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I want to calculate the volume of $V$ where V is the region

$$ V = \left\{ (x,y,z) : \left( \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \right)^2 = \frac{ x }{h} \right\} $$

Try:

The approach would be to compute

$$ \iiint_V dV$$

but I am having hard time trying to visualize this region, thus finding the limits of integration would be tough. Maybe we need to change coordinates. what method would it work best here?

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  • The change of coordinates $x'=\frac xa,\ y'=\frac yb,\ z'=\frac zc$ seems rather natural. And perhaps you'll need a second one after. –  Dec 06 '16 at 09:49
  • I suppose you want to calculate the volume enclosed by $V$ as $V$ is a surface... – b00n heT Dec 06 '16 at 09:56

2 Answers2

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An easy way to visualize the object is as follows:

  • Let us first take the square root (safe as the LHS is positive).
  • The equation admits a solution iff $x$ lies in the range $$\frac{x^4}{a^2}\leq\frac{x}{h}\iff \frac{x}{h}(\frac{hx^3}{a^2}-1)\leq 0\iff 0\leq x\leq \sqrt[3]{\frac{a^2}{h}}.$$ For each fixed $x_0$ in that range, let us rewrite the equation as $$\frac{y^2}{b^2}+\frac{z^2}{c^2}=\sqrt{\frac{x_0}{h}}-\frac{x_0^2}{a^2},$$ so you get an ellipse on the $(x_0,y,z$) plane, whose diagonals depend on $x_0$.

Patching all these ellipses together, you obtain an elliptic shaped cylinder of varying width (something like the following)

enter image description here

Consequently, I'd suggest you use some sort of cylindrical elliptic coordinate system, and not spherical coordinates.

Of course, if you first normalize the equation so that you get $$(x^2+y^2+z^2)^2=c\cdot x,$$ then you will get a cone (circles instead of ellipses) enwrapped around the same $x-$-axis, whose boundary is described by some curve, something like the following

%% ParametricPlot3D[{x, (x^2 - Sqrt[x])Cos[y], (x^2 - Sqrt[x]) Sin[y]}, {x, 0, 1}, {y, 0, 2 Pi}] %%

enter image description here

And note that if you do this substitution first, then your shape becomes nothing but a solid of revolution, so that its boundary area and its volume are easy to calculate.

b00n heT
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Take the substitution $$x=a \sin \theta \cos \phi$$ $$y=b \sin \theta \sin \phi$$ $$z=c \cos \theta $$ and this will yield the equation of the space as $$a \sin \theta \cos \phi=h$$

I think that these are ellipsoidal coordinates (although I am not totally sure about this name).

Now calculate $dx$ , $dy$ and $dz$, from the above set of transformation equations.

See if this helps you.

SchrodingersCat
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