Q 65. - Calculus

Question

Find the specified quantities for the solids described below:

The center of mass of the region from Exercise $55$, assuming that the density of the region is constant.

Step-by-Step Solution

Answer

The center of mass is given by $(\overline{x},\overline{y},\overline{z})=\left(0,0,\left(\frac{3R}{8}\right)(1+\cos \alpha )\right)$.

Step 1: Given Information

The density of region is constant.

Region above sphere is given by equation $\rho =R$ and below the cone is given by equation $\varphi =\alpha$.

Step 2: Evaluation of center of mass of x & y coordinate

For the given information, Center of ,mass if given by

$\overline{x}=\frac{M_{yz}}{M}=\frac{{\iiint }_{E}x\rho dxdydz}{\iiint \rho dxdydz}$

$\overrightarrow{y}=\frac{M_x}{M}=\frac{{\iiint }_{E}y\rho dxdydz}{{\iiint }_E\rho dxdydz}$

As density of region is constant $\rho =k$, the axis of center lies above $+z$ axis and base lies in $xy$ plane.

Hence, $\overline{x}=0, \overline{y}=0$

Step 3: Evaluation of center of mass of z coordinate

It can be given by $\overline{z}=\frac{M_{xy}}{M}\frac{{\iiint }_{E}z\rho dxdydz}{{\iiint }_E\rho dxdydz}$

$\overline{z}=\frac{M_{xy}}{M}=\frac{{\iiint }_{E}z\rho dV}{{\iiint }_E\rho dV}$

$=\frac{{\int }_{\varphi =0}^{\alpha }{\int }_{\theta =0}^{2\pi }{\int }_{\rho =0}^R\left(\rho \cos \varphi \right)k{\rho }^2\sin \varphi d\rho d\theta d\varphi }{{\int }^a{\int }^{2\pi }{\int }^{R}k{\rho }^2\sin \varphi d\rho d\theta d\varphi }$

$=\frac{{\left.{\left.{\left.\left(\frac{{\sin }^2\varphi }{2}\right)\right|}_{\varphi =0}^{\varphi =\alpha }k\left(\frac{{\rho }^4}{4}\right)\right|}_{\rho =0}^R\left(\theta \right)\right|}_{\theta =0}^{2\pi }}{{\left.{\left.{\left.k(-\cos \varphi )\right|}_{\varphi =0}^{\alpha }\left(\theta \right)\right|}_{\theta =0}^{2\pi }\left(\frac{{\rho }^3}{3}\right)\right|}_{\rho =0}^R}$

Putting limits yields

$\overline{z}=\frac{\left(\frac{{\sin }^2\alpha }{2}\right)k\left(\frac{R^4}{4}\right)\left(2\pi \right)}{k(1-\cos \alpha )\left(2\pi \right)\left(\frac{R^3}{3}\right)}$

$=\frac{{\sin }^2\alpha \times R^4\left(2k\pi \right)}{8}\times \frac{3}{\left(2k\pi \right)R^3(1-\cos \alpha )}$

$=\left(\frac{3R}{8}\right)(1+\cos \alpha )$

Hence, required Center of Mass is given by $(\overline{x},\overline{y},\overline{z})=\left(0,0,\left(\frac{3R}{8}\right)(1+\cos \alpha )\right)$