(a) Using spherical coordinates, find the volume cut from the sphere \(r \leq a\) by the cone \(\theta=\alpha<\pi / 2\) (b) Show that the \(z\) coordinate of the centroid of the volume in (a) is given by the formula \(\bar{z}=3 a(1+\cos \alpha) / 8\)

First, convert the given surfaces to spherical coordinates. The equation of the sphere in spherical coordinates is given by \(r \leq a\). The cone's equation in spherical coordinates is given by \(\theta = \alpha\, \alphaugh\leq \ \frac{𝜋}{2}\).

To find the volume, set up a triple integral using spherical coordinates. The volume element in spherical coordinates is \(r^2 \sin{\theta} \, dr \, d\theta \, d\phi\). The limits for \(r\) are from 0 to \(a\), for \(\theta\) are from 0 to \(\alpha\), and for \(\phi\) are from 0 to 2\(\pi\). Therefore, the integral becomes \[V = \int_0^{2\pi} \int_0^{\alpha} \int_0^a r^2 \sin{\theta} \, dr \, d\theta \, d\phi\]

Now, evaluate the triple integral step by step: \[\int_0^{2\pi} \int_0^{\alpha} \int_0^a r^2 \sin{\theta} \, dr \, d\theta \, d\phi\]. First, integrate with respect to \(r\): \[\int_0^a r^2 \, dr = \frac{a^3}{3}\]. Next, integrate with respect to \(\theta\): \[\int_0^{\alpha} \sin{\theta} \, d\theta = -\cos{\theta}\bigg|_0^{\alpha} = 1 - \cos{\alpha}\]. Finally, integrate with respect to \(\phi\): \[\int_0^{2\pi} d\phi = 2\pi\]. Therefore, the volume \(V\) is \[V = \frac{2\pi a^3 (1 - \cos{\alpha})}{3}\]

The centroid's \(z\)-coordinate is found by calculating \(\bar{z} = \frac{1}{V} \int_V z \, dV\). Since \(z = r\cos{\theta}\) in spherical coordinates, the integral for the \(z\)-coordinate of the centroid becomes: \[\bar{z} = \frac{1}{V} \int_0^{2\pi} \int_0^{\alpha} \int_0^a r^3 \cos{\theta} \sin{\theta} \, dr \, d\theta \, d\phi\].

Evaluate the integral step by step: First, integrate with respect to \(r\): \[\int_0^a r^3 \, dr = \frac{a^4}{4}\]. Next, integrate with respect to \(\theta\): \[\int_0^{\alpha} \cos{\theta} \sin{\theta} \, d\theta = \frac{-\cos^2{\theta}}{2}\bigg|_0^{\alpha} = \frac{1 - \cos^2{\alpha}}{2} = \frac{\sin^2{\alpha}}{2}\]. Lastly, integrate with respect to \(\phi\): \[\int_0^{2\pi} d\phi = 2\pi\]. Thus, the numerator for \(\bar{z}\) is \[2\pi \cdot \frac{a^4}{4} \cdot \frac{\sin^2{\alpha}}{2} = \frac{\pi a^4 \sin^2{\alpha}}{2}\] and the volume \(V\) is \[\frac{2\pi a^3 (1 - \cos{\alpha})}{3}\]. Therefore, \[\bar{z} = \frac{\frac{\pi a^4 \sin^2{\alpha}}{2}}{\frac{2\pi a^3 (1 - \cos{\alpha})}{3}} = \frac{3a \sin^2{\alpha}}{4(1 - \cos{\alpha})}\]. Simplify the expression to obtain \[\bar{z} = \frac{3a (1 + \cos{\alpha})(1 - \cos{\alpha})}{8 (1 - \cos{\alpha})}= \frac{3 a (1 + \cos{\alpha})}{8}\].