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 \(D\), \(\iint_Ddxdy\), exists. Its value is denoted by \(\mu(D)\):
\[\mu(D) = \iint_Ddxdy.\]
Example. Let us calculate the area of the disk \(D = \{(x,y)\mid x^2 + y^2 \leq a^2\}\). 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,
\[\begin{eqnarray*} \mu(D) &=& \iint_Ddxdy\\ &=&\int_0^a\left(\int_0^{2\pi}rd\theta\right)dr\\ &=&2\pi\int_0^a rdr\\ &=&2\pi\left[\frac{r^2}{2}\right]_0^a = \pi a^2. \end{eqnarray*}\]
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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
\[\mu(V) = \iiint_Vdxdydz\]
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,
\[\theta = \arccos \frac{z}{\sqrt{x^2 + y^2 + z^2}},\]
\(\varphi\) be the angle between the \(x\)-axis and the projection of \(\overrightarrow{OP}\) on the \(x\)-\(y\) plane,
\[\varphi = \mathrm{atan2}(y, x) = \left\{ \begin{array}{ll} \arctan(y/x) & \text{if $x > 0$},\\ \arctan(y/x) + \pi & \text{if $x < 0$ and $y \geq 0$},\\ \arctan(y/x) - \pi & \text{if $x < 0$ and $y < 0$},\\ +\frac{\pi}{2} & \text{if $x = 0$ and $y > 0$},\\ -\frac{\pi}{2} & \text{if $x = 0$ and $y < 0$},\\ \text{undefined} & \text{if $x = y = 0$}. \end{array}\right.\tag{Eq:atan2}\]
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:
\[\begin{eqnarray*} x &=& r\sin\theta\cos\varphi,\\ y &=& r\sin\theta\sin\varphi,\\ z &=& r\cos\theta \end{eqnarray*}\]
where \(r \in [0, \infty)\), \(\theta \in [0, \pi]\), \(\varphi \in [0, 2\pi]\). The triple \((r, \theta, \varphi)\) is called the spherical coordinates of \((x,y,z)\). The Jacobi determinant of the spherical coordinates is given by the following:
\[ \begin{eqnarray*} \begin{vmatrix} x_r & x_\theta & x_\varphi\\ y_r & y_\theta & y_\varphi\\ z_r & z_\theta & z_\varphi \end{vmatrix} &=& \begin{vmatrix} \sin\theta\cos\varphi & r\cos\theta\cos\varphi & -r\sin\theta\sin\varphi\\ \sin\theta\sin\varphi & r\cos\theta\sin\varphi & r\sin\theta\cos\varphi\\ \cos\theta & -r\sin\theta & 0 \end{vmatrix}\\ &=& r^2\sin\theta. \end{eqnarray*}\]
Accordingly, the change of variables to the spherical coordinates in a triple integral will be
\[ \iiint_Df(x,y,z)dxdydz = \iiint_Ef(x(r,\theta,\varphi),y(r,\theta,\varphi),z(r,\theta,\varphi))r^2\sin\theta drd\theta d\varphi.\]
Example. Let us calculate the volume of the sphere with radius \(a\) centered at the origin:
\[B = \{(x,y,z)\mid x^2 + y^2 + z^2 \leq a^2\}.\]
Using the spherical coordinates, we have \(dxdydz = r^2\sin\theta dr d\theta d\varphi\), and the ranges of \(r, \theta, \varphi\) are \([0, a]\), \([0,\pi]\), \([0, 2\pi]\), respectively. Thus,
