MATH 251 Lecture 33

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Confused about Notation?

Surface Integrals

$\underset{\mathrm{\Omega }}{\iint }f\mathrm{d}S$
where $\mathrm{\Omega }$ is a finite surface.

$\mathrm{\Omega }$ can be open (like a section of a surface) or closed (a surface that surrounds a solid)

Like with all integrals, we chop it up into pieces, and add up all the products of the area of the little piece and the value of the function:

1. Parameterize the surface (a surface is defined by two parameters, usually $u$ and $v$) such that $X(u,v)=(x(u,y),y(u,v),z(u,v))$
2. Calculate $\mathrm{d}S$, the "area element", in terms of parameters.
3. Evaluate the resulting double integral.

Example: CoM of Hemisphere Shell

Find the center of mass of the upper hemisphere $x^2+y^2+z^2=1$, where $z\ge 0$. Assume that the mass density (per unit area) is constant $\rho =1$

$\begin{array}{rl}\overline{x}& =0\\ \overline{y}& =0\\ \overline{z}& =\frac{1}{A}\underset{\mathrm{\Omega }}{\iint }z\mathrm{d}S=\frac{1}{2\pi }\underset{\mathrm{\Omega }}{\iint }z\mathrm{d}S\end{array}$

Parameterize the Region

We can use spherical coordinates $(1,\varphi ,\theta )$, where $\varphi \le \frac{\pi }{2}$, so

$\begin{array}{rl}x(\varphi ,\theta )& =\sin \varphi \cos \theta \\ y(\varphi ,\theta )& =\sin \varphi \sin \theta \\ z(\varphi ,\theta )& =\cos \varphi \end{array}$

Determine Area Element

For a rectangular region ${\mathrm{\Omega }}^{\prime }$ that maps to the hemisphere $\mathrm{\Omega }$, a small region gets mapped to a parallelogram.

$\begin{array}{rl}\mathrm{d}S& =\Vert \frac{\partial X}{\partial \varphi }\times \frac{\partial X}{\partial \theta }\Vert \mathrm{d}\varphi \mathrm{d}\theta \\ \frac{\partial X}{\partial \varphi }& =⟨\cos \varphi \cos \theta ,\cos \varphi \sin \theta ,-\sin \varphi ⟩\\ \frac{\partial X}{\partial \theta }& =⟨-\sin \varphi \sin \theta ,\sin \varphi \cos \theta ,0⟩\end{array}$

Now we compute their cross product

$\sin \varphi \left|\begin{array}{ccc}\hat{ı}& \hat{ȷ}& \hat{k}\\ \cos \varphi \cos \theta & \cos \varphi \sin \theta & -\sin \varphi \\ -\sin \theta & \cos \theta & 0\end{array}\right|=\sin \varphi ⟨\sin \varphi \cos \theta ,\sin \varphi \sin \theta ,\cos \varphi ⟩$

Take the length of the resulting vector

$|\sin \varphi |\sqrt{{\sin }^2\varphi {\cos }^2\theta +{\sin }^2\varphi {\sin }^2\theta +{\cos }^2\varphi }=\sin \varphi$

since $0\le \varphi \le \frac{\pi }{2}$. Therefore, $\mathrm{d}S=\sin \varphi \mathrm{d}\varphi \mathrm{d}\theta$.

Evaluate Double Integral

$\begin{array}{rl}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\underset{0}{\overset{\frac{\pi }{2}}{\int }}z(\varphi ,\theta )\sin \theta \mathrm{d}\varphi \mathrm{d}\theta & =\underset{0}{\overset{\frac{\pi }{2}}{\int }}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \varphi \sin \theta \mathrm{d}\varphi \mathrm{d}\theta \\ & =\dots =\pi \end{array}$

Multiply this value by the mass of the surface, $\frac{1}{2\pi }$, so the final center of mass is $(0,0,\frac{1}{2})$

Example: Over Paraboloid

Set up the integral of ${\mathrm{e}}^{\pi x}$ over the region $z=x^2+y^2$, where $0\le z\le 4$.

Parameterize the Region

$\begin{array}{rl}x& =u\\ y& =v\\ z& =u^2+v^2\end{array}$

Determine Area Element

$\begin{array}{rlr}\mathrm{d}S& =\Vert X_u\times X_v\Vert \mathrm{d}u\mathrm{d}v{X}_u& =⟨1,0,2u⟩\\ X_v& =⟨1,0,2v⟩\\ X_u\times X_v& =⟨-2u,-2v,1⟩\\ \Vert X_u\times X_v\Vert & =\sqrt{4u^2+4v^2+1}\end{array}$

Set Up Integral

$\underset{R}{\iint }{\mathrm{e}}^{\pi u}\sqrt{4u^2+4v^2+1}\mathrm{d}u\mathrm{d}v$

Round 2: Polar Coordinates

$\begin{array}{rl}x& =r\cos \theta \\ y& =r\sin \theta \\ z& =r^2\\ X_r& =⟨\cos \theta ,\sin \theta ,2r⟩\\ X_{\theta }& =⟨-r\sin \theta ,r\cos \theta ,0⟩\\ X_r\times X_{\theta }& =⟨-2r^2\cos \theta ,-2r^2\sin \theta ,r⟩\\ \Vert X_r\times X_{\theta }\Vert & =r\sqrt{4r^2+1}\end{array}$

Therefore, the integral is $\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{2}{\int }}{\mathrm{e}}^{\pi r\cos \theta }r\sqrt{4r^2+1}\mathrm{d}r\mathrm{d}\theta$

For Functions of X and Y

given a function $f(x,y,z)$ over a surface $g(x,y)$:

$\iint f(x,y,z)\mathrm{d}S=\underset{A}{\iint }(x,y,g(x,y))\sqrt{1+{g_x}^2+{g_y}^2}\mathrm{d}A$

Example

Given $f(x,y,z)=yz$ and the surface $x+y+z=4$ in the first octant, find $\underset{S}{\iint }f(x,y,z)\mathrm{d}S$

$\underset{S}{\iint }yz\mathrm{d}S=\underset{A}{\iint }y(4-x-y)\sqrt{3}\mathrm{d}A$