Applications of multiple integrals

We can use multiple integrals to compute areas and volumes of various shapes.

Area of a planar region

Definition (Area)

Let $D$ be a bounded closed region in ${\mathbb{R}}^2$. $D$ is said to have an area if the multiple integral of the constant function 1 over $D$, ${\iint }_{D}dxdy$, exists. Its value is denoted by $\mu (D)$:

$$\mu (D)={\iint }_{D}dxdy.$$

Example. Let us calculate the area of the disk $D=\{(x,y)\mid x^2+y^2\le a^2\}$. Using the polar coordinates, $x=r\cos \theta ,y=r\sin \theta$, $dxdy=rdrd\theta$, and the ranges of $r$ and $\theta$ are $[0,a]$ and $[0,2\pi ]$, respectively. Thus,

$$\begin{array}{rcl}\mu (D)& =& {\iint }_{D}dxdy\\ & =& {\int }_0^a\left({\int }_0^{2\pi }rd\theta \right)dr\\ & =& 2\pi {\int }_0^{a}rdr\\ & =& 2\pi {\left[\frac{r^2}{2}\right]}_0^a=\pi a^2.\end{array}$$

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Volume of a solid figure

Definition (Volume)

Let $V$ be a solid figure in the $(x,y,z)$ space ${\mathbb{R}}^3$. $V$ is said to have a volume if the multiple integral

$$\mu (V)={\iiint }_{V}dxdydz$$

exists. Its value $\mu (V)$ is called the volume of the solid figure (shape) $V$.

The following spherical coordinates (See Figure 1) are a 3-dimensional (${\mathbb{R}}^3$) analog of the polar coordinates in ${\mathbb{R}}^2$. For the point $P=(x,y,z)\in {\mathbb{R}}^3$, let $r$ be its distance from the origin,

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

$\theta$ be the angle between $\overrightarrow{OP}$ and the $z$-axis,

$$\theta =\arccos \frac{z}{\sqrt{x^2+y^2+z^2}},$$

$\phi$ be the angle between the $x$-axis and the projection of $\overrightarrow{OP}$ on the $x$-$y$ plane,

$$\begin{array}{}\text{(Eq:atan2)}& \phi =\mathrm{atan}2(y,x)=\{\begin{array}{ll}\arctan (y/x)& \text{if }x>0,\\ \arctan (y/x)+\pi & \text{if }x<0\text{ and }y\ge 0,\\ \arctan (y/x)-\pi & \text{if }x<0\text{ and }y<0,\\ +\frac{\pi }{2}& \text{if }x=0\text{ and }y>0,\\ -\frac{\pi }{2}& \text{if }x=0\text{ and }y<0,\\ \text{undefined}& \text{if }x=y=0.\end{array}\end{array}$$

Figure 1. The spherical coordinate system.

    (Figure by Andeggs, source)

Exercise. Plot the function on the right-hand side of (Eq:atan2) and study the atan2 function ($\arctan$ with two arguments). □

Then, we have the following transformation of coordinates:

$$\begin{array}{rcl}x& =& r\sin \theta \cos \phi ,\\ y& =& r\sin \theta \sin \phi ,\\ z& =& r\cos \theta \end{array}$$

where $r\in [0,\mathrm{\infty })$, $\theta \in [0,\pi ]$, $\phi \in [0,2\pi ]$. The triple $(r,\theta ,\phi )$ is called the spherical coordinates of $(x,y,z)$. The Jacobi determinant of the spherical coordinates is given by the following:

$$\begin{array}{rcl}\left|\begin{array}{ccc}{x}_r& x_{\theta }& x_{\phi }\\ y_r& y_{\theta }& y_{\phi }\\ z_r& z_{\theta }& z_{\phi }\end{array}\right|& =& \left|\begin{array}{ccc}\sin \theta \cos \phi & r\cos \theta \cos \phi & -r\sin \theta \sin \phi \\ \sin \theta \sin \phi & r\cos \theta \sin \phi & r\sin \theta \cos \phi \\ \cos \theta & -r\sin \theta & 0\end{array}\right|\\ & =& r^2\sin \theta .\end{array}$$

