I have the following domain: $$D = \{(x,y,z) \mid x^2 + y^2 + z^2 \le 4, x^2 + y^2 \ge 1\}$$ The integral is the following: $$\iiint_D(x^2+y^2)\,dx\,dy\,dz$$ Now, I made a picture and I got some intuition about the possible range of the three coordinates. Here are my thoughts: $$-2\le x \le2$$ $$\sqrt{1-x^2} \le y \le \sqrt{2-x^2}$$ $$-\sqrt{4-x^2-y^2} \le z \le \sqrt{4-x^2-y^2}$$ I know that only the range for $z$ is right. Because for $y$ I got only the positive part between the two circles, and $x$ actually doesn't take all the values between $-2$ and $2$. I should add this restrictions for $x$ $-2\le x \le-1$ and $1\le x \le2$ and for $y$ apart from what has been written above I got $$-\sqrt{2-x^2} \le y \le -\sqrt{1-x^2}$$
But, now, the question is how am I supposed to plug this additional restrictions into the integral? Thanks for you answer
In spherical coordinates $(r,\theta,\phi)$, we have $r\le 2$ and $r\sin(\theta)\ge 1$.
Therefore, the volume integral can be written
$$\int_D (x^2+y^2)\,dV=\int_0^{2\pi}\int_{\pi/6}^{5\pi/6} \int_{\csc(\theta)}^2 r^2\sin^2(\theta)\,r^2\sin(\theta)\,dr\,d\theta\,d\phi$$
Can you finish now?