}(\theta)}^{h_{2}(\theta)} \int_{u_{1}(r\cos\theta,r\sin\theta)}^{u_{2}(r\cos\theta,r\sin\theta)} f(r\cos\theta,r\sin\theta,z)r \, \mathrm{d}z \, \mathrm{d}r \, \mathrm{d}\theta

15.8 Triple Integrals in Spherical Coordinates

Some quick reminders about spherical coordinates

P(\rho,\theta,\phi) \qquad \rho\geq 0\qquad 0\leq \phi\leq \pi

Now, some conversions

\begin{align*} x&=\rho \sin\phi \cos\theta & y&=\rho \sin \phi \sin\theta & z&=\rho \cos \phi \\ \rho&=\sqrt{ x^{2}+y^{2}+z^{2} } & \theta &= \arccos\frac{x}{\rho \sin \phi} & \phi&=\arccos \frac{z}{\rho} \end{align*}

Now, with rectangular coordinates, we began by integrating rectangular prisms for triple integrals. In spherical coordinates, we may consider instead a spherical wedge $E$ defined by

E=\{ (\rho,\theta,\phi)\mid a\leq p\leq b,\;\alpha\leq \theta\leq \beta,\;c\leq \phi\leq d \}

The derivation will not be covered in detail here, since it is mostly similar to other derivations we have done. Review the textbook if you are interested.

Eventually, we find

Triple Integration in Spherical Coordinates

\underset{ E }{ \iiint }f(x,y,z)\,\mathrm{d}V = \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 \, \mathrm{d}\rho \, \mathrm{d}\theta \, \mathrm{d}\phi

Where $E$ is a spherical wedge described as above.

The formula can be extended to include more general spherical regions with

E=\{ (\rho,\theta,\phi)\mid\alpha\leq \theta\leq \beta,\;c\leq \phi\leq d,\;g_{1}(\theta,\phi)\leq \rho\leq g_{2}(\theta,\phi) \}

15.9 Change of Variables in Multiple Integrals

In other words, $u$ -substitution for multivariate integrals. Consider the change of variables given by a transformation $T$ from the $uv$ -plane to the $xy$ -plane

T(u,v)=(x,y)

Where $x,y$ are described by

x=g(u,v)\qquad y=h(u,v)

It is also sometimes denoted

x=x(u,v)\qquad y=y(u,v)

It is typically assumed that $T$ is a $C^1$ transformation, which means that $g$ and $h$ have continuous first-order partial derivatives.

This transformation $T$ is defined as a function whose domain and range are subsets of $\mathbb{R}^{2}$ .

If $T(u,v)=(x,y)$ , the point $(x,y)$ is the image of $(u,v)$ for $T$ .
If $T(u_{1},v_{1})=T(u_{2},v_{2})\implies (u_{1},v_{1})\neq(u_{2},v_{2})$ , i.e. not two points have the same image, $T$ is injective.
If $T$ is injective, $\exists T^{-1}$ , the unique inverse transformation from the $xy$ -plane to the $uv$ -plane.

The equations describing the relationship of $x,y$ with $u,v$ can typically be solved in terms of the set of variables describing the image to produce the image under transformation $T$ .

Consider a rectangle $S$ in the $uv$ -plane whose lower left corner is $(u_{0},v_{0})$ and has dimensions $\Delta u$ and $\Delta v$ . Consider that the image of $S$ under transformation $T$ of the region $R$ can be described by the vector

\mathbf{r}(u,v)=g(u,v)\mathbf{i}+h(u,v)\mathbf{j}

As $u,v$ range over the specified domain. Consider that

\mathbf{r}_{u}=g_{u}(u_{0},v_{0})\mathbf{i}+h_{u}(u_{0},v_{0})\mathbf{j}=\frac{ \partial x }{ \partial u }\mathbf{i}+\frac{ \partial y }{ \partial u } \mathbf{j}

And

\mathbf{r}_{v}=g_{v}(u_{0},v_{0})\mathbf{i}+h_{v}(u_{0},v_{0})\mathbf{j}=\frac{ \partial x }{ \partial v } \mathbf{i}+\frac{ \partial y }{ \partial v } \mathbf{j}

