TITLE: Tricky Double integral over a union of two regions.
QUESTION [5 upvotes]: Problem: Evaluate the integral, $$\int \int_D xydA$$ where $D$ is the union region shown below: 

I have solved the problem and have obtained the answer as $\frac{-1}{9}.$ Since there are no answers at the end of the book, I would like to confirm whether this is the correct answer or not. Thank you. 
The integral that I set up is as follows:
$$\int_{-1}^{1}\int _{0}^{\sqrt{1-x^2}}xydydx+\int_{-1}^{1}\int _{-x/3-1/\sqrt{3}}^{\sqrt{x^2-1}}xydydx$$

REPLY [2 votes]: The two sides of the semicircle cancel each other out, so the integral over that part comes to $0$. Therefore, we only need to worry about the bottom portion. This is given by
$$\int_{-\sqrt3}^0 \int^{\sqrt{1+y^2}}_{-y\sqrt3-1}xy\, dx\,dy = \frac{1}{2}\int_{-\sqrt3}^0 y(1+y^2) - y(1+y\sqrt3)^2 \, dy = -\frac{3}{4}$$
You seem to have three mistakes in your second integral. First, as mentioned, the upper bound for $y$ is only valid when $x\geq 1$. Second, your lower bound for $y$ is incorrect; the equation for that line is $y=-x/\sqrt3-1/\sqrt3$. This, the $x$ bounds should be $-1$ to $2$.