\(\def\ans#1{\bbox[border:1px solid green,6pt]{#1}}\)

Surface Integrals

So far, we've seen integrals over lines and paths, integrals over regions, and integrals over volumes. Now, we'll do integrals over surfaces in \(\mathbb{R}^3\).

Parametric Surfaces

First, we need to discuss how to describe a surface parametrically. Recall that curves in space can be parametrized in terms of one parameter t; surfaces in space will require two parameters u and v:

\[\vec{r}(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle\]

Express the following surface of a cylinder parametrically:

\[S = \{(x,y,z)\ :\ x=a\cos\theta, y=a\sin\theta, 0 \leq \theta \leq 2\pi, 0 \leq z \leq h\}\]

Solution

Let \(u=\theta\) and \(v=z\):

\[\ans{\begin{align} \vec{r}(u,v) &= \langle a \cos u, a \sin u, v \rangle\\ \\ &0 \leq u \leq 2\pi,\ 0 \leq v \leq h \end{align}}\]

Express the following plane parametrically:

\[3x-2y+z=2\]

Solution

Let \(u=x\) and \(v=y\):

\[\ans{\vec{r}(u,v) = \langle u,v,2-3u+2v \rangle}\]

Express the following surface of a paraboloid parametrically:

\[z=x^2+y^2\ \ \ \ 0 \leq z \leq 9\]

Option A

Let \(u=x\) and \(v=y\):

\[\ans{\begin{align} \vec{r}(u,v) &= \langle u,v,u^2+v^2 \rangle\\ \\ &0 \leq v \leq 9 \end{align}}\]

Option B

Let \(u=\theta\) and \(v=r = \sqrt{z}\):

\[\ans{\begin{align} \vec{r}(u,v) &= \langle v \cos u, v \sin u, v^2 \rangle\\ \\ &0 \leq u \leq 2\pi, \ 0 \leq v \leq 3 \end{align}}\]

Surface Integrals


If a surface is expressed parametrically as \(\vec{r} (u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle\), then \[\begin{align} \vec{r}_u &= \dfrac{\partial \vec{r}}{\partial u} = \left\langle \dfrac{\partial x}{\partial u},\dfrac{\partial y}{\partial u},\dfrac{\partial z}{\partial u} \right\rangle\\ \vec{r}_v &= \dfrac{\partial \vec{r}}{\partial v} = \left\langle \dfrac{\partial x}{\partial v},\dfrac{\partial y}{\partial v},\dfrac{\partial z}{\partial v} \right\rangle\\ \end{align}\] and the area of a differential rectangle on the surface is given by \(|\vec{r}_u \times \vec{r}_v| \ du\ dv\).

Definition

Let \(f\) be a continuous scalar-valued function on a smooth surface \(S\) given parametrically by \[\vec{r}(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle,\] where \(a \leq u \leq b\) and \(c \leq v \leq d\). If \(\vec{r}_u\) and \(\vec{r}_v\) are continuous on this region \(R\) \[R = \{(u,v)\ :\ a \leq u \leq b, c \leq v \leq d\}\] then the surface integral of \(f\) over \(S\) is \[\iint_S f(x,y,z) \ dS = \iint_R f\bigg( x(u,v),y(u,v),z(u,v) \bigg) \ |\vec{r}_u \times \vec{r}_v| \ dA.\]


Note

If \(f(x,y,z)=1\), this integral gives the surface area of \(S\):

\[\textrm{Surface area } = \iint_R |\vec{r}_u \times \vec{r}_v| \ dA\]

Examples

Find the surface area of the cylinder with radius 3 and height 5.

\[\begin{align} \vec{r}(u,v) &= \langle 3 \cos u, 3 \sin u, v \rangle\\ \\ &0 \leq u \leq 2\pi, \ 0 \leq v \leq 5 \end{align}\]

