# Green's Theorem

1. Dec 10, 2014

### Amy Marie

1. The problem statement, all variables and given/known data
Use Green's Theorum to evaluate the line integral c (x^2)y dx, where c is the unit circle centered at the origin.

2. Relevant equations

3. The attempt at a solution
Taking the partial derivative with respect to y and subtracting it from zero(I'm taking the dy in the original problem to be zero because there wasn't one), I set up the double integral:

∫y = -1 to y =1 ∫x = -sqrt(1 - (y^2)) to x = sqrt(1 - ( y^2)) -(x^2) dxdy

I'm just kind of confused because there is no dy in the problem, and I'm not sure why the answer was -π/4.
I might not have set it up right because when I put the x bounds into -(x^3)/3, I'm not sure how to then integrate with respect to y.

2. Dec 10, 2014

### Staff: Mentor

Since the path is a circle about the origin have you thought of using polar coordinates?

3. Dec 10, 2014

### Zondrina

You should first apply Green's theorem properly.

$$\oint_C \vec F \cdot d \vec r = \oint_C x^2y \space dx + 0 \space dy = \iint_D Q_x - P_y \space dA$$

Where $\vec F = P \hat i + Q \hat j$ and $d \vec r = \vec r'(t) \space dt = x'(t) \hat i + y'(t) \hat j$.

So what are $Q_x$ and $P_y$?

Now the curve $C$ is the unit circle $x^2 + y^2 = 1$ in a counter clockwise orientation. How do you parametrize a counter-clockwise circle around the origin? This should give you limits for $r$ and $\theta$.

4. Dec 10, 2014

### Amy Marie

Thank you...this helped!

5. Dec 10, 2014

### Amy Marie

Thank you...that worked!