While subject to the force:
$$F(x,y) = y^3i + (x^3 + 3xy^2)j$$ a particle travels counterclockwise once around circle of radius 3. Find work done using Green's theorem.
So $P = y^3$, $Q=x^3+3xy^2$, and $dP/dy =3y^2$ and $dQ/dx = 3x^2 + 3y^2$
So where do I go from here? Why do I have to convert to polar coordinates from here? How do I do this?
By Green's theorem $$\int_{C}\vec F\cdot \,d\vec r=\int\int_{A}\bigg(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\biggr)\,dA$$ where $A$ is the area that the circle of radius $3$ encloses. I.e. $A=\{(x,y)\in \mathbb{R^2}\,:\, x^2+y^2\leq 9\}$. Substituting $\frac{\partial Q}{\partial x},\frac{\partial P}{\partial y}$ the second integrals equals to $$\iint_{\{(x,y):\,x^2+y^2\leq 9\}}3 x^2\,dx\,dy $$
Now the easiest way to solve this is to use polar coordinates. Set $x=r\cos \theta$ and $y=r\sin \theta$. In polar coordinates the integral becomes \begin{align} \int_{0}^{2\pi}\int_{0}^{3}3\cdot r^3\cdot \cos ^2\theta\,dr\,d\theta&=\frac{81}{4}\cdot \int_{0}^{2\pi}\cos^2\theta\,d\theta\\ &=\frac{81}{4}\int_{0}^{2\pi}\frac{\cos 2\theta+1}{2}\,d\theta\\ &=\frac{81\pi}{4}+\int_{0}^{2\pi}\frac{\cos 2\theta}{2}\,d\theta\\ &=\frac{81\pi}{4}+\frac{\sin 2\theta}{4}\biggl|_{0}^{2\pi}\\ &=\frac{81\pi}{4} \end{align} where in the second equality we used the inequality $\cos 2\theta = 2\cos^2\theta-1$.