91Ó°ÊÓ

The hemisphere is defined as \(B=\{x \in \mathbb{R}^{3} \mid\|x\|0\}\). We can use spherical coordinates to express the volume integral. The spherical coordinates are given by: \(x = r\sin\theta\cos\phi\) \(y = r\sin\theta\sin\phi\) \(z = r\cos\theta\) The volume element in spherical coordinates is \(dV = r^2\sin\theta dr d\theta d\phi\).

In order to find the mass, we need to integrate the density function over the entire volume of the hemisphere: \(M = \int_V \varrho(x) dV\), with \(0 \le r \le R\), \(0 \le \theta \le \frac{\pi}{2}\), and \(0 \le \phi \le 2\pi\). Substitute the density function \(\varrho(x) = ar\cos\theta\), and the volume element \(dV = r^2\sin\theta dr d\theta d\phi\). \(M = a \int_{0}^{R} \int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} r^3\cos\theta\sin\theta dr d\theta d\phi\) Now we will evaluate each integral separately.

First evaluate the \(\phi\) integral: \(\int_{0}^{2\pi} d\phi = 2\pi\) Next, evaluate the \(\theta\) integral: \(\int_{0}^{\frac{\pi}{2}} \cos\theta\sin\theta d\theta = \frac{1}{2}\) (use substitution u = sin(θ), du = cos(θ)dθ) Finally, evaluate the \(r\) integral: \(\int_{0}^{R} r^3 dr = \frac{1}{4}R^4\) (using the power rule) Put everything together the mass: \(M = a \cdot 2\pi \cdot \frac{1}{2} \cdot \frac{1}{4}R^4 = \frac{1}{4}\pi aR^4\)

The third coordinate of the center of mass is given by: \(z_c = \frac{1}{M} \int\limits_{0}^{R}\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{2\pi} arz \cdot r^2\sin\theta dr d\theta d\phi\) \(z_c = \frac{4}{\pi aR^4} \int\limits_{0}^{R}\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{2\pi} ar^3\cos\theta \cdot r^2\sin\theta dr d\theta d\phi\) Now we will integrate each part separately using the same bounds as before. Evaluate the \(\phi\) integral: \(\int_{0}^{2\pi} d\phi = 2\pi\) Evaluate the \(\theta\) integral: \(\int_{0}^{\frac{\pi}{2}} \cos\theta\sin\theta d\theta = \frac{1}{2}\) Evaluate the \(r\) integral: \(\int_{0}^{R} r^5 dr = \frac{1}{6}R^6\) Putting everything together: \(z_c = \frac{4}{\pi aR^4} \cdot 2\pi \cdot \frac{1}{2} \cdot \frac{1}{6}R^6 = \frac{2}{3}R\) The mass of the hemisphere is \(M = \frac{1}{4}\pi aR^4\) and the third coordinate of its center of mass is \(z_c = \frac{2}{3}R\).