The formula is $({x^4 \over{a^4}} + {y^2 \over{b^2}} + {z^2 \over{c^2}})^2 = {x^2 \over{p^2}}$, where $a,b,c,p$ is positive. The left part is pretty similuar to ellipsoid, but there are few differences. I also have tried to use the difference of two squares and brackets expansion, but it didn't help.
And the second function - what is the volume of this body?
Thanks for any help!
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After anisotropic scaling you can consider the reduced equation
$$(x^4+y^2+z^2)=x^2.$$
It is symmetric by revolution and its section by the plane $z=0$, $$(x^4+y^2)^2=x^2,$$
or
$$y=\pm\sqrt{|x|-x^4}.$$
The study is not very difficult.
So the surface is an hourglass stretched along the three axis.
The volume is given by the integral of the area of the cross section by the $yz$ plane, $\pi y^2=|x|-x^4$ over the whole $x$ range; then apply the scaling factors.
As a first step, move the right-hand side to the left side, and use the conjugate rule. You will get $$ \Bigl(\frac{x^4}{a^4}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+\frac{x}{p}\Bigr)\cdot \Bigl(\frac{x^4}{a^4}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-\frac{x}{p}\Bigr)=0 $$ You will in general get two "blobs", symmetric with respect to $x$. In the case $a=b=c=p=1$, it looks like this:
To calculate the volume, we can consider $x\geq 0$ (and multiply the result by $2$). The only parenthesis that can be zero is the second, $$ \Bigl(\frac{x^4}{a^4}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-\frac{x}{p}\Bigr) $$ and if we cut it with planes orthogonal to the $x$-axis, we get ellipses, $$ \frac{y^2}{b^2}+\frac{z^2}{c^2}=\frac{x}{p}-\frac{x^4}{a^4} $$ for $0<x<a^{4/3}/p^{1/3}$ and nothing for other planes. The ellipses have half-axes $$ b\sqrt{x/p-x^4/a^4}\quad \text{and}\quad c\sqrt{x/p-x^4/a^4} $$ so the area $A(x)$ of each such ellips equals $$ A(x)=\pi b\sqrt{x/p-x^4/a^4}c\sqrt{x/p-x^4/a^4}=\pi bc(x/p-x^4/a^4). $$ Thus, the volume $V$ of both blobs equals (I leave it to you to perform the integral) $$ V=2\int_0^{a^{4/3}/p^{1/3}}A(x)\,dx=\cdots=\frac{3\pi a^{8/3}bc}{5p^{5/3}}. $$