Q 56. - Calculus

Question

Let $S=\left\{\left(x,y\right)|x^2+y^2\le 1,x\ge 0\right\}$

If the density at each point in S is proportional to the point’s distance from the origin, find the mass of S.

Step-by-Step Solution

Answer

Answer is $\begin{array}{rcl}& & \frac{\pi }{3}k\end{array}$ where k is the constant of proportionality.

Step 1. Given information

$S=\left\{\left(x,y\right)|x^2+y^2\le 1,x\ge 0\right\}$

Step 2. Explanation

$S=\left\{\left(x,y\right)|x^2+y^2\le 1,x\ge 0\right\}$

$\begin{array}{rcl}m& =& {\int }_0^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}k\sqrt{x^2+y^2}dydx\\ & =& 2{\int }_0^1{\int }_0^{\sqrt{1-x^2}}k\sqrt{x^2+y^2}dydx\end{array}$

Let $x=r \cos \theta , y= r \sin \theta$

$\begin{array}{rcl}m& =& 2k{\int }_0^{\pi /2}{\left[\frac{r^3}{3}\right]}_0^{1}d\theta \\ & =& 2k{\int }_0^{\pi /2}\left[\frac{1}{3}-\frac{0}{3}\right]d\theta \\ & =& \frac{2}{3}k{\int }_0^{\pi /2}d\theta \\ & =& \frac{2}{3}k[\theta {]}_0^{\pi /2}\\ & =& \frac{2}{3}k\left[\frac{\pi }{2}-0\right]\\ & =& \frac{\pi }{3}k\end{array}$

 k is the constant of proportionality. 

Mass is 9 units.