Intutition behind triple integrals between two surfaces I am attempting to learn multivariable calculus on my own. A problem in my book asks me to evaluate
$$
\iiint_\mathcal{W} xz \, \mathrm{d}V
$$
where $\mathcal{W}$ is the domain bounded by the cylinder $\displaystyle \frac{x^2}{4} + \frac{y^2}{9} = 1$ and the sphere $x^2 + y^2 + z^2 = 16$ in the first octant ($x, y, z \geq 0$).
The answer is $\displaystyle \frac{126}{5}$.
However, I cannot obtain this number nor can I understand the logic behind evaluating such an integral. If suppose, I take the $z$ variable, I can see that it can vary from $z = 0$ to $z = \displaystyle \sqrt{16 - x^2 - y^2}$. That reduces the integral to
$$
\frac{1}{2}\int_{lowX}^{highX} \int_{lowY}^{highY} x(16 - x^2 - y^2) \,\mathrm{d}y\,\mathrm{d}x
$$
My first question is what is this quantity, physically?
Next, as I look down from the $z$ axis --which I have eliminated, I see two curves on the $xy$ plane --an ellipse $\displaystyle \frac{x^2}{4} + \frac{y^2}{9} = 1$ and a circle $x^2 + y^2 = 16$. So, if I vary $y$ from $0$ to $4$, $x$ goes from $\displaystyle 2\sqrt{1-y^2/9}$ to $\sqrt{16-y^2}$. 
Hence, my next question is what is so special about $z=0$? Why would I look at the curves on that plane? Why not the plane $z=1$? Is it because this gives me the maximal domain on $x$ and $y$ --a way of saying telling me the extent to which the variables vary? Or am I wrong to reason along these lines?
P.S.:- I better be wrong, because along these lines I do not get $126/5$. I get $3592/81$.
 A: We are given this (a priori infinitely long) vertical cylinder whose base is an elliptical disc $E$ in the $(x,y)$-plane  with semi-axes $2$ and $3$. In addition we are given a sphere of radius $4$. Note that $E$ is completely inside this sphere, but the sphere cuts off the cylinder, making   round top and bottom surfaces of the resulting cylindrical body. Actually we don't want the full cylindrical body, but only the part ${\cal W}$ of it in the first octant. We are then told to compute the integral
$$\int_{\cal W} x z\>{\rm d}V=\int_{E'}\left( \int_0^{\sqrt{16-x^2-y^2}} x z\>dz\right){\rm d}(x,y)\ ,\tag{1}$$
whereby $E'$ denotes the part of $E$ in the first quadrant. On the RHS at each point $(x,y)\in E'$ a vertical stalk has been erected, with lower end at $z=0$ and upper end at $z=\sqrt{16-x^2-y^2}$ on the sphere. The body ${\cal W}$ is the union of these stalks. Now
$$\int_0^{\sqrt{16-x^2-y^2}}  z\>dz={z^2\over2}\biggr|_{z=0}^{z=\sqrt{16-x^2-y^2}}={1\over2}(16-x^2-y^2)\ .\tag{2}$$
At this point the spherical shaped top boundary of ${\cal W}$ is completely taken care of. We now
plug $(2)$  into the RHS of $(1)$, and obtain
$$\eqalign{\int_{\cal W} x z\>{\rm d}V&=\int_{E'} {x\over2} (16-x^2-y^2)\>{\rm d}(x,y)\cr  &=\int_0^3\left(\int_0^{{2\over3}\sqrt{9-y^2}}{x\over2}(16-x^2-y^2)dx\right)dy\ .\cr}$$
Note that the $\int_{E'}$ integral is completely in the $(x,y)$-plane. In order to compute it "by reduction" we have drawn at each level $y\in[0,3]$ a horizontal beam beginning at $x=0$ and ending at $x={2\over3}\sqrt{9-y^2}$ on the right half of the elliptical boundary. The reason for this choice of integration order is that we wanted to make good use of the factor $x$ in the integral. In this way no square roots will appear in the calculation.
Calculate the inner integral (with respect to $x$, while $y$ is held constant). You obtain a polynomial in $y$, which you then have to integrate from $0$ to $3$. The result is indeed ${126\over5}$.
A: Sometimes a figure is worth 1000 words:

The integral is:
$$\int\limits_{x=-2}^2 \int\limits_{y=-3\sqrt{1 - x^2/4}}^{+3\sqrt{1 - x^2/4}} \int\limits_{z = -\sqrt{16 - x^2 - y^2}}^{+\sqrt{16 - x^2 - y^2}} x z\ dx\ dy\ dz$$
Of course the value of this integral is $0$ because for any position with $+x$ there is a corresponding point with $-x$, cancelling each other out.
