Dented Sphere

The centre of mass of this dented sphere can be computed by computing the COM of the two pieces that make up this sphere. One piece is the part of the sphere below the plane $z = \frac{1}{2}$ . Let the mass per unit area of the sphere be unity. To compute the Z coordinate of the COM of this part:

$z_{c_1} = \frac{\int_{S} z \ dm}{\int_{S} dm}$

$\int_{S} z \ dm = \int_{a}^{\pi}\int_{0}^{2\pi} \cos{\theta} \sin{\theta} \ d\theta \ d\phi = -2\pi\left(\frac{3}{8}\right)$ $\int_{S}dm = \int_{a}^{\pi}\int_{0}^{2\pi}\sin{\theta} \ d\theta \ d\phi = 2\pi\left(\frac{3}{2}\right)$

Where $a = \arccos{0.5}$ .

$z_{c_1} = - \frac{1}{4}$

The second part of this sphere is that lies above the plane $z = 0.5$ . To compute the Z coordinate of the COM of this part:

$z_{c_2} = \frac{\int_{S} z \ dm}{\int_{S} dm}$

$\int_{S} z \ dm = \int_{0}^{a}\int_{0}^{2\pi} \cos{\theta} \sin{\theta} \ d\theta \ d\phi = 2\pi\left(\frac{3}{8}\right)$ $\int_{S}dm = \int_{0}^{a}\int_{0}^{2\pi}\sin{\theta} \ d\theta \ d\phi = 2\pi\left(\frac{1}{2}\right)$

$z_{c_2} = \frac{3}{4}$

This means that the COM of the segment of the sphere lying above the plane $z = 0.5$ lies at a distance of 0.25 units along the z-axis from that plane. Now if this part is mirrored (to obtain the death star) as the problem describes, then the COM of the smaller part is then 0.25 units below the plane $z = 0.5$ . So finally, this dented sphere has two parts with masses $3/2$ and $1/2$ respectively, and the Z - COM coordinate of each part relative to the origin is $-1/4$ and $1/4$ respectively. The factor $2\pi$ is ignored in all numbers specified in this paragraph as it would eventually cancel out.

From here, the Z-coordinate of COM of the overall dented sphere is:

$\boxed{z_{com} = \frac{\left(\frac{3}{2}\times\frac{-1}{4}\right) + \left(\frac{1}{2}\times\frac{1}{4}\right)}{ \frac{3}{2}+\frac{1}{2}} = \frac{-1}{8}}$