Double integral over odd region

Question

Let $(x,y)$ be Cartesian coordinates, let $\mathbf{i}$ and $\mathbf{j}$ be unit vectors in the $x$- and $y$-directions, let $(r,\theta)$ be polar coordinates, and let $\mathbf{e}_{\theta} = \left(-\sin\theta\right)\mathbf{i}+\left(\cos\theta\right)\mathbf{j}$. Let $R$ be the region of points $(x,y)$ satisfying \begin{equation*}\left\lvert x\right\rvert^{\frac{1}{2}} + \left\lvert y\right\rvert^{\frac{1}{2}} \leq 1\end{equation*} I need to find the integral \begin{equation*}\iint\limits_{R}x\left(x^2+y^2\right)^{\frac{1}{2}}\mathbf{e}_{\theta} \mathop{}\!\mathrm{d}A\end{equation*} I tried using polar coordinates as suggested by the problem statement, but (for the $\mathbf{j}$-component) I end up with the integral \begin{equation*}\int_{0}^{\frac{\pi}{2}}\frac{\cos^{2}\theta}{\left(\sqrt{\cos\theta}+\sqrt{\sin\theta}\right)^8}\end{equation*} which seems intractable. Any assistance would be much appreciated.

Answer 1

I understand that ${\bf e}_\theta$ is meant to be a function of $x$ and $y$, namely $${\bf e}_\theta(x,y)=\left(-{y\over r},{x\over r}\right)\ .$$ We therefore have to calculate $$\int_R x r\>\left(-{y\over r},{x\over r}\right)\>dA=\int_R(-xy, x^2)\>dA\ .$$ The first component of the result is $=0$, by symmetry. For the second component we restrict to the part $R'$ of $R$ lying in the first quadrant. We then have $$\eqalign{\int_R x^2\>dA&=4\int_{R'}x^2 dA=4\int_0^1 \int_0^{(1-\sqrt{x})^2}x^2\>dy\>dx\cr &=4\int_0^1x^2(1-2\sqrt{x}+x)\>dx\cr &={1\over21}\ .\cr}$$