Where did I go wrong with my application of Stoke's Theorem?

[Joosh]

Hello again, everyone. Have a multivariate calculus question this time around. If anyone can point me in the right direction and help me see where WebAssign finds me wrong, it would be greatly appreciated.

1. Homework Statement

Homework Equations

∫∫_ScurlF ⋅ dS = ∫_CF ⋅ dr

The Attempt at a Solution

 

[Charles Link]

Your picture is very difficult to read, but I solved it, and I think I can get you to the correct answer. This triangle has vertices at the 3 points, but what is the length of each side? (Hint: Each side does not have length 3). Also, this triangle lies in a plane. What is the unit vector $\hat{n}$ normal to that plane? Be sure and properly normalize it. Now $\nabla \times F \cdot \hat{n}$ is constant in the plane of the triangle, because the plane of the triangle has a simple equation (which you should be able to write out very quickly) and shows up in this $\nabla \times F \cdot \hat{n}$ expression . (Please write out this equation for the plane of the triangle, so that it can be verified that you got it correct. It helps to see the calculations, and your picture is hard to read.) Finally, what is the height (altitude) of the equilateral triangle?, which you need to compute the area of the triangle, etc. Try again; I think you will be able to correct a mistake or two that you made.

 

[Charles Link]

Alternatively, this problem can be worked by performing the surface integral over the 3 perpendicular faces (in the xy plane, the xz plane, and yz plane) instead of integrating over the single plane of the triangle (containing the 3 points). I computed the calculation this way as well and got the same answer. (For this case, the three surface integrals are essentially equivalent, and it just takes a little effort to set up the limits on a single relatively simple 2-D integral.) For the surface integral side of Stokes theorem, so long as F(x,y,z) is a well-behaved function, I believe the surface used in the surface integral can be any well-defined surface whose perimeter involves the line integral. Perhaps one of the other readers can add to this, but I believe I have this concept correct.

 

[Joosh]

I appreciate the pointers, Charles! However, I eventually figured what I did wrong. I simply made a mistake while plugging in my numbers.