Verify Green's theorem for the functions: $M(x,y)=2x^3+y^3$ and $N(x,y)=3xy^2$, and the region $D$ which is the annulus between the circles of radii $a$ and $b$ ($a<b$), centered at the origin. Be careful with the orientations.
I am aware of Green's theorem but am not sure how to approach and proceed with this problem. I don't know how to set up the integral and I am very confused. Help would be greatly appreciated! Please and thank you!
You need to show that $$ \int_{\partial D} (M\,dx + N\,dy) = \iint_{D} \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)\,dA $$
For the right-hand side: \begin{align*} \frac{\partial N}{\partial x} &=6xy \\ \frac{\partial M}{\partial y} &=3y^2 \\ \end{align*} So \begin{align*} \iint_{D} \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)\,dA &= \iint_D \left(6xy-3y^2\right)\,dA \\ &= \int_0^{2\pi} \int_a^b 6\left((r \cos \theta)(r\sin\theta) - 3(r\sin\theta)^2\right)\,r\,dr\,d\theta \end{align*} For the left-hand side, let $C_a$ and $C_b$ be the circles of radius $a$ and $b$ centered at the origin. Then as oriented curves, $\partial D = C_b - C_a$. We can parametrize $C_a$ by $x=a\cos\theta$, $y=a\sin\theta$, $0 \leq\theta\leq 2\pi$, and similarly for $C_b$. Therefore \begin{align*} \int_{\partial D} \left(M\,dx + N\,dy\right) &= \int_0^{2\pi}\left(\left(2(b\cos\theta)^3 +(b\sin\theta)^3\right)(-b\sin\theta) + \left(3(b\cos\theta)(b\sin\theta)^2\right)(b\cos\theta)\right)\,d\theta \\ &\qquad-\int_0^{2\pi}\left(\left((a\cos\theta)^3 +(a\sin\theta)^3\right)(-a\sin\theta) + \left(3(a\cos\theta)(a\sin\theta)^2\right)(a\cos\theta)\right)\,d\theta \end{align*} Evaluate each and show that they are equal.