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&#039;t look right but i don&#039;t know where i went wrong. I&#039;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