Calculate $\iint_{D_r} (Q_x (x,y) - P_y(x,y)) dx dy$

Question

Consider $g:\mathbb{R} \rightarrow \mathbb{R}$ twice derivable and the vector field $F(x,y) = (P(x,y),Q(x,y))$, with $P(x,y) = x\,g(x,y)$, $Q(x,y) = y\,g(x,y)$. If $D_r$ is a disc with center of origin and radius $r>0$, so $\iint_{D_r} (Q_x (x,y) - P_y(x,y)) dx dy$ is equal to:

(A) $\pi r^2$

(B) $2\pi r$

(C) $r$

(D) $1$

->(E) $0$

I tried to this form:

Answer 1

By Green's theorem $$\eqalign{\int_{D_r}(Q_x-P_y)\>{\rm d}(x,y)&=\int_{\partial D_r}P\>dx+Q\>dy\cr &=\int_{\partial D_r}g(x,y)\bigl(x\>dx+y\>dy)\cr &={1\over2}\int_{\partial D_r}g(x,y)\>d(x^2+y^2)=0\ .\cr}$$ If the last steps look like magic plug in a parametric representation $$\phi\mapsto (r\cos\phi,r\sin\phi)\qquad(0\leq\phi\leq 2\pi)$$ of $\partial D_r$. The pullback of $x\,dx+y\,dy$ then evaluates to $0$: $$x\>dx+y\>dy=x(\phi)x'(\phi)+y(\phi)y'(\phi)=r^2\bigl(\cos\phi(-\sin\phi)+\sin\phi\cos\phi\bigr)\equiv0\ .$$

By the way: The field $${\bf F}({\bf z})=\bigl(P({\bf z}),Q({\bf z})\bigr)=g({\bf z})\,{\bf z}$$ is orthogonal to $d{\bf z}$ in all points ${\bf z}\in\partial D_r$.

Answer 2

By the way, \begin{align*} &\iint Q_{x}-P_{y}\\ &=\int Pdx+Qdy\\ &=\int_{0}^{2\pi}(r\cos\theta)g(r\cos\theta,r\sin\theta)(-r\sin\theta)+(r\sin\theta)g(r\cos\theta,r\sin\theta)r\cos\theta d\theta \\ &=0. \end{align*}