1. Use Green's Theorem to compute  where C is the rectangle having vertices (0,0), (1,0), (1,5), and (0,5).

The proper conditions apply, since the outside of the rectangle is a simple closed curve (we assume it's oriented the right way) and the partials are all fine.

So by Green's (fun.) Theorem the integral is equal to .
 
 

2. Compute the curl of the vector field F(x,y,z) = x²i - e^xyzj + cos y k.

We set up the usual determinant  and work it out to get (-sin y + xy e^xyz)i + (0 - 0)j + (-yz e^xyz - 0)k. If you want to simplify this to (-sin y + xy e^xyz)i - yz e^xyz k you can, but obviously it's not much of an improvement.
 
 

3. Compute the divergence of the vector field F(x,y,z) = x²i - e^xyzj + cos y k.

We do the respective partials and add them up to get 2x - xz e^xyz + 0.
 
 

1. Compute  where C is the boundary of the region in the first quadrant between a circle of radius 1 and a circle of radius 2.

Even though the problem doesn't require it, it's much easier to compute this by Green's. It's also much easier to compute the integral that results if you switch it to polar. It all looks something like . This integral is strightforward and comes out to 15  / 8.
 
 

2. Compute the curl of the vector field F(x,y,z) = .

It comes out to 0 (the vector 0, not the number 0). All the coefficients derivatives involved are 0, because the partial derivatives involved cancel each other.  For instance, the coefficient of i includes -3xy/(x²+y²+z²)^(5/2) minus itself, producing 0.  The j and k coefficients work out similarly.

This is actually a good thing -- if this sort of function is supposed to correspond to stuff coming out of the sun, we'd rather it come in straight lines without any "swirliness".
 
 

3. Compute the divergence of the vector field F(x,y,z) = .

[Note: Vector fields of this sort may be used to model photon flow from a star or neutrino flow from a black hole. Wow.]

Remember the quotient rule? Probably the least error-prone way to work this is by thinking of the coefficients as (for instance in the case of i): . Then the derivative is . This simplifies a lot -- first divide top and bottom by (x² + y² + z² )^1/2 and then collect like terms to get . Then it gets cool. That was just the x term. When we add on the partials of the y and z terms (they look a lot like the x term, just with -2y² and -2z² respectively), everything cancels... at least as long as x, y, and z aren't all zero. So the divergence is 0 everywhere except at the origin, where its value is a function of K. As with #2, pretty reasonable considering the context.