find the volume of the solid enclosed by this equation. How to visualize it? How do we find the volume of the solid 
$$ ( \frac{x^2}{a^2} + \frac{y^2}{b^2} +\frac{z^2}{c^2})^2 = \frac{x}{h} $$ ?
I tried to use Wolframalpha to plot it but it doesn't give me the plot. 
 A: Rewrite the equation as
$$\frac{y^2}{b^2}+\frac{z^2}{c^2}=\sqrt{\frac xh}-\frac{x^2}{a^2}.$$
This represents an ellipse in the vertical plane at $x$, with an area proportional to the RHS. The latter expression is positive from $x=0$ until $x^3=a^4/h$.
The volume is an egg of Colombus* with elliptic cross-sections. Its measure will be $\pi bc$ times the integral of the RHS.
$$V=\pi\frac{a^2}{3h}bc.$$
(*With a pretty flat base.)
A: For first, let we set $x = a u,y=b v, z=c w,\delta=\frac{a}{h}$. 
It is sufficient to find the volume of the solid enclosed by:
$$ (u^2+v^2+w^2)^2 = \delta u.\tag{1}$$
If $(1)$ is fullfilled, then $u\geq 0$, and we can compute the volume by slicing the solid according the the value of $u$. The section is a circle having square radius:
$$ R_u^2 = \sqrt{\delta u} - u^2 $$
so, by Cavalieri's principle, the volume is given by:
$$ \int_{0}^{\delta^{1/3}}\pi(\sqrt{\delta u}-u^2)\,du = \frac{\pi \delta}{3}\tag{2} $$
and the volume of the original solid is just:
$$ V = abc\cdot\frac{\pi\delta}{3} = \color{red}{\frac{\pi}{3h}a^2bc}.\tag{3}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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Solid 'enclosed' by
$\ds{\pars{{x^{2} \over a^{2}} + {y^{2} \over b^{2}} + {z^{2} \over c^{2}}}^{2}
     ={x \over h}:\ {\large ?}}$.

\begin{align}&\color{#66f}{\large%
\left.\int_{{\mathbb R}^{3}}\dd x\,\dd y\,\dd z\,
\right\vert_{\pars{%
{x^{2} \over a^{2}} + {y^{2} \over b^{2}} + {z^{2} \over c^{2}}}^{2}\
<\ {x \over h}}}
=\left.\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{0}^{\infty}\dd x\,\dd y\,\dd z\,
\right\vert_{\pars{%
{x^{2} \over a^{2}} + {y^{2} \over b^{2}} + {z^{2} \over c^{2}}}^{2}\
<\ {x \over \verts{h}}}
\\[5mm]&=\left.\verts{abc}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}
r^{2}\sin\pars{\theta}\,\dd r\,\dd\theta\,\dd\phi\,
\right\vert_{r^{4}\ <\ \verts{a \over h}r\sin\pars{\theta}\cos\pars{\phi}}
\\[5mm]&=\left. 2\verts{abc}\int_{0}^{\pi}\int_{0}^{\pi}\int_{0}^{\infty}
r^{2}\sin\pars{\theta}\,\dd r\,\dd\theta\,\dd\phi\,
\right\vert_{r^{3}\ <\ -\verts{a \over h}\sin\pars{\theta}\cos\pars{\phi}}
\\[5mm]&=\left. 2\verts{abc}\int_{0}^{\pi/2}\int_{0}^{\pi}\int_{0}^{\infty}
r^{2}\sin\pars{\theta}\,\dd r\,\dd\theta\,\dd\phi\,
\right\vert_{r^{3}\ <\ \verts{a \over h}\sin\pars{\theta}\sin\pars{\phi}}
\\[5mm]&=2\verts{abc}\int_{0}^{\pi/2}\int_{0}^{\pi}
\int_{0}^{%
\braces{\verts{a}\sin\pars{\theta}\sin\pars{\phi}/\verts{h}}^{\,\,\,\,\,1/3}}
r^{2}\sin\pars{\theta}\,\dd r\,\dd\theta\,\dd\phi
\\[5mm]&=2\verts{abc}\int_{0}^{\pi/2}\int_{0}^{\pi}
{\verts{a}\sin\pars{\theta}\sin\pars{\phi}/\verts{h} \over 3}
\sin\pars{\theta}\,\dd\theta\,\dd\phi
\\[5mm]&={2 \over 3}\,a^{2}\verts{bc \over h}\ \overbrace{%
\bracks{\int_{0}^{\pi}\sin^{2}\pars{\theta}\,\dd\theta}}
^{\ds{=\ \dsc{\pi \over 2}}}\ \overbrace{%
\bracks{\int_{0}^{\pi/2}\sin\pars{\phi}\,\dd\phi}}^{\ds{=\ \dsc{1}}}\ =\
\color{#66f}{\large{1 \over 3}\,\pi a^{2}\verts{bc \over h}}
\end{align}
A: I used this for visualization with WolframAlpha:


Example with $a = b = c = h = 1$. 
