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:
$$\int_{-1}^{1}\int _{0}^{\sqrt{1-x^2}}xydydx+\int_{-1}^{1}\int _{-x/3-1/\sqrt{3}}^{\sqrt{x^2-1}}xydydx$$
On
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 $2$.
The first integral (for the semi-circle) is good and should evaluate to $0$ (which you could expect because of the factor $x$ and the symmetry w.r.t. the $y$-axis).
The second integral isn't correct. Note that if you take fixed limits for $x$, the region isn't contained between $x=-1$ and $x=1$, it runs further until $x=2$. I wouldn't recommend this order of integration because then you would have to split the integral in two parts since the limits for $y$ will be different for $-1 \le x \le 1$ (where the upper limit is either $y=0$ or the semi-circle since you could include that part in this integral) and for $1 \le x \le 2$ (where the upper limit would be the hyperbola).
If you take fixed limits for $y$ for the part of the region below the $x$-axis, you let $y$ run from $-\sqrt{3}$ to $0$ and now you don't have to split up for $x$, since $x$ runs from the line to the hyperbola for all $y$ in this interval. Can you set up the integral and take it from there?