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Chapter 13: Q 32. (page 1039)

Let $T_{2}$ be triangular region with vertices $(1, 0), (2, 1), and (2, - 1)$
If the density at each point in $T_{2}$ is proportional to the point鈥檚 distance from the $y$ -axis, find the mass of $T_{2}$ .

Short Answer

Expert verified

The mass is $m = \frac{5}{2} k$

Step by step solution

01

Given Information

The vertices of triangular region is $(1, 0), (2, 1), and (2, - 1)$

02

Calculation of Mass

Mass of the lamina is calculated by

$m = {鈭�}_{惟} 蚁 (x, y) d A$

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

$蚁 (x, y) = k x$

$m = {鈭�}_{1}^{2} {鈭�}_{- x + 1}^{x - 1} k x d y d x$

Solving inner integral first

$m = {鈭�}_{1}^{2} k x [y]_{- x + 1}^{x - 1} d x$

$m = {鈭�}_{1}^{2} k x [x - 1 - (- x + 1)] d x$

$m = 2 k {鈭�}_{1}^{2} (x^{2} - x) d x$

$m = 2 k {[\frac{x^{3}}{3} - \frac{x^{2}}{2}]}_{1}^{2}$

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

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

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

Hence, $m = \frac{5}{3} k$

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Most popular questions from this chapter

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