Applications of Double/Triple Integrals This is the question that I need to solve using mathematica:

The concentration of an air pollutant at a point $(x,y,z)$ is given by: $$p(x,y,z) = x^2y^4z^3 \text{ particles}/m^3$$ We're interested in studying the air quality in a region in 3-space which satisfies $x\ge0$ and is bounded by the $xy$-plane, the surface $z=\sqrt{4-x^2-y^2}$, the plane $y=-2x$, and the $xz$-plane.  Find the total amount of pollutant in this region.

I don't want a full answer to this question, obviously (I need to work it out myself), but I'm just having trouble getting started.  How do I use the bounds here to set up an integral for the problem?  What exactly do the bounds mean/represent in the first place?
 A: It's a bit of a puzzle to work out the region of integration and a nice way to express it using limits of integration.  A priori the fact that a bunch of "bounds" are thrown out by a problem does not tell us whether the region is actually finite (a finite volume in this case), and even if so you might not be able to package up the region with a single set of nested integral signs and their associated limits of integration.
Without a picture to guide us, I'd probably start from the beginning of the list, assuming some good faith on the part of the problem's constructor.  Now $x \ge 0$ is an easy one to express, by itself.  If we put $x$ as the variable for the outer integration, then the two nested integrals can have limits that depend on the $x$ value referenced in the outer integration.
"Bounded by the $xy$-plane" is a little ambiguous.  The $xy$-plane is where $z=0$, so as a boundary this means either restricting to $z \ge 0$ or $z \le 0$, one side or the other.  Let's keep an open mind as we parse the remaining pieces of the boundary puzzle.
Aha! The surface $z = \sqrt{4 - x^2 - y^2}$ clarifies things.  For one, the square root sign here explicitly means nonnegative values of $z$ on the boundary, so evidently it's the upper half space $z \ge 0$ we need.  Also this is a hemisphere, half the surface of a ball, so it does restrict integration to a finite volume.
I'll let you pursue the remaining pieces of the puzzle, but what we ultimately hope to do is express the region using nested limits of integration something like this:
$$ \int_0^c \int_{f(x)}^{g(x)} \int_{u(x,y)}^{v(x,y)} p(x,y,z) dz dy dx $$
where the functional notation I've used should be replaced by some expressions, and I've assumed $x$ for the outer integration, $z$ for the innermost integration.
I've left you a particularly tricky point to work out, perhaps with the aid of drawing a picture.  The problem says $y = -2x$ is part of the boundary, but which side of this plane is the region on?  Is $y$ supposed to be above or below $-2x$?  The final boundary piece is described as the $xz$-plane, which is where $y = 0$.  How does that fit with the other pieces?
If the order of integration were as I've suggested above, we'd have to find a constant $c$ that is the maximum of the range for $x$, and I've filled in $0$ as the minimum of that range.  The limits of integration on $y$ can then depend on the particular value of $x$, and further the limits of $z$ on both $x$ and $y$.
Evaluating the integral is then a matter of working from the innermost one out, doing the integration with respect to $x$ (or so I've assumed) last.
A: This problem is conducive to using spherical coordinates.  If you draw a picture, you will see that the effect of the plane $y=-2 x$ is to extend the range of the azimuthal angle to $[0,\pi/2+\arctan{2}]$.  (The azimuthal angle is measured with respect to the positive $x$ axis.)  Therefore, the integral you seek is
$$\int_0^2 dr \, r^2 \int_0^{\pi/2} d\theta \, \sin{\theta} \, \int_0^{\pi/2 + \arctan{2}} d\phi \, p(r \sin{\theta} \cos{\phi}, r \sin{\theta} \sin{\phi},r \cos{\theta}) \\ = \int_0^2 dr \, r^{11} \int_0^{\pi/2} d\theta \, \sin^7{\theta} \cos^3{\theta} \, \int_0^{\pi/2 + \arctan{2}} d\phi \, \cos^2{\phi} \sin^4{\phi}$$
Note that these integrals may be evaluated separately and then multiplied together for the final result.
A: $\int_{y=-4/\sqrt 5}^{y=0} dy  \int_{x=-y/2}^{x=\sqrt{4-y^2}} dx \int_{z=0}^{z=\sqrt{4-x^2-y^2}}x^2 y^4 z^3 dz$