Accordingly, the change of variables to the spherical coordinates in a triple integral will be

$${\iiint }_{D}f(x,y,z)dxdydz={\iiint }_{E}f(x(r,\theta ,\phi ),y(r,\theta ,\phi ),z(r,\theta ,\phi ))r^2\sin \theta drd\theta d\phi .$$

Example. Let us calculate the volume of the sphere with radius $a$ centered at the origin:

$$B=\{(x,y,z)\mid x^2+y^2+z^2\le a^2\}.$$

Using the spherical coordinates, we have $dxdydz=r^2\sin \theta drd\theta d\phi$, and the ranges of $r,\theta ,\phi$ are $[0,a]$, $[0,\pi ]$, $[0,2\pi ]$, respectively. Thus,

$$\begin{array}{rcl}{\iiint }_{B}dxdydz& =& {\int }_0^{a}dr{\int }_0^{\pi }d\theta {\int }_0^{2\pi }{r}^2\sin \theta d\phi \\ & =& {\int }_0^a{r}^{2}dr{\int }_0^{\pi }\sin \theta d\theta {\int }_0^{2\pi }d\phi \\ & =& {\left[\frac{r^3}{3}\right]}_0^a{[-\cos \theta ]}_0^{\pi }\cdot (2\pi )\\ & =& \frac{4\pi a^3}{3}.\end{array}$$

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Example. Let $C_1$ be a right circular cylinder with radius $a$ and height $h$, $C_2$ another right circular cylinder with radius $a$ and height $w$. Let $C$ be a solid made by perpendicularly intersecting $C_1$ and $C_2$ at the midpoints of their axes. Let us find the volume of $C$.

Let us place $C$ so that the axis of $C_1$ is on the $x$-axis and the axis of $C_2$ is on the $z$-axis, and the point of intersection is at the origin (Draw figures). Then

$$\begin{array}{rcl}{C}_1& =& \{(x,y,z)\mid y^2+z^2\le a^2,-\frac{h}{2}\le x\le \frac{h}{2}\},\\ C_2& =& \{(x,y,z)\mid x^2+y^2\le a^2,-\frac{w}{2}\le z\le \frac{w}{2}\},\\ C& =& C_1\cup C_2.\end{array}$$

The volume of $C$ is given by

$$\mu (C)=\mu (C_1\cup C_2)=\mu (C_1)+\mu (C_2)-\mu (C_1\cap C_2)$$

where

$$\begin{array}{rcl}\mu (C_1)& =& \pi a^{2}h,\\ \mu (C_2)& =& \pi a^{2}w,\\ C_1\cap C_2& =& \{(x,y,z)\mid x^2+y^2\le a^2,y^2+z^2\le a^2\}.\end{array}$$

By symmetry, $\mu (C_1\cap C_2)=8\mu (W)$ where

$$W=\{(x,y,z)\mid x^2+y^2\le a^2,y^2+z^2\le a^2,x\ge 0,y\ge 0,z\ge 0\}.$$

Let us define the region $D$ as

$$D=\{(x,y)\mid x^2+y^2\le a^2,x\ge 0,y\ge 0\}.$$

Then,

$$\begin{array}{rcl}\mu (C_1\cap C_2)& =& 8{\iiint }_{W}dxdydz=8{\iint }_{D}dxdy{\int }_0^{\sqrt{a^2-y^2}}dz\\ & =& 8{\iint }_D\sqrt{a^2-y^2}dxdy=8{\int }_0^{a}dy{\int }_0^{\sqrt{a^2-y^2}}\sqrt{a^2-y^2}dx\\ & =& 8{\int }_0^a(a^2-y^2)dy=8{[a^{2}y-\frac{y^3}{3}]}_0^a\\ & =& \frac{16a^3}{3}.\end{array}$$

Thus, we have

$$\mu (C)=\pi a^2(h+w)-\frac{16a^3}{3}.$$

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Definition (Solid of revolution)

Let $f(x)$ be a continuous function on $[a,b]$ such that $f(x)>0$ for all $x\in [a,b]$. The solid in ${\mathbb{R}}^3$ defined by the set

$$V=\{(x,y,z)\mid a\le x\le b,y^2+z^2\le \{f(x){\}}^2\}$$

is called a solid of revolution. That is, the solid of revolution is obtained by rotating the graph of $y=f(x)$ around the $x$-axis. Accordingly, the $x$-axis is called the axis of revolution.

