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For practice, I'm trying to solve problem 5 from this MIT practice set: problem 5.

The problem involves determining the bounds of integration for computing the volume of the region described by this surface:

$$z=(x^2+y^2+z^2)^2$$

The solution naturally converts the problem to spherical coordinates:

$$\int_0^{2\pi} \int_0^{\pi/2} \int_0^{\cos^{1/3} (\phi)} \rho^2 \sin( \phi) d \rho d \phi d\theta $$

I understand most of the solution.

I have a few questions:

1) one part I do not understand is why the bounds of $\phi$ are from $0$ to $\pi/2$. Is there anyone who can help me understand this?

2) My second question is that I want ot know if my solution is correct, even though it is different: My solution to this problem was actually to convert this problem to computing a double integral, u\sing the divergence theorem. By letting $F=(0,0,z)$, so that its divergence = 1, the divergence theorem gives us: $$Volume = = \iiint dV$$ $$ \iiint (div \vec F * dV) = \iiint dV$$, \since $div \vec F = 1$. And, the divergence theorem says: $$\iiint (div \vec F * dV) = \int \int \vec F \cdot d\vec S $$ Converting to spherical coordinates, $d\vec S = \rho ^2 \sin \phi (\cos \theta \sin \phi, \sin \theta \sin \phi, \cos \phi) d \phi d\theta$:
$$\int \int \vec F \cdot d\vec S = \int \int (0,0, \rho \cos \phi) \cdot (\cos \theta \sin \phi, \sin \theta \sin \phi, \cos \phi) \rho ^2 \sin \phi d\theta d\phi $$ and substituting in the equation $\rho ^3 = \cos \phi$, we get the following integral:

$$ \int_{0}^{\pi} \int_{0}^{2 \pi} \cos^3 (\phi) \sin (\phi) d\theta d\phi $$

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    $\begingroup$ \iiint produces $\iiint$ which is correctly spaced, and \sin,\cos,\tan produce the properly upright versions $\sin,\cos,\tan$. $\endgroup$
    – YiFan Tey
    Commented Oct 27, 2019 at 23:01
  • $\begingroup$ Fixed thanks @YiFan. Makes a huge difference. $\endgroup$
    – makansij
    Commented Oct 27, 2019 at 23:05
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    $\begingroup$ This is not a quadratic surface! The equation has degree $4$, so it's a quartic :) $\endgroup$ Commented Oct 27, 2019 at 23:52

1 Answer 1

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For 1), note that $0\le z\le 1$, so we know that $\phi$ must be between $0$ and $\pi/2$. Does it attain both bounds? Well, yes, at the top point of the region (namely, $(0,0,1)$) we have $\phi = 0$. The more interesting thing to notice is that the tangent plane of this surface at the origin is $z=0$ (why?), so $\phi$ ranges up to $\pi/2$. (If the surface were conical, instead, at the origin, we would need to think about the angle of the cone.)

For 2), your solution is not correct. Let me first say that you should call this a surface integral rather than a double integral. But your formula for $d\vec S$ holds precisely when $\rho$ is constant and the surface is spherical. If you actually parametrize the surface using $\phi$ and $\theta$, you will have $$(x,y,z) = \big((\cos\phi)^{1/3}\sin\phi\cos\theta, (\cos\phi)^{1/3}\sin\phi\sin\theta,(\cos\phi)^{1/3}\cos\phi\big),$$ and I leave it to you to derive the correct formula for $d\vec S$. (Yuck.)

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  • $\begingroup$ For 1), I'm still figuring out "(why?)", but otherwise I get it, thanks! For 2), you're saying that this: $d\vec S = \rho ^2 \sin \phi (\cos \theta \sin \phi, \sin \theta \sin \phi, \cos \phi) d \phi d\theta$ only holds when $\rho$ is constant, is that right? So if I wanted to compute $d \vec S$ from scratch, in theory I would need to compute $d \vec S = \frac{\partial \vec r }{\partial \phi} x \frac{\partial \vec r }{\partial \theta} d \theta d\phi$, where $\vec r$ is exactly what you gave above. Right? $\endgroup$
    – makansij
    Commented Oct 28, 2019 at 0:55
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    $\begingroup$ Yup @nundo, precisely. $\endgroup$ Commented Oct 28, 2019 at 1:07
  • $\begingroup$ OKay, thanks, you're really helping me a lot here. One more thing: I derived that formula for $d \vec S$ using the Lamé coefficients that I got from another source. Now I want you to confirm that this is incorrect because these Lamé coefficients for spherical coordinates are only valid when $\rho$ is independent of $\phi$ and $\theta$: $h_{\theta} = \rho sin \phi$ , and $h_{\phi}=r $,and $h_{r}=1$. Right? $\endgroup$
    – makansij
    Commented Oct 28, 2019 at 1:15
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    $\begingroup$ I assume $r=\rho$? Yeah, that's on the surface of a sphere centered at the origin. I don't know what your level of interest in math is, but you might find some of my YouTube lectures helpful or interesting. The link is in my profile. $\endgroup$ Commented Oct 28, 2019 at 1:18
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    $\begingroup$ Well, only if the sphere is centered at the origin is $\rho$ constant. If you don't already know the answer, work out the equation for the sphere of radius $a$ centered at $(0,0,a)$. $\endgroup$ Commented Oct 28, 2019 at 2:54

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