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Chapter 13: Q 27. (page 1038)

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 $y$ -axis, find the moments of inertia about the $x$ - and $y$ -axes. Use these answers to find the radii of gyration of $T$ about the $x$ - and $y$ -axes.

Short Answer

Expert verified

Radius of gyration is $R_{y} = \sqrt{\frac{3}{5}} and R_{x} = \sqrt{\frac{1}{5}}$

Mass of lamina is $m = \frac{2}{3} k$

Moments are $I_{x} = \frac{2}{15} k$ about $y$ axis and about $x$ axis is $I_{y} = \frac{2}{5} k$

Step by step solution

01

Given Information

Density at every point is proportional to point's distance from $y$ axis.

02

Moment of Inertia about axis

$x$ It is given by

$I_{y} = {鈭�}_{惟} x^{2} 蚁 (x, y) d A$

Putting limits

$I_{y} = {鈭�}_{0}^{1} {鈭�}_{- x}^{x} x^{2} 蚁 (x, y) d y d x$

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

$I_{y} = k {鈭�}_{0}^{1} {鈭�}_{- x}^{x} x^{3} d y d x$

Solving inner integral first

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

$I_{y} = k {鈭�}_{0}^{1} [x - (- x)] x^{3} d x$

$I_{y} = k {鈭�}_{0}^{1} 2 x^{4} d x$

Integrating wrt $x$

$I_{y} = 2 k {[\frac{x^{5}}{5}]}_{0}^{1} = \frac{2}{5} k$

03

Moment of Inertia about y axis

Similarly

$I_{x} = {鈭�}_{惟} y^{2} 蚁 (x, y) d A$

$I_{x} = {鈭�}_{0}^{1} {鈭�}_{- x}^{x} y^{2} 蚁 (x, y) d y d x$

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

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

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

role="math" localid="1653986937609" $I_{x} = k {鈭�}_{0}^{1} [\frac{2 x^{4}}{3}] d x$

$I_{x} = {[\frac{2 x^{5}}{15}]}_{0}^{1} k = \frac{2}{15} k$

04

Calculating Mass of Lamina

Mass of lamina is given by $m = {鈭�}_{惟} 蚁 (x, y) d A$

$m = {鈭�}_{0}^{1} {鈭�}_{- x}^{x} k x d y d x$

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

$m = k {鈭�}_{0}^{1} 2 x^{2} d x$

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

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

05

Find radius of gyration

About both axis, it is given by

$R_{y} = \sqrt{\frac{I_{y}}{m}} and R_{x} = \sqrt{\frac{I_{x}}{m}}$

$R_{y} = \sqrt{\frac{\frac{2}{5} k}{\frac{2}{3} k}} and R_{x} = \sqrt{\frac{\frac{2}{15} k}{\frac{2}{3} k}}$

$R_{y} = \sqrt{\frac{3}{5}} and R_{x} = \sqrt{\frac{1}{5}}$

$R_{y} = \sqrt{\frac{3}{5}} and R_{x} = \sqrt{\frac{1}{5}}$

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Most popular questions from this chapter

Describe the three-dimensional region expressed in each iterated integral in Exercises 35鈥�44.

${鈭�}_{0}^{2} {鈭�}_{0}^{1 - 3 x / 2} {鈭�}_{2 x + (4 / 3) y - 4}^{0} f (x, y, z) d z d y d x$

Evaluate the triple integrals over the specified rectangular solid region.

$\begin{aligned} {鈭�}_{R} 鈥� (x + 2 y + 3 z) d V, where \\ R = {(x, y, z) 鈭� 0 鈮� x 鈮� 4, 1 鈮� y 鈮� 5, and 2 鈮� z 鈮� 7} \end{aligned}$

Describe the three-dimensional region expressed in each iterated integral in Exercises 35鈥�44.

${鈭�}_{0}^{3} {鈭�}_{0}^{3} {鈭�}_{0}^{3 - y} f (x, y, z) d z d y d x$

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

$鈭� {鈭�}_{R} (3 - x + 4 y) d A w h e r e R = {(x, y) | 0 鈮� x 鈮� 1, - 1 鈮� y 鈮� 3}$

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}

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