Compute volume of the function $(x^{2}+y^{2})^{2}+z^{4}=y$

Attempted solution:

$y$ is the sum of a square and a fourth power, clearly $y$ is positive.

since the object consists of positive y, the object will occupy 2 quadrants in its proj. on xy plane and it will occupy 2 quadrants on zy plane. the object will have azimuth and co-latitude and distance: $0\le\phi\le\pi,\\ 0\le \theta\le\pi\\ \rho\ge0 $

Now to convert directly to spherical coordinates is my problem, as I understand it, it is easier to convert from Cartesian to cylindrical, and then from cylindrical to spherical.

the conversion from Cartesian to cylindrical is as follows: $$x=rcos\theta,\\y=rsin\theta,\\z=z$$ Thus the function $(x^2+y^2)^2+z^4=y$ becomes $$(r^2cos^2\theta+r^2sin^2\theta)^2+z^4=rsin\theta$$ $$(r^2)^2*(1)+z^4=rsin\theta\\r^4+z^4=rsin\theta$$

Now To convert from cylindrical to spherical we may use: $$r=\rho sin\phi,\\\theta=\theta,\\z=\rho cos\phi$$

We get $$\rho^4 sin^4\phi+\rho^4cos^4\phi=\rho sin\theta sin\theta\\\rho^3(sin^4\phi+cos^4\phi)=sin^2\theta\\\rho=\sqrt[3]{\frac{sin^2\theta}{sin^4\phi+cos^4\phi}}$$

the largest value of $sin^nu$ and $cos^nu$ is 1, thus the largest value of $\rho$ is $\sqrt[3]{\frac{1}{2}}$

Thus the limits of integration are
$$0\le \phi \le\pi,\\0\le\theta\le\pi.\\0\le\rho\le\sqrt[3]{\frac{1}{2}}$$

What knowledge I would like mathstack's to share;

Is this an okay method to convert to spherical coordinates? Am I missing an easier way to convert directly from Cartesian to spherical coordinates? How do I set up the integral, since I want to integrate with respect to Rho, Theta and Phi?

please DO NOT solve the triple integral, that would be missing the point.


refer to this plot:

enter image description here

My attempt at integrating:

volume of the blob is:

$$\int\int\int_A\:dV\\=\int_0^{\sqrt[3]{1/2}}\int_0^\pi\int_0^\pi\:\rho^2 \:sin\phi\: d\theta\: d\phi\: d\rho$$

Check back I will attempt to evaluate

$$=\int_0^{\sqrt[3]{1/2}}\int_0^\pi\:\rho^2 \:[-cos\phi]^{\pi}_{0}\: d\phi\: d\rho$$

$$=\int_0^{\sqrt[3]{1/2}}\int_0^\pi\:\rho^2 \:[-cos\pi+cos(0)]\: d\phi\: d\rho$$

$$=\int_0^{\sqrt[3]{1/2}}\int_0^\pi\:2*\rho^2\: d\phi\: d\rho$$

$$=2*\int_0^{\sqrt[3]{1/2}}\int_0^\pi\:\rho^2\: d\phi\: d\rho=2*\int_0^{\sqrt[3]{1/2}}\:[\phi*\rho^2]^\pi_0\: d\rho=2\pi*\int_0^{\sqrt[3]{1/2}}\:\rho^2\:d\rho$$


Thus the volume bound by the surface is $\pi/3$

If anyone would care to check my evaluation that would be greatly appreciated!

  • $\begingroup$ Anyone care to submit some input? $\endgroup$ – helpmeh Nov 25 '16 at 14:56
  • 1
    $\begingroup$ A couple comments/questions: the equation you give is not a "function", since $y$ appears on both sides. Also, what do you mean by "volume of the function"? Do you mean the equation describes a closed surface, and you want the volume of the interior? $\endgroup$ – Nick Dec 1 '16 at 17:19
  • $\begingroup$ Yes! thank you for clearing that up, the equation describes a roughly blob like shape, i guess it really is a implicit function. I'll graph it in maple, and post a screenshot here $\endgroup$ – helpmeh Dec 1 '16 at 17:21
  • $\begingroup$ And we definitely want to find the volume of the interior, would i be aple to just solve the spherical equation for Rho, and triple integrate? $\endgroup$ – helpmeh Dec 1 '16 at 17:30
  • $\begingroup$ I think so, but I do notice a mistake you made in converting. I will post below. $\endgroup$ – Nick Dec 1 '16 at 17:32

I think your method is correct (of converting first to cylindrical, and then to spherical), but you did make one mistake. Here I will convert directly to spherical from Cartesian using the transformation:

$$ \begin{align*} x &= \rho \sin \phi \cos \theta \\ y &= \rho \sin \phi \sin \theta \\ z &= \rho \cos \phi \end {align*} $$

So the equation $y = (x^2+y^2)^2 + z^4$ becomes:

$$ \rho \sin \phi \sin \theta = \rho^4 \left( \sin^4 \phi + \cos^4 \phi \right) $$

Which you can solve for $\rho$:

$$ \rho = \sqrt[3]{\frac{\sin \phi \sin \theta}{\sin^4\phi + \cos^4\phi}} $$

Note it is $\sin\phi\sin\theta$ in the numerator, and not $\sin^2\theta$.

Now to compute the volume, you should be able to integrate the volume differential $(\rho^2 \sin\phi)$ over the region where $\rho$ is between 0 and the expression above (and the appropriate bounds for $\theta$ and $\phi$).

  • $\begingroup$ notice theta only covers 2 quadrants in the expression's projection on x-y plane and z-x plane, thus the region of integration in terms of theta is $0\le\theta\le\pi$ ill attempt to set up the integral, are all cases of spherical systems integrated with volume differential? $\endgroup$ – helpmeh Dec 1 '16 at 17:48
  • $\begingroup$ Right, it doesn't contain the origin, so my $\phi,\theta$ bounds are not quite right. $\endgroup$ – Nick Dec 1 '16 at 17:49
  • 2
    $\begingroup$ In ANY coordinate system, you need the volume differential. In cartesian coordinates, it is just $dx dy dz$, but this changes when you convert. For example, in spherical coordinates, $dx dy dz$ becomes $\rho^2 \sin \phi d \rho d \theta d \phi$. $\endgroup$ – Nick Dec 1 '16 at 17:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.