Chapter 13: Q 33. (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’s distance from the $y$ -axis, find the center of mass of $T_{2}$ .

Short Answer

Expert verified

The center of mass is $(\bar{x}, \bar{y}) = (\frac{17}{10}, 0)$

Step by step solution

Given Information

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

Density is $ρ (x, y) = k x$

Find x-

The formula is:

$\bar{x} = \frac{\iint_{Ω} x ρ (x, y) d A}{\iint_{Ω} ρ (x, y) d A} and \bar{y} = \frac{\iint_{Ω} y ρ (x, y) d A}{\iint_{Ω} ρ (x, y) d A}$

Limits of $y$ varies from $- x + 1$ to $x + 1$

$x$ varies from $1 - 2$

Formula becomes $\bar{x} = \frac{\int_{1}^{2} \int_{- x + 1}^{x - 1} x k x d y d x}{\int_{1}^{2} \int_{- x + 1}^{x - 1} k x d y d x}$

$= \frac{\int_{1}^{2} \int_{- x + 1}^{x - 1} k x^{2} d y d x}{\int_{1}^{2} \int_{- x + 1}^{x - 1} k x d y d x}$

$\bar{x} = \frac{\int_{1}^{2} x^{2} [y]_{- x + 1}^{x - 1} d x}{\int_{1}^{2} x [y]_{- x + 1}^{x - 1} d x}$

$= \frac{\int_{1}^{2} x^{2} [(x - 1) - (- x + 1)] d x}{\int_{1}^{2} x [(x - 1) - (- x + 1)] d x}$

$= \frac{\int_{1}^{2} x^{2} [2 x - 2] d x}{\int_{1}^{2} x [2 x - 2] d x}$

$\bar{x} = \frac{\int_{1}^{2} (x^{3} - x^{2}) d x}{\int_{1}^{2} (x^{2} - x) d x}$

$= \frac{{[\frac{x^{4}}{4} - \frac{x^{3}}{3}]}_{1}^{2}}{{[\frac{x^{3}}{3} - \frac{x^{2}}{2}]}_{1}^{2}}$

$= \frac{[(\frac{2^{4}}{4} - \frac{2^{3}}{3}) - (\frac{1}{4} - \frac{1}{3})]}{[(\frac{2^{3}}{3} - \frac{2^{2}}{2}) - (\frac{1}{3} - \frac{1}{2})]} = \frac{7}{10}$

Find y-

The formula is

$\bar{y} = \frac{\iint_{Ω} y ρ (x, y) d A}{\iint_{Ω} ρ (x, y) d A}$

Using values of $ρ (x, y)$ and values of $x, y$

$\bar{y} = \frac{\int_{1}^{2} \int_{- x + 1}^{x - 1} y k x d y d x}{\int_{1}^{2} \int_{- x + 1}^{x - 1} k x d y d x}$

$\bar{y} = \frac{\int_{1}^{2} k x {[\frac{y^{2}}{2}]}_{- x + 1}^{x - 1} d x}{\int_{1}^{2} k x [y]_{- x + 1}^{x - 1} d x}$

$= \frac{\int_{1}^{2} k x [\frac{(x - 1)^{2} - (- x + 1)^{2}}{2}] d x}{\int_{1}^{2} 2 k (x^{2} - x) d x}$

$= \frac{\int_{1}^{2} k x [0] d x}{\int_{1}^{2} 2 k (x^{2} - x) d x} = 0$

Centroid is $(\bar{x}, \bar{y}) = (\frac{17}{10}, 0)$

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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’s distance from the $y$ -axis, find the center of mass of $T_{2}$ .

Short Answer

Step by step solution

Given Information

Find x-

Find y-

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Let T2be triangular region with vertices (1,0),(2,1),and(2,-1)If the density at each point in T2is proportional to the point’s distance from the y-axis, find the center of mass of T2.

Short Answer

Step by step solution

Given Information

Find x-

Find y-

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Decision Maths

Geometry

Mechanics Maths

Discrete Mathematics

Logic and Functions

Study anywhere. Anytime. Across all devices.

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’s distance from the $y$ -axis, find the center of mass of $T_{2}$ .