Q 49. - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

Step 1. Given information

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

Step 2. Explanation

Mass of lamina is equal to $\iint \rho \left(x,y\right) dA$

Here, $\rho \left(x,y\right)$ is the uniform density of the lamina.

As $\rho \left(x,y\right)$ is proportional to point's distance from $y-$axis, take $\rho \left(x,y\right)=k\sqrt{x^2+y^2}$

So,

$\begin{array}{rcl}m& =& {\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}k\sqrt{x^2+y^2}dydx\\ & =& 2k{\int }_{-1}^1{\left[\frac{1}{2}y\sqrt{x^2+y^2}+\frac{1}{2}{x}^2\ln \left(y+\sqrt{x^2+y^2}\right)\right]}_0^{\sqrt{1-x^2}}dx\\ & =& 2k{\int }_{-1}^1\left[\frac{1}{2}\sqrt{1-x^2}+\frac{1}{2}{x}^2\ln \left(\frac{1+\sqrt{1-x^2}}{x}\right)\right]dx\\ & =& 2k{\left[\frac{x}{2}\sqrt{1-x^2}+x\ln \left(\frac{1+\sqrt{1-x^2}}{x}\right)+\frac{3}{2}{\sin }^{-1}x\right]}_{-1}^1\\ & =& \frac{2}{3}k\mathrm{\pi }\end{array}$

k is the constant of proportionality.