stokes theorem - finding the surface

[SigepBrandon]

Problem:

Let S be the part of the plane $\displaystyle z=f(x,y)=4x-8y+5$ above the region $\displaystyle (x-1)^{2}+(y-3)^{2}\leq 9$ oriented with an upward pointing normal. Use Stokes' Theorem to evaluate $\displaystyle {\oint_{\delta s}^{}} (2z\widehat{i}+x\widehat{j}+\widehat{k})\cdot d\overrightarrow{s}.$

Stokes' states $\displaystyle {\oint_{\delta s}^{}} F\cdot d\overrightarrow{s}$

F i'm taking to be $\displaystyle (2z\widehat{i}+x\widehat{j}+\widehat{k})$, but the surface? I know the plane is well.. a plane, and the region is a circle of radius 3 at (1,3) I'm just not sure how to put them together. After that I know I'll need to parametrize and take the derivative of the parametrization to dot F with ds, but if someone could point me in the right direction I'd greatly appreciate it.
-bs

 

[SigepBrandon]

ok. So i took z=4x-8y+5, and using the region => x= r*cos(theta)+1, y=r*sin(theta)+3

substituting those into F eventually gave me-
$\displaystyle <8r*(sin(\theta )+cos(\theta))+4, r*cos(\theta),1>$ for r=[0,3] theta=[0.2pi] dotted with the normal <2j+k> and ending with the double integral...

$\displaystyle \int_{0}^{2\pi}\int_{0}^{3} 2r*cos(\theta)+1 dr d\theta$

does that even make sense? thanks-
bs.

 

[galactus]

The plane and the circle create a slanted ellipse.

If we have a vector field $\displaystyle F=Mi+Nj+Pk$, the curl is given by:

$\displaystyle \text{curl F}=\left(\frac{dP}{dy}-\frac{dN}{dz}\right)i+\left(\frac{dM}{dz}-\frac{dP}{dx}\right)j+\left(\frac{dN}{dx}-\frac{dM}{dy}\right)k$

The curl is then $\displaystyle [0,2,1]$.

The plane $\displaystyle -4x+8y+z=5$ has normal $\displaystyle [-4,8,1]$

The area of the portion of the surface z=f(x,y) projected onto a region R is $\displaystyle \int\int_{R}\sqrt{\left(\frac{{\partial}z}{{\partial}x}\right)^{2}+\left(\frac{{\partial}z}{{\partial}y}\right)^{2}}dA$

$\displaystyle \frac{{\partial}z}{{\partial}x}=-4, \;\ \frac{{\partial}z}{{\partial}y}=8$

In this case, the circle projects an ellipse onto the plane and this works out to be $\displaystyle \sqrt{16+64+1}\cdot \text{circle area}= \;\ 9\cdot 9\pi=81\pi$

But, to use the surface integral, we get $\displaystyle \int_{0}^{2\pi}\int_{0}^{3}9rdrd\theta=81\pi$

So, we can use:

\(\displaystyle \underbrace{\sqrt{(-4)^{2}+8^{2}+1^{2}}\cdot \pi (3)^{2}}}_{\text{area of ellipse}}\cdot \underbrace{[0,2,1]\cdot [-4,8,1]}_{\text{dot product of curl and normal}}=81\pi\cdot 17=1377\pi\)

 

[SigepBrandon]

thanks cody,

I am rushing trying to get all this done and obviously haven't grasped these last couple sections. Thanks again for the guidance.