Question

If the density at each point in $\mathcal{T}$ is proportional to the point's distance from the y-axis, find the mass of$\mathcal{T}.$

Short answer

The mass of the triangular lamina is

$m=\frac{2}{3}k$

Step-by-step solution

Given information

The objective of this problem is to find the mass of the triangular region.

calculation

For an example, take a triangle with vertices $(0,0),(1,1)$ and $(1,-1)$.

Mass of the lamina is can be calculated by the formula

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

$\rho (x,y)$is the uniform density of the lamina. If$\rho (x,y)$is proportional to the point's distance from the y - axis.

$$\begin{gathered} \rho (x,y)=kx \\ m={\int }_0^1{\int }_{-x}^{x}kxdydx \end{gathered}$$

Integrate with respect to y first.

$$\begin{gathered} m={\int }_0^{1}kx[y{]}_{-x}^{x}dx \\ m={\int }_0^{1}kx[x-(-x)]dx \\ m=2k{\int }_0^1{x}^{2}dx \end{gathered}$$

Integrate with respect to x.

$m=2k{\left[\frac{x^3}{3}\right]}_0^1$

Substitute the limits

$m=\frac{2}{3}k$

Thus, the mass of the triangular lamina is

$m=\frac{2}{3}k$