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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?

Chill2Macht
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Desh
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  • cylindrical coordinates seem to be a good choice here – tired Dec 14 '16 at 16:10
  • Although this may be difficult to bash out, what if you tried rearranging the equation as $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\sqrt{\frac{z}{k}-\frac{z^4}{c^4}}$, which is just an equation for an ellipse, and integrating over $z$? – Calconym Dec 14 '16 at 16:21
  • @Calconym How to integrate it over $z$? – Desh Dec 14 '16 at 16:40
  • This is an equation for an ellipse in terms of $x,y$, whose size is dependent on $z$. We know the area of an ellipse of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is just $\pi ab$, so we can rearrange the equation as $\frac{x^2}{a^2\sqrt{\frac{z}{k}-\frac{z^4}{c^4}}}+\frac{y^2}{b^2\sqrt{\frac{z}{k}-\frac{z^4}{c^4}}}=1$. Applying the formula, we get an area in terms of $z$ of $\pi ab\sqrt{\frac{z}{k}-\frac{z^4}{c^4}}$. Then applying Cavalieri's principle, we can effectively find the sum of all of these areas wherever $\sqrt{\frac{z}{k}-\frac{z^4}{c^4}}>0$ in terms of constants a, b, k, and c. – Calconym Dec 14 '16 at 16:53
  • If you need any graphical clarification I can post a diagram of the curve. – Calconym Dec 14 '16 at 16:54

1 Answers1

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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.

Michael Seifert
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