Parametric Surface and Surface Integral

Integral Formulas Recap §

Mass of a straight wire:

$$\int f(x)dx$$

Mass of a curvy wire:

$$\int f(x,y)dS$$

Mass of planar lamina:

$$\iint f(x,y)dA$$

Surface Integral:

$$\iint f(x,y,z)dS$$

Evaluating a Surface Integral §

Let $S$ be a surface given by $z=g(x,y)$ and let $R$ be it sprojection onto the xy-plane. If $g,g_x,g_y$ are continuous on $R$ and $f$ is continuous on $S$, then the surface integral of $f$ over $S$ is

$${\iint }_{S}f(x,y,z)dS={\iint }_{R}f(x,y,g(x,y))\sqrt{1+[g_x(x,y){]}^2+[g_y(x,y){]}^2}dA$$

Relationship holds true for $y=g(x,z)$ and $x=g(y,z)$, just substitute variables as needed.

Applications §

• mass of shell
• moment of inertia
• center of mass
• pressure and gravitation
• electrical charge
• Gauss’s Law

Examples §

Example 1 §

Evaluate the surface integral ${\iint }_{S}f(x,y,z)dS$ where S is the parabolic cylinder $y=3x^2$, $0\le x\le 4,0\le z\le 4$

Solution §

Surface is in xz-plane.

$$y=g(x,z)=3x^2$$ $$\begin{array}{rl}\iint f(x,y,z)dS& =\iint f(x,y,z)\cdot \sqrt{1+g_x^2+g_z^2}dS\\ & ={\int }_{x=0}^{x=4}{\int }_{z=0}^{z=4}3x\cdot \sqrt{1+(6x{)}^2+0}dzdx\\ & =1539.89\end{array}$$

Example 2 §

Evaluate ${\iint }_S(x-2y+z)dS$ where $S:z=2,x^2+y^2\le 1$

Solution §

$$z=2$$ $$\begin{array}{rl}{\iint }_{S}f(x,y,z)dS& =\iint f(x,y,z)\cdot \sqrt{1+g_x^2+g_y^2}\\ & ={\int }_{x=-1}^{x=-1}{\int }_{y=-\sqrt{1-x^2}}^{y=\sqrt{1-x^2}}(x-2y+2)\cdot \sqrt{1+0+0}dydx\end{array}$$

OR

$$\begin{array}{rl}{\iint }_{S}f(x,y,z)dS& =\iint f(x,y,z)\cdot \sqrt{1+g_x^2+g_y^2}\\ & ={\int }_{\theta =0}^{\theta =2\pi }{\int }_{r=0}^{r=1}(r\cos \theta -2r\sin \theta +2)rdrd\theta \\ & =2\pi \end{array}$$

Example 3 §

Evaluate $\iint f(x,y,z)dS$ where $f(x,y,z)=\frac{xy}{z}$, $S:z=x^2+y^2,4\le x^2+y^2\le 16$ in Quadrant 1

Solution §

$$\begin{array}{rlr}\iint f(x,y,z)\cdot \sqrt{1+g_x^2+g_y^2}dS& ={\int }_0^{\frac{\pi }{2}}{\int }_{r=2}^{r=4}\frac{\cos \theta +\sin \theta }{r}\sqrt{1+4x^2+4y^2}drd\theta & \\ & ={\int }_0^{\frac{\pi }{2}}{\int }_2^4\sin \theta \cos \theta \sqrt{1+4r^2}drd\theta & =18.91\end{array}$$

Parameter Surface Integral §

Parametric Equation §

$$\begin{array}{c}x=x(u,v)\\ y=y(u,v)\\ z=z(u,v)\\ \vec{r}(u,v)=<x(u,v),y(u,v),z(u,v)>\end{array}$$

Surface Integral w/ Parametric Surface §

If S is a surface given as $\vec{r}(u,v)=<x(u,v),y(u,v),z(u,v)$ over region $D$ in the uv plane, then the surface integral of $f(x,y,z)$ over $S$ is

$${\iint }_{S}f(x,y,z)dS={\iint }_{D}f(x(u,v),y(u,v),z(u,v))|{\vec{r}}_u\times {\vec{r}}_v|dA$$

Examples §

Example 4 §

Use parametric surface to evaluate

$${\iint }_{S}f(x,y,z)dS,S:y=3x^2,0\le x\le 4,0\le z\le 4,f(x,y,z)=3x$$

Solution §

$$\begin{array}{rl}x& =u\\ y& =3u^2\\ z& =v\\ \vec{r}(u,v)& =<u,3u^2,v>\end{array}$$ $$\begin{array}{c}0\le u\le 4\\ 0\le v\le 4\end{array}$$ $$\begin{array}{rl}{\int }_{u=0}^{u=4}{\int }_{v=0}^{v=4}3u||<1,6u,0>\times <0,0,1>||dvdu& ={\int }_0^4{\int }_0^{4}3u\sqrt{1+36u^2}dudv\\ & =1539.89\end{array}$$

Example 5 §

Find ${\iint }_S(y=5)dS$ where $\vec{r}(u,v)=<u,v,2v>$, $0\le u\le 1$, $0\le v\le u$

Solution §

$$\begin{array}{rl}{\int }_{u=0}^{u=}{\int }_{v=0}^{v=u}(v+5)\sqrt{5}dvdu& =\frac{8\sqrt{5}}{3}\end{array}$$