These are the tangent vectors at $(x_{0},y_{0})$ to the image curve in the $xy$ -plane. The image region $R=T(S)$ can thus be approximated by a parallelogram whose sides are determined by the secant vectors

\mathbf{a}=\mathbf{r}(u_{0}+\Delta u,v_{0})-\mathbf{r}(u_{0},v_{0})\qquad \mathbf{b}=\mathbf{r}(u_{0},v_{0}+\Delta v)-\mathbf{r}(u_{0},v_{0})

Consider that

\mathbf{r}_{u}=\lim_{ \Delta u \to 0 } \frac{\mathbf{r}(u_{0}+\Delta u),v_{0}-\mathbf{r}(u_{0},v_{0})}{\Delta u}

And similarly for $\mathbf{r}_{v}$ . Thus,

\mathbf{a}\approx\Delta u \; \mathbf{r}_{u} \qquad \mathbf{b}\approx \Delta v\;\mathbf{r}_{v}

Hence, the area of $R$ can be approximated by

\lvert (\Delta u\;\mathbf{r}_{u})\times (\Delta v \;\mathbf{r}_{v}) \rvert =\boxed{ \lvert \mathbf{r}_{u}\times \mathbf{r}_{v} \rvert \, \Delta u\,\Delta v }

Note that

\mathbf{r}_{u}\times \mathbf{r}_{v}=\begin{vmatrix*} \dfrac{ \partial x }{ \partial u } & \dfrac{ \partial x }{ \partial v } \\ \dfrac{ \partial y }{ \partial u } & \dfrac{ \partial y }{ \partial v } \end{vmatrix*}\mathbf{k}

The determinant here is commonly denoted as the Jacobian.

The Jacobian of a transformation

T

given by

x=g(u,v)

and

y=h(u,v)

\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix*} \dfrac{ \partial x }{ \partial u } & \dfrac{ \partial x }{ \partial v } \\ \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 }

Thus, the area of $R$ , $\Delta A$ , can be approximated as

\Delta A\approx \left\lvert \frac{\partial(x,y)}{\partial(u,v)} \right\rvert \Delta u\,\Delta v

Where the Jacobian is evaluated at $(u_{0},v_{0})$ .

Thus, we come to the topic at hand: changing variables of a double integral.

Change of Variables in a Double Integral

Suppose that $T$ is a $C^1$ transformation whose Jacobian is nonzero that $T$ maps a region $S$ in the $uv$ -plane onto a region $R$ in the $xy$ -plane. Suppose that $f$ is continuous on $R$ and that $R$ and $S$ are type I or type II plane regions. Suppose also that $T$ is injective, except perhaps on the boundary of $S$ . Then

\underset{ R }{ \iint }f(x,y)\,\mathrm{d}A = \underset{ S }{ \iint }f(x(u,v),y(u,v))\left\lvert \frac{\partial(x,y)}{\partial(u,v)} \right\rvert \,\mathrm{d}u\,\mathrm{d}v

As an exercise, consider the transformation $T$ that maps from polar coordinates to rectangle coordinates, i.e. $x=g(r,\theta)=r\cos\theta$ and $y=h(r,\theta)=r\sin\theta$ . Show that we derive the same formula for double integration over polar coordinates.

What about triple integrals? We can define the Jacobian of $T$ in 3 dimensions similarly.

\frac{\partial(x,y,z)}{\partial(u,v,w)}=\begin{vmatrix*} \dfrac{ \partial x }{ \partial u } & \dfrac{ \partial x }{ \partial v } & \dfrac{ \partial x }{ \partial w } \\ \dfrac{ \partial y }{ \partial u } & \dfrac{ \partial y }{ \partial v } & \dfrac{ \partial y }{ \partial w } \\ \dfrac{ \partial z }{ \partial u } & \dfrac{ \partial z }{ \partial v } & \dfrac{ \partial z }{ \partial w } \end{vmatrix*}

The formula for changing variables for triple integrals follows similarly. As an exercise, verify this for triple integrals in spherical or cylindrical coordinates.