Q 46. - Calculus

Question

$\mathcal{C}=\left\{(x,y)\mid x^2+y^2\le 1\right\}$

If the density at each point in $C$ is proportional to the point’s distance from the $y$-axis, find the mass of $C$.

Step-by-Step Solution

Answer

The mass of region is $m=\frac{4}{3}k$

Step 1: Given Information

We are given that $C=\left\{(x,y)\mid x^2+y^2\le 1\right\}$

Step 2: Calculating Mass of Region

Mass of region is given by $m={\iint }_{\Omega }\rho (x,y)dA$

$\rho (x,y)=kx$

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

According to figure

$m=4\times \text{ mass of a quadrant }$

$m=4{\int }_0^1{\int }_0^{\sqrt{1-x^2}}kxdydx$

$m=4{\int }_0^1{\int }_0^{\sqrt{1-x^2}}kxdydx$

Calculating inner integral first

$m=4{\int }_{01}^{1}kx[y{]}_0^{\sqrt{1-x^2}}dx$

$m=4k{\int }_0^{1}x\sqrt{1-x^2}dx$

Put $1-x^2=t^2$

$\Rightarrow -2 x d x=2 t d t$

$m=4k{\int }_0^1{t}^{2}dt=4k{\left[\frac{t^3}{3}\right]}_0^1=4k\left[\frac{1}{3}\right]$

Mass of region is $m=\frac{4}{3}k$