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

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:

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


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:



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 \(|\vec{r}_u \times \vec{r}_v| \ du\ dv\).


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 \[\iint_S f(x,y,z) \ dS = \iint_R f\bigg( x(u,v),y(u,v),z(u,v) \bigg) \ |\vec{r}_u \times \vec{r}_v| \ dA.\]


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

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


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

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


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

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:

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



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


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

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:

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


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 \((0,0,2)\).


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.


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

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.