Q 32. - Calculus

Question

Let $T_2$ be triangular region with vertices $(1,0),(2,1)\text{, and }(2,-1)$

If the density at each point in $T_2$ is proportional to the point’s distance from the $y$-axis, find the mass of $T_2$.

Step-by-Step Solution

Answer

The mass is $m=\frac{5}{2}k$

Step 1: Given Information

The vertices of triangular region is $(1,0),(2,1)\text{, and }(2,-1)$

Step 2: Calculation of Mass

Mass of the lamina is calculated by

$m={\iint }_{\Omega }\rho (x,y)dA$

$\rho (x,y)$ is uniform density and is proportional to point's distance from $y$ axis.

$\rho (x,y)=kx$

$m={\int }_1^2{\int }_{-x+1}^{x-1}kxdydx$

Solving inner integral first

$m={\int }_1^{2}kx[y{]}_{-x+1}^{x-1}dx$

$m={\int }_1^{2}kx[x-1-(-x+1\left)\right]dx$

$m=2k{\int }_1^2\left(x^2-x\right)dx$

$m=2k{\left[\frac{x^3}{3}-\frac{x^2}{2}\right]}_1^2$

$m=2k\left[\left(\frac{2^3}{3}-\frac{2^2}{2}\right)-\left(\frac{1}{3}-\frac{1}{2}\right)\right]$

$m=2k\left[\left(\frac{8}{3}-\frac{4}{2}\right)-\left(\frac{1}{3}-\frac{1}{2}\right)\right]$

$m=2k\left[\frac{5}{6}\right]$

Hence, $m=\frac{5}{3}k$