Question 5.7.5: Find the centroid of the "ice-cream-cone-shaped" region belo...

The region is sketched in Figure 5. Since x^{2}+y^{2}+z^{2}=\rho^{2} and z=\rho \cos \varphi, the sphere \left\{(x, y, z): x^{2}+y^{2}+z^{2}=z\right\} can be written in spherical coordinates as

\rho^{2}=\rho \cos \varphi, \quad \text { or } \quad \rho=\cos \varphi.

Since x^{2}+y^{2}=\rho^{2} \sin ^{2} \varphi \cos ^{2} \theta+\rho^{2} \sin ^{2} \varphi \sin ^{2} \theta=\rho^{2} \sin ^{2} \varphi, the equation of the cone is

\rho^{2} \cos ^{2} \varphi=\rho^{2} \sin ^{2} \varphi, \quad \text { or } \quad \cos ^{2} \varphi=\sin ^{2} \varphi,

and since 0 \leq \varphi \leq \pi, the equation is

Thus, assuming that the density of the region is the constant k, we have
\begin{aligned}\mu &=\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{0}^{\cos \varphi} k \rho^{2} \sin \varphi  d \rho  d \varphi  d \theta \\&=k \int_{0}^{2 \pi} \int_{0}^{\pi / 4} \sin \varphi\left\{\left.\frac{\rho^{3}}{3}\right|_{0} ^{\cos \varphi}\right\}  d \varphi  d \theta=k \int_{0}^{2 \pi} \int_{0}^{\pi / 4} \frac{\cos ^{3} \varphi}{3} \sin \varphi  d \varphi  d \theta \\&=k \int_{0}^{2 \pi}\left\{-\left.\frac{\cos ^{4} \varphi}{12}\right|_{0} ^{\pi / 4}\right\} d \theta=\frac{1}{12} k \int_{0}^{2 \pi} \frac{3}{4} d \theta=\frac{\pi k}{8}\end{aligned}
Since x=\rho \sin \varphi \cos \theta, we have
\begin{aligned}M_{y z} &=k \int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{0}^{\cos \varphi}(\rho \sin \varphi \cos \theta) \rho^{2} \sin \varphi  d \rho  d \varphi  d \theta \\&=k \int_{0}^{2 \pi} \cos \theta  d \theta \int_{0}^{\pi / 4} \int_{0}^{\cos \varphi} \rho^{3} \sin ^{2} \varphi  d \rho  d \varphi=0\end{aligned}

since

(an unsurprising result because of symmetry). Similarly, since y=\rho \sin \varphi \sin \theta,

(again because of symmetry). Finally, since z=\rho \cos \varphi,

Thus \bar{x}=\bar{y}=0 and \bar{z}=M_{x y} / \mu=(7 \pi k / 96) /(\pi k / 8)=7 / 12.