Thread: Line integral for a plane (and Stoke's Theorem)

1. Line integral for a plane (and Stoke's Theorem)

Hi, I am trying to answer the following question:

Part a is easy, just bookwork. However, I'm finding verifying the theorem more challenging, particularly the line integral side of the formula.

I think my biggest problem is evaluating the line integral for the slanted planar surface. I can't think of how a parametrization could be used, and as far as other methods go I'm totally in the dark as to how to do this.

I'd really, really appreciate help with this problem.
Jon

2. The contour is the intersection of the plane 3x+ 2y+ z= 6 with the coordinate planes. That is a triangle with sides in the three coordinate planes. When y and z are 0, 3x= 6 so x= 2. (2, 0, 0) is a vertex. When x and z are 0, 2y= 6 so y= 3. (0, 3, 0) is a vertex. When x and y are 0, z= 6. (0, 0, 6) is the final vertex.

The contour, then, is the line from (2, 0, 0) to (0, 3, 0), then the line from (0, 3, 0) to (0, 0, 6), followed by the line from (0, 0, 6) back to (2, 0, 0). Because that is not "smooth" (there are sharp corners at each vertex) you cannot find a single differentiable parameterization for the entire path. The best thing to do is do each straight line separately. Also remember that the can be many different parameterizations for the same line or curve, depending upon how you want to choose the parameter. The simplest thing to do is to take t going from 0 to 1 on each segment.

On the first segment, (2, 0, 0) to (0, 3, 0), while t goes from 0 to 1, x most go from 2 to 0: x= 2- 2t will work. y must go from 0 to 3: y= 3t does that. And, of course, z= 0 at every point.

x= 2- 2t, y= 3t, z= 0 is a good parameterization for the first segement.

Can you find parameterizations for the segment from (0, 3, 0) to (0, 0, 6) and then from (0, 0, 6) to (2, 0, 0)?