# Center of mass of a trick sphere-cone intersection

B is the solid region occupying the space situated inside the sphere of radius R centered at the origin and above the cone of equation $z = \sqrt{x^2 + y^2}$. The B density is proportional to the distance to the z = 0 plane.

a) Using spherical coordinates, determine the coordinates of the center of mass of the solid B. Your answer will depend on the parameter R.

I think that I can describe $B$ like that

$B = \{(\rho, \theta, \phi)| 0\le\rho\le R, 0 \le\theta \le2\pi, 0 \le\phi\le \pi/4\}$.

As the $B$ density is proportional to the distance to the $z = 0$ plane, I imagine that I should have $\rho(x, y, z) = z$ hence $z = \rho\cos(\phi)$ I have:

$m = \int\int\int_{E} \rho(x, y, z) dV = \int_{0}^{R}\int_{0}^{2\pi}\int_{0}^{\pi/4} \rho\cos(\phi)\rho^2\sin(\phi)d\rho d\theta d\phi =,,,= \frac{R^{4}\pi}{8}$.

By symmetry:

$M_{yz}=M_{xz}=0$ so $\overline{x}=\overline{y}=0$ and $\overline{z}=\frac{(16 - 4\sqrt{2})R}{15}$.

Therefore the center of mass is $(\overline{x}, \overline{y}, \overline{z}) = (\frac{M_{yz}}{m}. \frac{M_{xz}}{m}, \frac{M_{xy}}{m}) = (0, 0, \frac{(16 - 4\sqrt{2})R}{15})$

b) For what values of R the center of mass is it located above the plane $z = 1$?

For this to happen I think that I need $\overline{z}\ge1$, so $\frac{(16 - 4\sqrt{2})R}{15}\ge1$ and consequently $R \ge 1,45$

If someone could have a look in my work and say if that makes any sense will be very grateful. I'm not so sure of my calculations :/

• Where did you find this problem ? – Tony Piccolo Feb 14 '15 at 16:25
• @TonyPiccolo My multivariable calculus teacher gave me. – user78723 Feb 14 '15 at 16:46

There is no need to convert to angular coordinates, it just makes the problem harder. Everything rotationally symmetric around the $z$ axis, so to find the volume from $z=a$ to $z=b$, just evaluate

$$\text{Volume} = \int_{a}^b \pi r(z)^2 \rho(z) ~{\rm d}z$$

where $\rho(z)$ is the density , given as just $kz$, and $r(z)$ is the radius of the shape at the current $z$ height.

The sphere and the cone intersect at $z = \frac 1{\sqrt{2}} R$. The radius of the cone until that point is $r(z) = z$, so the volume of the lower cone portion of the solid is:

$$\text{Cone} = \int_{0}^{R/\sqrt{2}} \pi z^2 kz ~{\rm d}z = \frac{\pi k R^4}{16}$$

The upper spherical portion has a radius satisfying $r(z)^2 + y^2 = R^2$, so

$$\text{Sphere} = \int_{R/\sqrt{2}}^R \pi \sqrt{R^2 - z^2} kz ~{\rm d}z = \frac{\pi k R^3}{3\cdot 2^{3/2}}$$

Comparing the 2, the bottom is more massive when $3\sqrt{2}R > 8$. When that is the case, the center of mass may be found by splitting the cone into 2 parts, above and below $z=L$:

$$\text{Cone bottom} = \int_{0}^{L} \pi z^2 kz ~{\rm d}z = \frac{\pi k L^4}{4}$$ $$\text{Cone top} = \int_{L}^{R/\sqrt{2}} \pi z^2 kz ~{\rm d}z = \frac{\pi k (R^4/4 - L^4)}{4}$$

And we want to find the $L$ where $\text{Sphere} + \text{Cone Top} = \text{Cone Bottom}$ :

$$\frac{\pi k R^3}{3\cdot 2^{3/2}} + \frac{\pi k (R^4/4 - L^4)}{4} = \frac{\pi k L^4}{4}$$ $$L^4 = \frac{12R^4 + 16\sqrt{2}R^3}{96}$$

To find which $R$ result in $L > 1$, it is not enlightening to do analytically, but numerically it comes out to approximately:

$$R > 1.351940150423244$$

• Hi @DanielV, What have you found as mass and center of mass coordinates? And sorry, but I really need to answer in spherical coordinates. – user78723 Feb 15 '15 at 13:14