What is the volume bound by $x^2+y^2+z^4=1$?

Question

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.

Answer 1

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.

Answer 2

Slice it into pieces of constant $z$. Between $z$ and $z+dz$, there is a circle of radius $\sqrt{1-z^4}$ and area $pi(1-z^4)$. Integrate this from $z=-1$ to $z=1$.

Answer 3

Plotting the surface in Mathematica is straightforward:

ContourPlot3D[x^2 + y^2 + z^4 == 1, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}]