finding center of mass of a homogenous hemisphere shell

Question

Exercise:

Find the center of gravity of a hemispherical shell of radius $a$. (Assume the density is $1$, and place it so its base is on the $xy$-plane.

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Attempt:

By symmetry, the center of mass (CoM) is along the $z$-axis.

Obviously, the mass ($M$) is $2\pi a^2$

$\text{CoM}_z = \frac{1}{M}\iint_S{zdS}$

Using polar coordinates: $z = \cos \phi$ and $dS = a^2\sin \phi d\phi d \theta$

So: $\text{CoM}_z = \frac{1}{2\pi a^2}\int^{2\pi}_0\int^{\frac{\pi}{2}}_0{(a \cos \phi)(a^2 \sin \phi d\phi d\theta)} = a\int^{\frac{\pi}{2}}_0{\cos \phi \sin \phi d\phi} = a[\frac{1}{3}\sin^3 \phi]^\frac{\pi}{2}_0 = \frac{1}{3}a $

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Answer:

$$\text{CoM} = (0, 0, \frac{1}{3}a)$$

Eyeballing it, the answer seems reasonable.

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Is my solution correct?

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Note: This is an exercise for self-study. If a mod could add the homework-and-exercise tag, that would be appreciated.

Answer 1

It should be $\frac 12a$ not $\frac 13a$

You can see that $$\int\cos\phi\sin\phi d\phi=\color{red}{\frac 12}\sin^{\color{red}{2}}\phi$$