Application Green's theorem

Question

Let C the circle $x^{2}+(y-1)^{2}=1$. Calculate the integral $$\int_{C}{xy^{2}dy-yx^{2}dx}$$

I take $x=\cos(t)$ and $y=\sin(t)+1$, with $t\in[0,2\pi]$, then $$\int_{C}{xy^{2}dy-yx^{2}dx}=\int_{0}^{2\pi}{[\cos^{2}(t)(\sin(t)+1)^{2}+(\sin(t)+1)\cos^{2}(t)\sin(t)]dt}$$ $$=\int_{0}^{2\pi}{[cos^{2}(t)+2\sin^{2}(t)\cos^{2}(t)+3\sin(t)cos^{2}(t)]dt}=\dfrac{3\pi}{2}$$

But now, I wanna calculate the integral with Green's theorem. Let $D$, the interior of the circle $x^{2}+(y-1)^{2}=1$, then $$\int_{C}{Pdx+Qdy}=\int_{C}{(-yx^{2})dx+(xy^{2})dy}$$ $$\iint_{D}{x^{2}+y^{2}}dxdy$$

But now How calculate this, I referer to about limit of integral, (take x=cos(t) and y=sin(t)). Regards

Answer 1

Use the Green's theorem: $$ \oint_{C}\Big(Mdy-Ndx\Big)=\int\int_{x^{2}+(y-1)^{2}\leq1}\Big(\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}\Big)dxdy $$ with $M=xy^{2}$ and $N=yx^{2}$. To compute this integral we can use $x=r\cos(\theta)$ and $y=r\sin(\theta)$ (polar coordinates). Hence you circle converts to $r=2\sin(\theta)$. Therefore, $$ \int\int_{x^{2}+(y-1)^{2}\leq1}(x^{2}+y^{2})dxdy=\int_{0}^{\pi}\int_{0}^{2\sin(\theta)}(r^{2})rdrd\theta. $$

Answer 2

Hint: change the variables in the double integral as \begin{align} x=&r\cos t,\\ y-1=&r\sin t, \end{align} where $t\in[0,2\pi]$. Then you get a simple trigonometric integrand and easy limits in the iterated integral (in $r$ and $t$) as well.

P.S. Do not forget the Jacobian determinant $r$ when changing the variables.