# How to find the center of mass in this problem

How can I find the centre of mass of the surface of the sphere $x^2+y^2+z^2=a^2$ that is contained in the cone $z\tan(\gamma)=\sqrt{x^2+y^2}$, $0 \lt \gamma \lt$ $\pi/2$ a constant, where the density is proportional to distance from the z axis.

I know my answer will be of the form $CM=(Xc,Yc,Zc)$

where the components are found using $$Xc=1/M \iiint_{R} K\sqrt{x^2+y^2}x$$ $$Yc=1/M \iiint_{R} K\sqrt{x^2+y^2}y$$ $$Zc=1/M \iiint_{R} K\sqrt{x^2+y^2}z$$

But I am confused on the bounds of integration.

Mainly because of the rather odd seeming cone, and I dont get how to account for the variation of $\gamma$.

Can anyone help?

• The equation $x\tan(\gamma)=\sqrt{x^2+y^2}$ does not define a cone, since there is no $z$ in it. Do you mean $z\tan(\gamma)=\sqrt{x^2+y^2}$? – Rory Daulton Feb 14 '16 at 1:28
• Yes my mistake, I will update. Thanks – Quality Feb 14 '16 at 1:29
• Do you want the mass of the 2-dimensional surface of the sphere contained in the cone or of the 3-dimensional spherical region contained in the cone? – Rory Daulton Feb 14 '16 at 1:32
• I want the centre of the mass of the surface of the sphere contained within the cone. That is all I was given – Quality Feb 14 '16 at 1:34
• The surface of a sphere between two $z$ values is proportional to their difference; that is, the rate $\mathrm dS/\mathrm dz$ of surface per distance along the $z$ axis is constant. Thus, the centre of mass of the surface between your $z$ values $a$ and $a\left(1+\tan^2\gamma\right)^{-1/2}$ is at the midpoint $z=\frac a2\left(1+\left(1+\tan^2\gamma\right)^{-1/2}\right)$. The $x$ and $y$ coordinates of the centre of mass are $0$ by symmetry. (I converted this from an answer to a comment because it seems it doesn't answer your question.) – joriki Feb 14 '16 at 2:19

## 1 Answer

What you want to do when computing such integrals is to describe the solid in a plane. More specifically, in spherical coordinates, you are interested in the projection of the solid in the $(\rho,\phi)$ 2-dimensional space.

In this spirit, you can rewrite your solid as follows $$R=\{(\rho,\theta,\phi)\;|\;(\rho,\phi)\in D, \;0\le \theta \le 2\pi \}$$

The reason why $0\le \theta \le 2\pi$ is that once you have $D$, you have to rotate it $2\pi$ in order to generate the solid.

Then $$D=\{ (\rho,\phi)\;|\;0\le \rho \le a,0\le \phi \le \gamma \}$$

The variations of $\phi$ can be justified as follows. The upper limit of $\phi$ is the cone, so we have to find its spherical equation:

\begin{align} z\tan\gamma=\sqrt{x^2+y^2} \quad &\Rightarrow \quad \rho \cos \phi \tan\gamma= \sqrt{(\rho\sin\phi\cos\theta)^2+(\rho\sin\phi\sin\theta)^2}\\ & \Rightarrow \quad \rho \cos \phi \tan\gamma= \rho\sin\phi \\ & \Rightarrow \quad \tan\gamma=\tan\phi\\ & \Rightarrow \quad \gamma = \phi \end{align}

As for $\rho$, draw $D$ in the $(\rho,\phi)$ 2-dimensional space, draw any spherical radius $\rho$, you will se that when $\rho=0$, you are already in $D$, and when $\rho=a$, you exit $D$.

So, say you want $X_c$. As you proposed: $$X_c=\frac{\iiint_R kx \sqrt{x^2+y^2}\; dV}{\iiint_R k \sqrt{x^2+y^2}\; dV}$$ which yields in spherical coordinates: $$X_c=\frac{\int_0^{2\pi}\int_0^a \int_0^{\gamma}k(\rho \sin\phi\cos\theta) (\rho\sin\phi)\,\rho^2\sin\phi\;d\phi d\rho d\theta}{\int_0^{2\pi}\int_0^a \int_0^{\gamma}k(\rho\sin\phi)\,\rho^2\sin\phi\;d\phi d\rho d\theta}=0$$

• Could you please elaborate and add some more details and il be happy to accept – Quality Feb 14 '16 at 4:38
• I detailed the variations of $\phi$, I suppose its the difficult part. Do you understand those of $\theta$ and $\rho$ ? – Kuifje Feb 14 '16 at 4:53
• i think I somewhat get most of it, but id like to be able to actually see the calculations/result etc – Quality Feb 14 '16 at 21:27
• calculations/result of what exactly? Above, you have details regarding the bounds of integration, as asked in your initial question. If you need more clarifications on a specific point, please let me know, I will be glad to help. – Kuifje Feb 14 '16 at 21:47
• I just dont understand the problem really at all. I dont know what integrals to evaluate – Quality Feb 14 '16 at 21:59