# Problem using Green's Theorum

1. Nov 3, 2009

### Titans86

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

F = [−y^3, x^3], C the circle x^2 + y^2 = 25

Book gives answer as Pi*1875*1/2, I get Pi*1875

3. The attempt at a solution

$$\int\int(3x^2 + 3y^2)dxdy$$

$$\int\int(75(cos^2\vartheta + sin^2\vartheta))rdrd\vartheta$$

$$75\int[1/2 r^2]^{5}_{0}d\vartheta$$

$$\frac{1875}{2}\intd\vartheta$$

$$[\frac{1875}{2}\vartheta]^{2\pi}_{0}$$

$$=1875\pi$$ Where did I go wrong?

2. Nov 3, 2009

### rock.freak667

When you have

$$\int\int(3x^2 + 3y^2)dxdy$$

and then convert to polar coordinates x=rcos(theta) and y=rsin(theta)

r is not 5. r is a variable you are integrating.

3. Nov 3, 2009

### Titans86

Of course!!! Thank you for the quick reply.
Regards,