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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.

Thanks!

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

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

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

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

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

$$=2\pi*[\phi^3/3]^{\sqrt[3]{1/2}}_0\\=2\pi*\frac{1/2}{3}=\pi/3$$

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

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

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  • $\begingroup$ Anyone care to submit some input? $\endgroup$ – helpmeh Nov 25 '16 at 14:56
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    $\begingroup$ 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
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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 $\phi$).

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  • $\begingroup$ 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
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    $\begingroup$ 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

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