Solution

\[\begin{align} \vec{r}_u &= \langle -3 \sin u, 3 \cos u, 0 \rangle\\ \vec{r}_v &= \langle 0,0,1 \rangle\\ \\ \vec{r}_u \times \vec{r}_v &= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\\ -3 \sin u & 3 \cos u & 0\\ 0 & 0 & 1 \end{vmatrix}\\ \\ &= \langle 3 \cos u, 3 \sin u, 0 \rangle\\ \\ |\vec{r}_u \times \vec{r}_v| &= 3 \end{align}\]

Therefore, the surface area is

\[SA = \int_0^{2\pi} \int_0^5 3 \ dv\ du = \int_0^{2\pi} 15 \ du = \ans{30\pi}\]

Try it yourself:

(click on the problem to show/hide the answer)

Find the surface area of the plane \(3x-2y+z=2\) over the region \(0 \leq x \leq 2\), \(0 \leq y \leq 1\).
\(\displaystyle\int_0^2 \displaystyle\int_0^1 \sqrt{14}\ dv\ du = 2\sqrt{14}\)

Find the surface area of the following part of a paraboloid:

\[z=x^2+y^2, \ \ \ \ 0 \leq z \leq 9\]

Solution

Recall:

\[\begin{align} \vec{r}(u,v) &= \langle v \cos u, v \sin u, v^2 \rangle\\ \\ &0 \leq u \leq 2\pi, \ 0 \leq v \leq 3 \end{align}\]

Then

\[\begin{align} \vec{r}_u &= \langle -v \sin u, v \cos u, 0 \rangle\\ \vec{r}_v &= \langle \cos u, \sin u, 2v \rangle\\ \\ \vec{r}_u \times \vec{r}_v &= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\\ -v \sin u & -v \cos u & 0\\ \cos u & \sin u & 2v \end{vmatrix}\\ \\ &= \langle 2v^2 \cos u, 2v^2 \sin u, -v \rangle\\ \\ |\vec{r}_u \times \vec{r}_v| &= v\sqrt{4v^2+1} \end{align}\]

Therefore, the surface area is

\[\begin{align} SA &= \int_0^3 \int_0^{2\pi} v\sqrt{4v^2+1} \ du\ dv\\ &= 2\pi \int_0^3 v\sqrt{4v^2+1} \ dv\\ &= \dfrac{\pi}{6} \bigg[(4v^2+1)^{3/2}\bigg]_0^3\\ &= \ans{\dfrac{\pi}{6}\left(37^{3/2}-1\right)} \end{align}\]

Evaluate the integral \(\displaystyle\iint_S \dfrac{y}{z} \ dS\) where \(S\) is the following parametric surface:

\[\vec{r}(u,v) = \langle u^2, u \sin v, u \cos v \rangle, \ \ \ \ 0 \leq u \leq 1, \ \ 0 \leq v \leq \dfrac{\pi}{3}\]

Solution

First, note that \(y=u \sin v\) and \(z=u \cos v\), so \[\dfrac{y}{z} = \dfrac{\sin v}{\cos v}.\]

\[\begin{align} \vec{r}_u &= \langle 2u, \sin v, \cos v \rangle\\ \vec{r}_v &= \langle 0, u \cos v, -u \sin v \rangle\\ \\ \vec{r}_u \times \vec{r}_v &= \langle -u, 2u^2 \sin v, 2u^2 \cos v \rangle\\ \\ |\vec{r}_u \times \vec{r}_v| &= u\sqrt{1+4u^2} \end{align}\]

Therefore, this surface integral is

\[\begin{align} \iint_S \dfrac{y}{z} \ dS &= \int_0^{\pi/3} \int_0^1 \dfrac{\sin v}{\cos v} \cdot u \sqrt{1+4u^2}\ du\ dv\\ &= \int_0^{\pi/3} \left[\dfrac{1}{12} (4u^2+1)^{3/2} \bigg|_0^1 \right] \ \dfrac{\sin v}{\cos v} \ dv\\ &= \dfrac{1}{12} (5^{3/2}-1) \int_0^{\pi/3} \dfrac{\sin v}{\cos v} \ dv\\ &= \dfrac{1}{12} (5^{3/2}-1) \bigg[-\ln |\cos v|\bigg]_0^{\pi/3}\\ &= \ans{-\dfrac{1}{12}(5^{3/2}-1)\ln \dfrac{1}{2}} \end{align}\]

