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.
 A: It makes no sense sense to integrate over $x$ between the limits you established for $z$. 
You can let, for an eighth of your region, to have $0\leq z\leq (1-x^4-y^2) ^{1/6}$, $0\leq y\leq(1-x^4) ^{1/2}$, $0\leq x\leq1$. This gives you
$$
8\int_0^1\int_0^{\sqrt{1-x^4}} (1-x^4-y^2) ^{1/6}\,dy\,dx. 
$$
But I don't think you can integrate this easily. 
A: 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
? 



A: Let us first consider the problem of finding the volume $V$ of the portion of the given region in the first octant; then by symmetry, the final result will be $8V$.
Thus, suppose $R$ is this intersection of the region and the first octant.  Then if we make the substitution $x = t^{1/2}$, $y = u$, $z = v^{1/3}$, we get
$$V = \iiint_R dx\,dy\,dz = \frac{1}{6} \iiint_O t^{-1/2} v^{-2/3} dt\,du\,dv,$$
where $O$ is the portion of the unit ball in the first octant.  Now, if we convert this latter integral into spherical coordinates, we get
$$V = \frac{1}{6} \int_0^{\pi/2} \int_0^{\pi/2} \int_0^1 (\rho \sin\phi \cos\theta)^{-1/2} (\rho \cos\phi)^{-2/3} \cdot \rho^2 \sin \phi \,d\rho \,d\phi \,d\theta \\
= \frac{1}{6} \left(\int_0^1 \rho^{5/6} d\rho\right) \left(\int_0^{\pi/2} \sin^{1/2}\phi \cos^{-2/3}\phi \,d\phi\right) \left(\int_0^{\pi/2} \cos^{-1/2}\theta \,d\theta \right).$$
We can now evaluate each integral, in the latter two cases using the identity that if $a, b > -1$, then
$$\int_0^{\pi/2} \sin^a \theta \cos^b \theta \,d\theta = \frac{\Gamma(\frac{a+1}{2}) \Gamma(\frac{b+1}{2})} {2 \Gamma(\frac{a+b+2}{2})}.$$
