Question

Show that the first moment of$\in$is$M_{YZ}=\frac{1}{16}\mathrm{\pi k}$

Short answer

It is done by using the formula$M_{YZ}=\int \int {\int }_{\in }x\rho (x,y,z)dxdydz$

Step-by-step solution

Given Information

$\in$is the portion of unit sphere centered at the origin of first octant.

Simplification and determination of limits

To determine the first moment of solid w.r.t $yz$plane, formula to be used:

$M_{YZ}=\int \int {\int }_{\in }x\rho (x,y,z)dxdydz$

Here, $k$is uniform density of $\in$

$\Rightarrow \rho \left(x,y,z\right)=k$

We know the relation:

$x=\rho \sin \varphi \cos \theta ,y=\rho \sin \varphi \sin \theta ,z=\rho \cos \varphi$

Here, $\rho =\sqrt{x^2+y^2+z^2},\tan \theta =\frac{y}{z},\cos \varphi =\frac{z}{\rho }$

Also $dxdydz={\rho }^2\sin \varphi d\rho d\varphi d\theta$

$0<\theta <\frac{\mathrm{\pi }}{2},0<\varphi <\frac{\mathrm{\pi }}{2},0<\rho <1$are limits of spherical coordinates $\rho ,\varphi ,\theta$

Calculating the First Moment

First moment of $\in =\int \int {\int }_{V}x\rho \left(x,y,z\right)dxdydz$

Converting into spherical coordinates

$={\int }_{\varphi =0}^{\frac{\mathrm{\pi }}{2}}{\int }_{\theta =0}^{\frac{\mathrm{\pi }}{2}}{\int }_{\rho =0}^1{\rho }^2\sin \varphi \left(\rho \sin \varphi \cos \theta \right)kd\rho d\varphi d\theta$

$=k\left[{\int }_{\varphi =0}^{\frac{\mathrm{\pi }}{2}}{\sin }^2\varphi d\varphi \right]\times \left[{\int }_{\theta =0}^{\frac{\mathrm{\pi }}{2}}\cos \theta d\theta \right]\times \left[{\int }_{\rho =0}^1{\rho }^{3}d\rho \right]$

Solving the integral

$=k\left(\frac{1}{2}\times \frac{\mathrm{\pi }}{2}\right){\left(\frac{{\rho }^4}{4}\right)}_0^1{\left(\sin \theta \right)}_0^{\frac{\mathrm{\pi }}{2}}$

By application of limits, we get

$M_{YZ}=k\left(\frac{\mathrm{\pi }}{4}\right)\left(\frac{1}{4}\right)\left(1\right)$

Hence, first moment is$M_{YZ}=\frac{\mathrm{\pi k}}{16}$