\[\begin{eqnarray*} \iiint_Bdxdydz &=& \int_0^adr\int_0^{\pi}d\theta\int_0^{2\pi}r^2\sin\theta d\varphi\\ &=&\int_0^ar^2dr\int_0^{\pi}\sin\theta d\theta\int_0^{2\pi}d\varphi\\ &=&\left[\frac{r^3}{3}\right]_{0}^{a}\left[-\cos\theta\right]_{0}^{\pi}\cdot(2\pi)\\ &=&\frac{4\pi a^3}{3}. \end{eqnarray*}\]
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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
\[\begin{eqnarray*} C_1 &=& \left\{(x,y,z) \mid y^2 + z^2 \leq a^2, ~ -\frac{h}{2} \leq x \leq \frac{h}{2}\right\},\\ C_2 &=& \left\{(x,y,z) \mid x^2 + y^2 \leq a^2, ~ -\frac{w}{2} \leq z \leq \frac{w}{2}\right\},\\ C &=& C_1 \cup C_2. \end{eqnarray*}\]
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
\[\begin{eqnarray*} \mu(C_1) &=& \pi a^2 h,\\ \mu(C_2) &=& \pi a^2 w,\\ C_1 \cap C_2 &=& \left\{(x,y,z) \mid x^2 + y^2 \leq a^2, ~ y^2 + z^2 \leq a^2\right\}. \end{eqnarray*}\]
By symmetry, \(\mu(C_1 \cap C_2) = 8\mu(W)\) where
\[W = \left\{(x,y,z) \mid x^2 + y^2 \leq a^2, ~ y^2 + z^2 \leq a^2, ~ x \geq 0, ~ y \geq 0, ~ z \geq 0\right\}.\]
Let us define the region \(D\) as
\[D = \left\{(x,y) \mid x^2 + y^2 \leq a^2, ~ x \geq 0, ~ y \geq 0\right\}.\]
Then,
\[\begin{eqnarray*} \mu(C_1\cap C_2) &=& 8\iiint_Wdxdydz = 8\iint_Ddxdy\int_0^{\sqrt{a^2-y^2}}dz\\ &=&8\iint_D\sqrt{a^2 - y^2}dxdy = 8\int_0^ady\int_0^{\sqrt{a^2 - y^2}}\sqrt{a^2 - y^2}dx\\ &=&8\int_0^{a}(a^2 - y^2)dy = 8\left[a^2y - \frac{y^3}{3}\right]_{0}^a\\ &=& \frac{16a^3}{3}. \end{eqnarray*}\]
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
\[V = \{(x,y,z) \mid a \leq x \leq b, ~ y^2 + z^2 \leq \{f(x)\}^2\}\]
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
\[\mu(V) = \pi\int_a^b\{f(x)\}^2dx.\]
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
\[\begin{eqnarray*} \mu(V) &=& \iiint_Vdxdydz = \int_a^bdx\iint_{D(x)}dydz\\ &=& \int_a^bdx\int_0^{f(x)}rdr\int_0^{2\pi}d\theta\\ &=& \int_a^b2\pi\left[\frac{r^2}{2}\right]_{r=0}^{r=f(x)}dx\\ &=&\pi\int_a^b\{f(x)\}^2dx. \end{eqnarray*}\]
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Area of a surface in space
Consider a surface in \(\mathbb{R}^3\) defined by parameterized functions:
\[(u, v) \mapsto \Phi(u,v) = (x(u,v), y(u,v), z(u,v)).\]
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 \(\mathbb{R}^2\), \(\Phi(u,v)\) moves in a region \(S = \Phi(D)\) 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 \(\Delta\) such that
\[\Delta: \left\{ \begin{array}{c} a = a_0 < a_1 < a_2 < \cdots < a_{n-1} < a_n = b,\\ c = c_0 < c_1 < c_2 < \cdots < c_{m-1} < c_m = d. \end{array} \right.\]
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
\[\sum_{i=0}^{n-1}\sum_{j=0}^{m-1}\mu(\Phi(D_{ij}))\]
as \(|\Delta| \to 0\) (Recall that \(|\Delta|\) indicates the mesh of the partition).
