Green's Theorem Integration Question

[sunnyskies]

Here's the question:

Evaluate
$\int_{C} e^x cos y dx - e^xsinydy$
where $A = (ln 2, 0)$to $D =(0,1)$ and then from $D$ to $B = (-ln2, 0)$. Hint: Apply Green's theorem to the integral around the closed curve $ADBA$.

So using Green's Theorem, I got that the integral is equal to

$\int_{C}\frac{\partial}{\partial x}(-e^xsiny) - \frac{\partial}{\partial x}(e^xcosy)dxdy = 0$.

But surely the answer can't be 0? What am I doing wrong?

 

[HallsofIvy]

I assume that by "where A= (ln(2), 0) to D= (0, 1) and then from D to B= (-ln(2), 0)" you mean that C is the line from A to D, followed by the line from D to B.

Yes, by Green's theorem, the integral of that function, around the closed path, "A to D to B to A" is 0. But the problem only asks you to integrate for A to D to B, NOT back to A.

The point of the hint is that, since the integral around the closed path is 0, the integral "from A to D to B" must be the negative of the integral from B to A and so equal to the integral from A to B. You will still need to find the path integral from A= (ln 2, 0) to B= (-ln 2, 0) directly. Since y= 0 on that path, it should be relatively easy.