I'm trying to find $I=\int_E x^2 y$ where $E=\{(x,y)\in\Bbb R^2:x^2+y^2\le1, y\le x, x\ge0\}$.
According to my book, $I=\displaystyle \int_0^1\left(\int_{-\sqrt{1-x^2}}^x x^2 y dy\right)dx=\frac1{30}$ but I'm not getting why $x$ is the right upper limit for all abscissae. I mean, I've drawn $E$, and I thought the region delimited by $x=\frac1{\sqrt{2}}$, $y=0$ and $y=\sqrt{1-x^2}$ needed, indeed, $\sqrt{1-x^2}$ as upper limit. Where do I go wrong? In any case, shouldn't $I$ be (also) equal to $\displaystyle \int_0^1\left(\int_{-\sqrt{1-x^2}}^0 x^2ydy\right)dx+\int_0^{\frac1{\sqrt2}}\left(\int_{0}^x x^2ydy\right)dx+\int_{\frac1{\sqrt2}}^1\left(\int_{0}^{\sqrt{1-x^2}} x^2ydy\right)dx $
?
The book answer gives you this region
However, what you want is this
It's easier to do this in polar coordinates, since you have $0 \le r \le 1$ and $-\frac{\pi}{2} \le \theta \le \frac{\pi}{4}$. Therefore
$$ \int_E x^2 y\ dA = \int_{-\pi/2}^{\pi/4} \int_0^1 r^3\cos^2 \theta \sin \theta \ r \ dr \ d\theta $$