# A star is modeled as a ball of radius a with density given by density=k(1-(rho/alpha)^3)....

## Question:

A star is modeled as a ball of radius a with density given by density=k(1-({eq}\rho/\alpha {/eq}){eq}^3 {/eq}). Determine total mass of star and the coordinates of the center of mass of the star.

## Triple Integrals:

{eq}\\ {/eq}

Integrals defined over a region in space have to be integrated thrice. Triple integrals are vastly used in computing mass and volumes of solids.

Spherical coordinate system {eq}(\rho, \theta, \phi) {/eq} can be suitably selected while evaluating integrals over spherical regions.

It is defined as: {eq}x = \rho\cos{\theta}\sin{\phi}, \ y = \rho\sin{\theta}\sin{\phi}, \ z = \rho\cos{\phi} {/eq}

{eq}\\ {/eq}

Radius of a star is {eq}a {/eq} and its density is: {eq}\displaystyle k\left(1-\left(\frac{\rho}{a}\right)^3\right) {/eq}

where, {eq}k {/eq} is a constant.

{eq}\\ {/eq}

Consider an elemental volume of the star: {eq}\mathrm{d}V {/eq}. Mass of this volume is given by: {eq}\displaystyle \mathrm{d}m = \text{density } \times \mathrm{d}V = k\left(1-\left(\frac{\rho}{a}\right)^3\right) \ \mathrm{d}V {/eq}

Mass of the entire star is obtianed by integrating this elemental mass over the region of the star. Therefore:

{eq}\begin{align*} M = \iiint_D \underbrace{k\left(1-\left(\frac{\rho}{a}\right)^3\right)}_{\text{density}}\ \mathrm{d}V \end{align*} {/eq}

{eq}D {/eq} is the spherical region (ball) of the star. It is convenient to switch to spherical coordinates in the region of integration.

Over the region {eq}D {/eq}: {eq}0 \leq \rho \leq a, \ 0 \leq \theta \leq 2\pi, \ 0 \leq \phi \leq \pi {/eq}. Therefore:

{eq}\begin{align*} M &= \int_{0}^{2\pi}\int_{0}^{a}\int_{0}^{\pi}k\left(1-\left(\frac{\rho}{a}\right)^3\right) \cdot \underbrace{\rho^2\sin{\phi} \mathrm{d}\phi\mathrm{d}\rho\mathrm{d}\theta}_{\mathrm{d}V} \\ &= \int_{0}^{2\pi}\int_{0}^{a} k\rho^2\left(1-\left(\frac{\rho}{a}\right)^3\right) \Big[-\cos{\phi}\Big]_{0}^{\pi} \ \mathrm{d}\rho\mathrm{d}\theta \\ &= 2k \int_{0}^{2\pi}\int_{0}^{a} \rho^2 - \frac{\rho^5}{a^3} \ \ \mathrm{d}\rho\mathrm{d}\theta \\ &= 2k \int_{0}^{2\pi} \left[ \frac{\rho^3}{3} - \frac{\rho^6}{6a^3}\right]_{0}^{a} \ \mathrm{d}\theta \\ &= \frac{ka^3}{3} \int_{0}^{2\pi} \ \mathrm{d}\theta \\ \therefore \ M &= \frac{2\pi ka^3}{3} \end{align*} {/eq}