Triple integral (determining boundaries): Find the volume of the given solid. Above the paraboloid $z = x^2 + y^2$ and below the half cone $z=\sqrt{x^2+y^2}$.
I have great difficulty in determining the boundaries of the integrals. Could anyone provide me with a strategy to do this in this problem and for complex integrals such as these in general?
The correct answer is:
$$\displaystyle \int_0^1 \int_0^{\sqrt{1-y^2}} \int_{x^2+y^2}^{\sqrt{x^2+y^2}} xyz\; dz \; dx \; dy$$
After which polar coordinates are used, however this procedure is clear to me.

In cartesians, you are finding the amount of space between the two surfaces, which is expressed as a function of $z$.  So you have to integrate
$$\int_{x^2+y^2}^{\sqrt{x^2+y^2}} dz\:$$
over the region in $(x,y)$ where $\sqrt{x^2+y^2} >x^2+y^2$.  This region happens to be the circle $x^2+y^2$=1.  Therefore, you get to integrate the above integral over the disk $x^2+y^2 \le 1$, and you should get as the volume
$$\int_{-1}^1 dy \: \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} dx \: \int_{x^2+y^2}^{\sqrt{x^2+y^2}} dz\:$$
(I do not know where you got the $x y z$ from, as this is a volume calculation.)  You then use polar coordinates after doing the trivial integral over $z$ (or, equivalently, cylindrical coordinates), with $x = r \cos{\theta}$, $y=r \sin{\theta}$, and $dx\, dy = r \,dr \, d\theta$.  We then get for the volume
$$\int_0^{2 \pi} d\theta \: \int_0^1 dr \: r (r-r^2) = 2 \pi \left ( \frac{1}{3} - \frac{1}{4} \right ) = \frac{\pi}{12}$$
