MATH 251 Lecture 33

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Confused about Notation?

Surface Integrals

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

$\Omega$ 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 $z\geq 0$. Assume that the mass density (per unit area) is constant $\rho =1$

${\begin{aligned}{\bar {x}}&=0\\{\bar {y}}&=0\\{\bar {z}}&={\frac {1}{A}}\iint _{\Omega }z\,\mathrm {d} S={\frac {1}{2\pi }}\iint _{\Omega }z\,\mathrm {d} S\end{aligned}}$

Parameterize the Region

We can use spherical coordinates $(1,\phi ,\theta )$, where $\phi \leq {\tfrac {\pi }{2}}$, so

${\begin{aligned}x(\phi ,\theta )&=\sin {\phi }\cos {\theta }\\y(\phi ,\theta )&=\sin {\phi }\sin {\theta }\\z(\phi ,\theta )&=\cos {\phi }\end{aligned}}$

Determine Area Element

For a rectangular region $\Omega '$ that maps to the hemisphere $\Omega$, a small region gets mapped to a parallelogram.

${\begin{aligned}\mathrm {d} S&=\left\|{\frac {\partial X}{\partial \phi }}\times {\frac {\partial X}{\partial \theta }}\right\|\,\mathrm {d} \phi \,\mathrm {d} \theta \\{\frac {\partial X}{\partial \phi }}&=\left\langle \cos {\phi }\cos {\theta },\cos {\phi }\sin {\theta },-\sin {\phi }\right\rangle \\{\frac {\partial X}{\partial \theta }}&=\left\langle -\sin {\phi }\sin {\theta },\sin {\phi }\cos {\theta },0\right\rangle \end{aligned}}$

Now we compute their cross product

$\sin {\phi }{\begin{vmatrix}{\hat {\imath }}&{\hat {\jmath }}&{\hat {k}}\\\cos {\phi }\cos {\theta }&\cos {\phi }\sin {\theta }&-\sin {\phi }\\-\sin {\theta }&\cos {\theta }&0\end{vmatrix}}=\sin {\phi }\left\langle \sin {\phi }\cos {\theta },\sin {\phi }\sin {\theta },\cos {\phi }\right\rangle$

Take the length of the resulting vector

$\left|\sin {\phi }\right|{\sqrt {\sin ^{2}{\phi }\cos ^{2}{\theta }+\sin ^{2}{\phi }\sin ^{2}{\theta }+\cos ^{2}{\phi }}}=\sin {\phi }$

since $0\leq \phi \leq {\tfrac {\pi }{2}}$. Therefore, $\mathrm {d} S=\sin {\phi }\,\mathrm {d} \phi \,\mathrm {d} \theta$.

Evaluate Double Integral

${\begin{aligned}\int _{0}^{\frac {\pi }{2}}\int _{0}^{\frac {\pi }{2}}z(\phi ,\theta )\sin {\theta }\,\mathrm {d} \phi \,\mathrm {d} \theta &=\int _{0}^{\frac {\pi }{2}}\int _{0}^{\frac {\pi }{2}}\cos {\phi }\sin {\theta }\,\mathrm {d} \phi \,\mathrm {d} \theta \\&=\ldots =\pi \end{aligned}}$

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

Example: Over Paraboloid

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

Parameterize the Region

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

Determine Area Element

${\begin{aligned}\mathrm {d} S&=\left\|X_{u}\times X_{v}\right\|\mathrm {d} u\,\mathrm {d} vX_{u}&=\left\langle 1,0,2u\right\rangle \\X_{v}&=\left\langle 1,0,2v\right\rangle \\X_{u}\times X_{v}&=\left\langle -2u,-2v,1\right\rangle \\\left\|X_{u}\times X_{v}\right\|&={\sqrt {4u^{2}+4v^{2}+1}}\end{aligned}}$

Set Up Integral

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

Round 2: Polar Coordinates

${\begin{aligned}x&=r\cos {\theta }\\y&=r\sin {\theta }\\z&=r^{2}\\X_{r}&=\left\langle \cos {\theta },\sin {\theta },2r\right\rangle \\X_{\theta }&=\left\langle -r\sin {\theta },r\cos {\theta },0\right\rangle \\X_{r}\times X_{\theta }&=\left\langle -2r^{2}\cos {\theta },-2r^{2}\sin {\theta },r\right\rangle \\\left\|X_{r}\times X_{\theta }\right\|&=r{\sqrt {4r^{2}+1}}\end{aligned}}$

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

For Functions of X and Y

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

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

Example

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

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