Use Stokes Theorem to evaluate the integral

[gtfitzpatrick]

Homework Statement

Use Stokes Theorem to evaluate the integral$\oint_{C} F.dr$ where F(x,y,z) = $e^{-x} i + e^x j + e^z k$ and C is the boundary of that part of the plane 2x+y+2z=2 in the first octant

Homework Equations

$\oint_{C} F.dr = \int\int curlF . dS$

The Attempt at a Solution

So first out i calculated the curl and i got $e^x$ K

Also z=1-x-$\frac{1}{2}$y
and$\frac{\partial z}{\partial x} = -1$
and$\frac{\partial z}{\partial y} = -\frac{1}{2}$
and $\sqrt{(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 + 1}$ = $\sqrt{\frac{9}{4}}$ = $\frac{3}{2}$

To get my limits. when Z=0 the image of the plane on the xy plane is a triangle and so my limits will be x=0 to 1 and y=0 to 2-2x

so putting all this together i get

$\int^{1}_{0}\int^{2-2x}_{0} (e^x k). (\frac{2i+j+2k}{3})(\frac{3}{2}) dydx$

$\int^{1}_{0}\int^{2-2x}_{0} (e^x)dydx$
i have worked out these integrals and i get 2($e^1 +2[\itex])<br /> this doesn't look right but i don't know where i went wrong. I've gone over it twice<br /> anyone throw some light on where I am going wrong here?<br /> Thanks for reading!$

 

[who_]

Everything looks right. I don't think you evaluated the integral correctly though. I get 2e - 4.

 

[chiro]

Hey gtfitzpatrick.

I haven't done these kinds of problems in a while, but I'm wondering if you are trying to normalize the curl, do you have to divide by 3/2?

I can see you have normalized the plane with the division by 3 (SQRT(2^2 + 2^2 + 1)) and I see how you derived the limits for the triangle in the first octant, but the only thing I'm wondering about is this 3/2 factor.

 

[who_]

Hey gtfitzpatrick.

I haven't done these kinds of problems in a while, but I'm wondering if you are trying to normalize the curl, do you have to divide by 3/2?

I can see you have normalized the plane with the division by 3 (SQRT(2^2 + 2^2 + 1)) and I see how you derived the limits for the triangle in the first octant, but the only thing I'm wondering about is this 3/2 factor.

The 3/2 factor comes from the dA factor.

 

[chiro]

The 3/2 factor comes from the dA factor.

Thanks for that.

 

[gtfitzpatrick]

Everything looks right. I don't think you evaluated the integral correctly though. I get 2e - 4.

Thanks a million,yes your right i got a sign wrong, it should be 2e-4