What is the volume bound by $x^2+y^2+z^4=1$? I saw this question on an old exam for my calculus course, and I can't get anything (Mathematica, Matlab, etc.) to plot me a graph of the function. I'm not quite sure how to setup the triple integral to calculate the volume of this solid.

Technique due to Dirichlet, the volume in the first octant ($x,y,z \geq 0$) is
$$ \frac{\Gamma(3/2) \Gamma(3/2) \Gamma(5/4)}{\Gamma(1 + (1/2) + (1/2) + (1/4))}   $$
so the whole thing is $8$ times that. 
Meanwhile, $$ \Gamma(3/2) = (1/2) \sqrt \pi  $$ and
$$ \Gamma(9/4) = \frac{5}{4} \Gamma(5/4),   $$ reciprocal is $4/5,$ so I get one-eighth of the volume as
$$  \frac{ \pi}{5} $$ and all as
$$  \frac{8 \pi}{5} $$
Turns out the volume inside
$$ x^2 + y^2 + |z|^n = 1  $$ is
$$ \frac{2 \pi n}{n+1}  $$
Should be no trouble integrating using polar coordinates, but have not tried yet... Yep, easy in polar.