Example. Let $V$ be the solid of revolution of $y=f(x)$ around the $x$-axis. We can see that its volume is given by

$$\mu (V)=\pi {\int }_a^b\{f(x){\}}^{2}dx.$$

This can be proved as follows. 

Let $D(x)$ be the circle centered at the origin with radius $f(x)$ in the $(y,z)$ plane. By using the polar coordinates $y=r\cos \theta$, $z=r\sin \theta$, we have $dydz=rdrd\theta$ where the ranges of $r$ and $\theta$ are $[0,f(x)]$ and $[0,2\pi ]$, respectively. Therefore

$$\begin{array}{rcl}\mu (V)& =& {\iiint }_{V}dxdydz={\int }_a^{b}dx{\iint }_{D(x)}dydz\\ & =& {\int }_a^{b}dx{\int }_0^{f(x)}rdr{\int }_0^{2\pi }d\theta \\ & =& {\int }_a^{b}2\pi {\left[\frac{r^2}{2}\right]}_{r=0}^{r=f(x)}dx\\ & =& \pi {\int }_a^b\{f(x){\}}^{2}dx.\end{array}$$

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Area of a surface in space

Consider a surface in ${\mathbb{R}}^3$ defined by parameterized functions:

$$(u,v)\mapsto \mathrm{\Phi }(u,v)=(x(u,v),y(u,v),z(u,v)).$$

We assume that $x(u,v),y(u,v),z(u,v)$ are of class $C^1$. When $(u,v)$ moves in a bounded closed region in ${\mathbb{R}}^2$, $\mathrm{\Phi }(u,v)$ moves in a region $S=\mathrm{\Phi }(D)$ on a (possibly curved) surface. Let us find the area of $S$.

First, consider the rectangular region $\tilde{D}=[a,b]\times [c,d]$ such that $D\subset \tilde{D}$, and its partition $\mathrm{\Delta }$ such that

$$\mathrm{\Delta }:\{\begin{array}{c}a=a_0<a_1<a_2<\cdots <a_{n-1}<a_n=b,\\ c=c_0<c_1<c_2<\cdots <c_{m-1}<c_m=d.\end{array}$$

Hence, the region $\tilde{D}$ is partitioned into $nm$ small rectangular regions

$${\tilde{D}}_{ij}=[a_i,a_{i+1}]\times [c_j,c_{j+1}](i=0,1,\cdots ,n-1;j=0,1,\cdots ,m-1).$$

Let $D_{ij}=D\cap {\tilde{D}}_{ij}$. The desired area is obtained as the limit of the sum

$$\sum _{i=0}^{n-1}\sum _{j=0}^{m-1}\mu (\mathrm{\Phi }(D_{ij}))$$

as $|\mathrm{\Delta }|\to 0$ (Recall that $|\mathrm{\Delta }|$ indicates the mesh of the partition).

We approximate the surface region $\mathrm{\Phi }(D_{ij})$ by the parallelogram spanned by the two tangent vectors at $\mathrm{\Phi }(a_i,c_j)$:

$$\begin{array}{rcl}\mathbf{u}& =& (x_u(a_i,c_j),y_u(a_i,c_j),z_u(a_i,c_j))\mathrm{\Delta }{a}_i,\\ \mathbf{v}& =& (x_v(a_i,c_j),y_v(a_i,c_j),z_v(a_i,c_j))\mathrm{\Delta }{c}_j\end{array}$$

where $\mathrm{\Delta }{a}_i=a_{i+1}-a_i$ and $\mathrm{\Delta }{c}_j=c_{j+1}-c_j$, and $x_u(a_i,c_j)=\frac{\partial x}{\partial u}(a_i,c_j)$, etc.