Explicitly Defined Surfaces

If a surface is defined explicitly as \(z=g(x,y)\), we can parameterize it using \(u=x\) and \(v=y\): \[\vec{r} (u,v) = \langle u,v,g(u,v) \rangle\] Then \(\vec{r}_u = \langle 1,0,z_x \rangle\) and \(\vec{r}_v = \langle 0,1,z_y \rangle\):

\[\begin{align} \vec{r}_u \times \vec{r}_v &= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\\ 1 & 0 & z_x\\ 0 & 1 & z_y \end{vmatrix}\\ \\ &= \langle -z_x,-z_y,1 \rangle\\ \\ &\implies |\vec{r}_u \times \vec{r}_v| = \sqrt{z_x^2+z_y^2+1} \end{align}\]

The surface integral can thus be written as

\[\iint_S f(x,y,z) \ dS = \iint_R f\bigg(x,y,g(x,y)\bigg) \sqrt{z_x^2+z_y^2+1} \ dA\]

Again, if \(f(x,y,z)=1\), this integral gives the surface area.

Evaluate the integral \(\displaystyle\iint_S xy \ dS\) where \(S\) is the triangular surface with vertices \((1,0,0)\), \((0,2,0)\), and \((0,0,2)\).


Solution

First, find the equation of this plane by taking the cross product of two vectors in the plane:

\[\begin{align} \vec{PQ} &= \langle -1,0,2 \rangle\\ \vec{PR} &= \langle -1,2,0 \rangle\\ \\ \vec{PQ} \times \vec{PR} = \langle -4,-2,-2 \rangle \textrm{ or } \langle 2,1,1 \rangle \end{align}\]

Therefore, the equation of this plane is \(2(x-1)+1(y-0)+1(z-0)\) or \(z=-2x-y+2\), so \(z_x=-2\) and \(z_y=-1\).

\[\iint_S xy\ dS = \int_0^1 \int_0^{-2x+2} xy \sqrt{(-2)^2+(-1)^2+1} \ dy\ dx \approx \ans{0.408}\]

Surface Integrals in Vector Fields: Flux

There is a special surface integral known as the flux, which measures how much of a vector field \(\vec{F}\) passes through a surface. Physical applications of this include fluid flow, heat flow, electrical current flow, etc. (you may run across heat flux or current flux at some point in the future).


\[\textrm{Flux } = \ans{\iint_S \vec{F} \cdot \hat{n} \ dS = \iint_R \vec{F} \cdot (\vec{r}_u \times \vec{r}_v) \ dA}\]

because \(\hat{n} = \dfrac{\vec{r}_u \times \vec{r}_v}{|\vec{r}_u \times \vec{r}_v|}\).



If the surface is defined explicitly by \(z=g(x,y)\), then \[\ans{\iint_S \vec{F} \cdot \hat{n} \ dS = \iint_R -f z_x - g z_y + h \ dA}\] where \(\vec{F} = \langle f,g,h \rangle\).

Rainfall on a Sloped Roof

If \(\vec{F} = \langle 0,0,-1 \rangle\) and \(z = 4-2x-y\) for \(0 \leq x \leq 2, 0 \leq y \leq 4-2x\), find the flux of \(\vec{F}\) through this surface.


Solution

\[\begin{align} z_x &= -2\\ z_y &= -1\\ \\ \iint_S \vec{F} \cdot \hat{n}\ dS &= \int_0^2 \int_0^{4-2x} -1 \ dy\ dx\\ &= \int_0^2 2x-4 \ dx\\ &= x^2-4x \bigg|_0^2 = \ans{-4} \end{align}\]

This answer gives the amount of rain that passes through this surface in one unit of time (note that it is negative, since the rain is moving downward.