# Line integral

1. Sep 22, 2009

### manenbu

1. The problem statement, all variables and given/known data

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

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

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

2. Relevant equations

Green's theorem.

3. 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?

Last edited: Sep 22, 2009
2. Sep 22, 2009

### Dick

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

3. Sep 22, 2009

### manenbu

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

4. Sep 22, 2009

### 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.

5. Sep 22, 2009

### 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.

6. Sep 22, 2009