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 square of the point’s distance from the $y$-axis, find the mass of $T_2$.

Short answer

The mass of lamina is$m=\frac{17}{6}k$.

Step-by-step solution

Given Information

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

Calculating Mass of Lamina

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

Also density is $\rho (x,y)=k{x}^2$

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

Solving inner integral first

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

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

$m=2k{\int }_1^2\left(x^3-x^2\right)dx$

$m=2k{\left[\frac{x^4}{4}-\frac{x^3}{3}\right]}_1^2$

Putting limits

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

$m=2k\left[\frac{17}{12}\right]$

Hence,$m=\frac{17}{6}k$