Green's Theorem: Solving A Complex Integral

[manenbu]

Homework Statement

Solve:
$$\oint x^{99}y^{100}dx + x^{100}y^{99}dy$$

Assuming that it satisfies the conditions for Green's theroem, and:

$$y = \sin{t} + 2, x = \cos{t}, 0 \leq t \leq 2\pi$$

Homework Equations

Green's theorem.

The Attempt at a Solution

$$\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} = 100(xy)^{99}$$
Which means that the integral is 0.
Is this right? It's that simple, or am I missing something here?

 

[Dick]

Did you check that the contour was closed before you applied Greens theorem?

 

[manenbu]

Sorry, it should be $2\pi$. Yes, it is closed.

 

[arildno]

Rewrite the line integral's integrand as:
$$\sin^{99}t\cos^{99}(t)(\cos^{2}t-\sin^{2}t)=\frac{1}{2^{99}}\sin^{99}(2t)\cos(2t)$$

Observe that Green's theorem can be verified in this particular case.

 

[manenbu]

I'm sorry, but I didn't understand how did you make it look like that?
y = sint + 2, how did you get rid of the "+2"?

About Green's theorem - that was how I did it the first time. Differentiating each of the terms and observing that they are equal, thus I get $/iint_{D}0 dA$.
Someone else also pointed it out to me, that since this is a conservative field (with the potential being sin(xy)/100), a line integral of a conservative field in a closed circuit equals 0. Which is the same I get using Green's theorem.

I would still like your explanation as for how you got the integrand in this form.

 

[arildno]

Oops, I forgot about that!
Nonsense on my part..

The integrand will be slightly more complicated, I'll post a proper line integral solution later on