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 y-axis, find the mass of S.

Short answer

Answer is $\frac{2}{3}k,k$is the constant of proportionality.

Step-by-step solution

Step 1. Given information

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

Step 2. Explanation

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

Mass $={\int }_0^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}kxdydx$

$=2{\int }_0^1{\int }_0^{\sqrt{1-x^2}}kxdydx$

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

role="math" localid="1653727977284" $\begin{array}{rcl}\mathrm{Mass}& =& 2{\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^{1}kr\cos \theta rdrd\theta \\ & =& 2k{\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^1{r}^2\cos \theta drd\theta \\ & =& 2k{\int }_0^{\frac{\mathrm{\pi }}{2}}{\left[\frac{r^3}{3}\right]}_0^1\cos \theta d\theta \\ & =& \frac{2}{3}k{\left[\sin \theta \right]}_0^{\frac{\mathrm{\pi }}{2}}\\ & =& \frac{2}{3}k\left[\sin \left(\frac{\mathrm{\pi }}{2}\right)-\sin 0\right]\\ & =& \frac{2}{3}k\end{array}$

Mass is equal to $\frac{2}{3}k,k$is the constant of proportionality.