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Chapter 13: Q 29. (page 1039)

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

Short Answer

Expert verified

Mass is $\frac{k}{3}$ .

Moment of Mass are $M_{y} = \frac{k}{4}$ and $M_{x} = \frac{k}{6}$ .

Center of mass are $\overset{炉}{x} = \frac{3}{4}, \overset{炉}{y} = \frac{1}{2}$

Step by step solution

01

Given Information

Vertices of triangular region are $(0, 0), (1, 1), and (1, - 1)$

Density is proportional to point's distance from $x$ axis.

$蚁 (x, y) = k y$

02

Mass of Triangular Region

Mass of triangle = Twice the mass of upper triangle.

$m = 2 {鈭�}_{0}^{1} {鈭�}_{0}^{x} k y d y d x [m = {鈭�}_{惟} 蚁 (x, y) d A]$

$m = 2 k {鈭�}_{0}^{1} {[\frac{y^{2}}{2}]}_{0}^{x} d x$

$m = k {鈭�}_{0}^{1} x^{2} d x$

$m = k {[\frac{x^{3}}{3}]}_{0}^{1}$

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

03

First Moment of Mass about y axis

$M_{y} = {鈭�}_{惟} x 蚁 (x, y) d A$

Putting limits

$M_{y} = {鈭�}_{0}^{1} {鈭�}_{- x}^{x} x k y d y d x [蚁 (x, y) = k y]$

Do half the limit and twice the integral as per the given figure.

$M_{y} = 2 {鈭�}_{0}^{1} {鈭�}_{0}^{x} x k y d y d x$

$M_{y} = 2 k {鈭�}_{0}^{1} {[\frac{y^{2}}{2}]}_{0}^{x} x d x$

$M_{y} = k {鈭�}_{0}^{1} x^{3} d x$

$M_{y} = {[\frac{x^{4}}{4}]}_{0}^{1} k = \frac{k}{4}$

04

First Moment of Mass about x axis

$M_{x} = {鈭�}_{惟} y 蚁 (x, y) d A$

Putting limits

$M_{x} = {鈭�}_{0}^{1} {鈭�}_{- x}^{x} k y^{2} d y d x [蚁 (x, y) = k y]$

Solving inner integral first

$M_{x} = k {鈭�}_{0}^{1} {[\frac{y^{3}}{3}]}_{- x}^{x} d x$

$M_{x} = k {鈭�}_{0}^{1} [\frac{(x)^{3} - (- x)^{3}}{3}] d x$

$M_{x} = \frac{2}{3} k {鈭�}_{0}^{1} x^{3} d x$

$M_{x} = \frac{2}{3} k {[\frac{x^{4}}{4}]}_{0}^{1} = \frac{k}{6}$

05

Center of Mass

The coordinates are $\overset{炉}{x} = \frac{M_{y}}{m}, \overset{炉}{y} = \frac{M_{x}}{m}$

$\overset{炉}{x} = \frac{\frac{k}{k}}{\frac{k}{3}}, \overset{炉}{y} = \frac{\frac{6}{k}}{\frac{k}{3}}$

$\overset{炉}{x} = \frac{3}{4}, \overset{炉}{y} = \frac{1}{2}$

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Most popular questions from this chapter

Use Definition $13.4$ to evaluate the double integrals in Exercises $29 鈥� 32$ .

${鈭�}_{R} x y d A$

where

R = {(x, y) 鈭� 0 鈮� x 鈮� 2 & 1 鈮� y 鈮� 4}

Evaluate each of the integrals in exercise 33-36 as iterated integrals and then compare your answers with those you found in exercise 29-32

$鈭� {鈭�}_{R} x^{2} y^{3} d A W h e r e R = {(x, y) | 1 鈮� x 鈮� 3, 0 鈮� y 鈮� 2}$

Evaluate each of the double integral in the exercise 37-54 as iterated integrals

$鈭� {鈭�}_{R} \sin (x + 2 y) d A W h e r e R = {(x, y) | 0 鈮� x 鈮� 蟺, 0 鈮� y 鈮� \frac{蟺}{2}}$

$How many summands are in {鈭�}_{i = 2}^{15} {鈭�}_{j = 3}^{17} {鈭�}_{k = 4}^{19} \frac{i}{j + k} ?$

Evaluate Each of the integrals in exercises 33-36 as an iterated integral and then compare your answer with thoise you found in exercise 29-32

$鈭� {鈭�}_{R} x y d A W h e r e R = {(x, y) | 0 鈮� x 鈮� 2, 1 鈮� y 鈮� 4}$

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