Find the region defined by $x$, $y$ and $z$ $$D = \{(x,y,z) \in R^3 \mid 1 \le x^2+y^2+z^2 \le 9, z \le 0, |x| \le y, y \ge 0 \}$$
Find
$$\iiint \sqrt{x^2+y^2+z^2}e^{-(x^2+y^2+z^2)} \,dx\,dy\,dz$$
My problem is defining the region $D$, I should be able to solve the triple integration fine.
 A: I'm assuming you've seen spherical coordinates, in which case it would be
$$
D = \{ (\rho, \phi, \theta) \mid 1 \le \rho \le 3,  \ \pi/2 \le \phi \le \pi,  \ \pi/4 \le \theta \le 3\pi/4\}
$$
Explanation: As you can tell, your region $D$ is a specific part of a sphere. By definition, $\rho^2 = x^2+y^2+z^2$, so $1 \le \rho \le 3$ follows immediately.
Note that $\theta$ only describes the region in the $xy$-plane. So we ignore $z \le 0$ til we get to the $\phi$ variable. The inequalities in the $xy$-plane are $|x| \le y$, which is equivalent to $x \le y$ and $-x \le y$. 
The plot of $x \le y$ is:

and the  plot of $-x \le y$ is:

The plot of $|x| \le y$ is the region that is shaded in both of them:

As you can see, $\theta$ goes from the line $y=x$ to the line $y=-x$, which corresponds to $\pi/4$ to $3\pi/4$.
Finally, $z \le 0$. This is going to be given in terms of $\phi$. Note $0 \le \phi \le \pi$ always. And $\phi$ is the angle from the positive $z$-axis downwards (see http://mathworld.wolfram.com/SphericalCoordinates.html).
So if $z \le 0$, we take $\pi/2 \le \phi \le \pi$.