We approximate the surface region \(\Phi(D_{ij})\) by the parallelogram spanned by the two tangent vectors at \(\Phi(a_i, c_j)\):
\[\begin{eqnarray*} \mathbf{u} &=& (x_u(a_i,c_j), y_u(a_i,c_j), z_u(a_i,c_j))\Delta a_{i},\\ \mathbf{v} &=& (x_v(a_i,c_j), y_v(a_i,c_j), z_v(a_i,c_j))\Delta c_{j} \end{eqnarray*}\]
where \(\Delta a_i = a_{i+1} - a_{i}\) and \(\Delta c_j = c_{j+1} - c_{j}\), and \(x_u(a_i, c_j) = \frac{\partial x}{\partial u}(a_i, c_j)\), etc.
Quiz. Why is it reasonable to approximate \(\Phi(D_{ij})\) 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
\[\mathbf{u}\times \mathbf{v} = (u_2v_3 - u_3v_2, v_1u_3 - v_3u_1, u_1v_2 - u_2v_1).\]
Thus, the area of \(\Phi(D_{ij})\) is approximated as
\[\mu(\Phi(D_{ij})) \approx \sqrt{(y_uz_v - z_uy_v)^2 + (z_ux_v - x_uz_v)^2 + (x_uy_v - y_ux_v)^2}\Delta a_i \Delta c_j.\]
It is understood that all the derivatives are evaluated at \((a_i, c_j)\).
Remark. The vector product
\[\mathbf{u}\times\mathbf{v} = (y_uz_v - z_uy_v, z_ux_v - x_uz_v, x_uy_v - y_ux_v)\]
is normal to the surface \(S\) at \(\Phi(a_i, c_j)\). □
By summing these over all small rectangles \(D_{ij}\), and taking the limit of \(|\Delta| \to 0\), 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
\[(u, v) \mapsto \Phi(u,v) = (x(u,v),y(u,v), z(u,v))\]
where \((u,v)\) moves in a bounded closed region \(D\). Then the area of \(S\) is given by
\[\mu(S) = \iint_D\sqrt{(y_uz_v - z_uy_v)^2 + (z_ux_v - x_uz_v)^2 + (x_uy_v - y_ux_v)^2}dudv.\]
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 \(\Gamma\)
\[\Gamma = \{(x, y, z) \mid (x,y) \in D, ~ z = f(x,y)\}\]
is given by
\[\mu(\Gamma) = \iint_D\sqrt{\{f_x(x,y)\}^2 + \{f_y(x,y)\}^2 + 1}~dxdy.\]
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
\[S = \{(x,y,z) \mid z = a^2 - (x^2 + y^2), ~ z \geq 0\}\]
where \(a > 0\).
Since \(z \geq 0\), \((x,y)\) can move in the region \(D = \{(x,y) \mid x^2 + y^2 \leq a^2\}\). Let \(f(x,y) = a^2 - (x^2 + y^2)\). Then,
\[\begin{eqnarray*} f_x(x,y) &=& -2x,\\ f_y(x,y) &=& -2y. \end{eqnarray*}\]
Using the polar coordinates \(x = r\cos\theta, y = r\sin\theta\) with appropriate ranges for \(r\) and \(\theta\),
\[\begin{eqnarray*} \mu(S) &=& \iint_D\sqrt{4x^2 + 4y^2 + 1}dxdy\\ &=& \int_0^{2\pi}d\theta\int_0^ar\sqrt{4r^2 + 1}dr\\ &=&2\pi\left[\frac{1}{12}(4r^2 + 1)^{\frac{3}{2}}\right]_{r=0}^{r=a}\\ &=& \frac{\pi}{6}\{(\sqrt{4a^2 + 1})^3 - 1\}. \end{eqnarray*}\]
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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
\[\mu(S) = 2\pi\int_a^b|f(x)|\sqrt{1 + \{f'(x)\}^2}dx.\]
Proof. The surface \(S\) is given by the following parametric representation:
\[(x, \theta) \mapsto (x, f(x)\cos\theta, f(x)\sin\theta), ~ (a \leq x \leq b, ~ 0\leq \theta \leq 2\pi).\]
Now, apply the Theorem (Surface area), and we are done. ■
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