Suppose $E$ is the sphere $x^2 + y^2 + z^2 = 1$ whose density at each point is proportional to the distance from the origin. Find an expression for the mass of $E$ as a Triple Integral and explain why it's difficult to compute

I believe it is difficult to compute because the region is a sphere and not a box but I'm not exactly sure how to write the triple integral

  • 2
    $\begingroup$ $x^2+y^2=1$ is a circle (or a cylindrical surface in 3D), not a sphere! maybe you say $x^2+y^2+z^2\le 1$. In this case the problem is not so difficult and, thanks to the spherical symmetry, it is not really ''multivariable'' $\endgroup$ – Emilio Novati Dec 3 '18 at 20:43
  • $\begingroup$ Use spherical coordinates. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 3 '18 at 20:44
  • $\begingroup$ is E supposed to be the sphere $x^2+y^2 + z^2 = 1$ or is it the cylinder $x^2 + y^2 = 1$ in which case what is the range of $z$? $\endgroup$ – Doug M Dec 3 '18 at 20:44
  • $\begingroup$ My bad its supposed to be x^2+y^2+z^2=1 $\endgroup$ – Emily Dec 3 '18 at 20:46
  • $\begingroup$ Yes, that's exactly what I meant $\endgroup$ – Emily Dec 3 '18 at 20:49

For an arbitrary density $\rho$, the mass is expressible as a triple integral in spherical polar coordinates, viz. $\int_0^{2\pi}d\phi\int_0^\pi d\theta\sin\theta\int_0^1 \rho(r,\,\theta,\,\phi)r^2 dr$. If $\rho$ only depends on $r$ we can first integrate out the angles, giving $4\pi\int_0^1\rho(r)r^2 dr$. The choice $\rho=kr$ from your question gives $4\pi k\int_0^1 r^3 dr=\pi k$.

  • $\begingroup$ Sorry, I'm still confused here. Is this after converting to spherical coordinates. How would you express the triple integral of mass before converting to spherical coordinates? $\endgroup$ – Emily Dec 3 '18 at 20:59
  • $\begingroup$ @Emily What do you mean by "before"? No one coordinate system is "before", making others "after"; they're all equally valid, but for specific problems one may be more useful than another. $\endgroup$ – J.G. Dec 3 '18 at 21:05
  • $\begingroup$ I guess I meant to say is the integral you wrote in spherical coordinates? The question I have first wants me to express the integral without using spherical coordinates and explain why it's hard to compute $\endgroup$ – Emily Dec 3 '18 at 21:12
  • $\begingroup$ Well, if you express it without SCs, what do you express it with? Do they mean Cartesian ones, viz. $\int_{-1}^1 dx\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dy\int_{\-sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}} dz \rho$? I think you can see why that's hard: the limits of one variable depend on the value of another. $\endgroup$ – J.G. Dec 3 '18 at 21:14


If $\delta=kr$ is the density at distance $r$ from the center, than the mass of a spherical shell from $r$ and $r+dr$ is $m=\delta 4 \pi r^2 dr$.

Can you do from this?


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