Quiz. Why is it reasonable to approximate $\mathrm{\Phi }(D_{ij})$ by the parallelogram spanned by the above vectors $\mathbf{u}$ and $\mathbf{v}$? □

Now, recall that the area of the parallelogram spanned by $\mathbf{u}=(u_1,u_2,u_3)$ and $\mathbf{v}=(v_1,v_2,v_3)$ is given by the length of their vector product

$$\mathbf{u}\times \mathbf{v}=(u_2{v}_3-u_3{v}_2,v_1{u}_3-v_3{u}_1,u_1{v}_2-u_2{v}_1).$$

Thus, the area of $\mathrm{\Phi }(D_{ij})$ is approximated as

$$\mu (\mathrm{\Phi }(D_{ij}))\approx \sqrt{(y_u{z}_v-z_u{y}_v{)}^2+(z_u{x}_v-x_u{z}_v{)}^2+(x_u{y}_v-y_u{x}_v{)}^2}\mathrm{\Delta }{a}_i\mathrm{\Delta }{c}_j.$$

It is understood that all the derivatives are evaluated at $(a_i,c_j)$.

Remark. The vector product

$$\mathbf{u}\times \mathbf{v}=(y_u{z}_v-z_u{y}_v,z_u{x}_v-x_u{z}_v,x_u{y}_v-y_u{x}_v)$$

is normal to the surface $S$ at $\mathrm{\Phi }(a_i,c_j)$. □

By summing these over all small rectangles $D_{ij}$, and taking the limit of $|\mathrm{\Delta }|\to 0$, we have just proved the following theorem.

Theorem (Surface area)

Let $S$ be the surface determined by the parametric representation with the $C^1$ functions

$$(u,v)\mapsto \mathrm{\Phi }(u,v)=(x(u,v),y(u,v),z(u,v))$$

where $(u,v)$ moves in a bounded closed region $D$. Then the area of $S$ is given by

$$\mu (S)={\iint }_D\sqrt{(y_u{z}_v-z_u{y}_v{)}^2+(z_u{x}_v-x_u{z}_v{)}^2+(x_u{y}_v-y_u{x}_v{)}^2}dudv.$$

Corollary

Let $z=f(x,y)$ be a $C^1$ function defined on a neighbor of the bounded closed region $D$ in ${\mathbb{R}}^2$. The surface area of the graph $\mathrm{\Gamma }$

$$\mathrm{\Gamma }=\{(x,y,z)\mid (x,y)\in D,z=f(x,y)\}$$

is given by

$$\mu (\mathrm{\Gamma })={\iint }_D\sqrt{\{f_x(x,y){\}}^2+\{f_y(x,y){\}}^2+1}dxdy.$$

Proof. Consider the "parametric" representation

$$(x,y)\mapsto (x,y,f(x,y))$$

and apply the above theorem (Surface area). ■

Example. Let us find the surface area of

$$S=\{(x,y,z)\mid z=a^2-(x^2+y^2),z\ge 0\}$$

where $a>0$.

Since $z\ge 0$, $(x,y)$ can move in the region $D=\{(x,y)\mid x^2+y^2\le a^2\}$. Let $f(x,y)=a^2-(x^2+y^2)$. Then,

$$\begin{array}{rcl}{f}_x(x,y)& =& -2x,\\ f_y(x,y)& =& -2y.\end{array}$$

Using the polar coordinates $x=r\cos \theta ,y=r\sin \theta$ with appropriate ranges for $r$ and $\theta$,

$$\begin{array}{rcl}\mu (S)& =& {\iint }_D\sqrt{4x^2+4y^2+1}dxdy\\ & =& {\int }_0^{2\pi }d\theta {\int }_0^{a}r\sqrt{4r^2+1}dr\\ & =& 2\pi {[\frac{1}{12}(4r^2+1{)}^{\frac{3}{2}}]}_{r=0}^{r=a}\\ & =& \frac{\pi }{6}\{(\sqrt{4a^2+1}{)}^3-1\}.\end{array}$$

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Another application is the area of a surface of revolution.

Corollary (Surface area of revolution)

Let $y=f(x)$ be a $C^1$ function on $[a,b]$. Let $S$ be the surface obtained by rotating the graph of $y=f(x)$ around the $x$-axis. Then, its area is given by

$$\mu (S)=2\pi {\int }_a^b|f(x)|\sqrt{1+\{f^{\prime }(x){\}}^2}dx.$$

Proof. The surface $S$ is given by the following parametric representation:

$$(x,\theta )\mapsto (x,f(x)\cos \theta ,f(x)\sin \theta ),(a\le x\le b,0\le \theta \le 2\pi ).$$

Now, apply the Theorem (Surface area), and we are done. ■