<s>[INST] Find volume of body bounded by $\left(\frac{x^2}{a^2} + \frac{y^2}{b^2}\right)^2 + \frac{z^4}{c^4} = \frac{z}{k}$ 
Find the volume of body bounded by 
  $$\left(\frac{x^2}{a^2} + \frac{y^2}{b^2}\right)^2 + \frac{z^4}{c^4} = \frac{z}{k}.$$

What substitution is better in this case?
 [/INST] We change coordinates as follows:
$$
\xi = \frac{x}{a} \qquad \eta = \frac{y}{b} \qquad \zeta = \frac{k^{1/3}}{c^{4/3}} z
$$
In terms of these new coordinates, the bounding surface is given by
$$
\left(\xi^2 + \eta^2 \right)^2 + \frac{c^{4/3}}{k^{4/3}} \zeta^4 = \frac{c^{4/3}}{k^{4/3}} \zeta.
$$
Define $\kappa = c^{2/3}/k^{2/3}$, so we then have
$$
\xi^2 + \eta^2 = \kappa \sqrt{\zeta - \zeta^4}.
$$
This is now a solid of rotation in $(\xi, \eta, \zeta)$ space, with a radial coordinate $\rho^2 = \xi^2 + \eta^2$.  The volume of the solid of rotation in these coordinates will be
$$
\tilde{V} = \int \pi \rho(z)^2 \,  d\zeta = \pi \kappa \int_0^1 \sqrt{\zeta - \zeta^4} d\zeta.
$$
This last integral can be expressed in terms of the beta function by substituting $u = \zeta^3$:
$$
\int_0^1 \sqrt{\zeta - \zeta^4} d\zeta= \int_0^1 \sqrt{u^{1/3} - u^{4/3}} \frac{du}{3 u^{2/3}} =  \frac{1}{3} \int_0^1 u^{-1/2}\sqrt{1 - u} \, du = \frac{1}{3} B\left( \frac{1}{2}, \frac{3}{2} \right) = \frac{\pi}{6}.
$$
Thus, the volume of the solid in $(\xi, \eta, \zeta)$-space is
$$
\tilde{V} = \frac{\pi^2}{6} \frac{c^{2/3}}{k^{2/3}}
$$
and since the volume element in $(x,y,z)$ space is related to that in $(\xi, \eta, \zeta)$ space by
$$
dV = dx \, dy \, dz = (a d\xi)(b d\eta) (\frac{c^{4/3}}{k^{1/3}} d\zeta) = \left(\frac{a b c^{4/3}}{k^{1/3}} \right) d\tilde{V},
$$ we conclude that
$$
\boxed{ V = \frac{\pi^2}{6} \frac{a b c^2}{k}.}
$$
I think.  There may be an error or two with my powers in here, but I'm pretty confident in the method.
</s>