# Triple Integrals Help

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

• $x^2+y^2=1$ 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'' – Emilio Novati Dec 3 '18 at 20:43
• Use spherical coordinates. – GNUSupporter 8964民主女神 地下教會 Dec 3 '18 at 20:44
• 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 $z$? – Doug M Dec 3 '18 at 20:44
• My bad its supposed to be x^2+y^2+z^2=1 – Emily Dec 3 '18 at 20:46
• Yes, that's exactly what I meant – 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 $$4\pi k\int_0^1 r^3 dr=\pi k$$.
• 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. – 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 $$m=\delta 4 \pi r^2 dr$$.