Compute the volume of the region $x^4 + y^2 + z^6 = 1.$ 
Compute the volume of the region $x^4 + y^2 + z^6 = 1.$

I can let $f(y) = 1-y^2 = x^4 + z^6.$ Then $z= \sqrt[6]{f(y)-x^4}.$ I suppose I can use symmetry to calculate the volume of one portion of the integral, but I'm not sure how to come up with the integral.

Technique due to Dirichlet. In the positive octant, volume is
$$ \frac{\Gamma \left( 1 + \frac{1}{2} \right)\Gamma \left( 1 + \frac{1}{4} \right)\Gamma \left( 1 + \frac{1}{6} \right)}{\Gamma \left(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{6} \right)}  $$
and the full volume is $8$ times that. If you can read the scans below, a crucial step is that
$$ \frac{1}{p} \; \Gamma \left( \frac{1}{p}   \right)  =  \Gamma \left( 1+ \frac{1}{p}   \right)    $$
For your problem, the volume in the positive octant is about $0.770182765,$ so the volume of the whole solid is about
$$ 6.1614621 $$
From gp-pari:
?  gamma( 3/2)*  gamma( 5/4) *  gamma( 7/6)  / gamma(23/12)
%6 = 0.7701827651926944255287920651
? 
? 8 *  gamma( 3/2)*  gamma( 5/4) *  gamma( 7/6)  / gamma(23/12)
%7 = 6.161462121541555404230336521
? 



