Chapter 13: Q 36. (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 square of the point’s distance from the $y$ -axis, find the center of mass of $T_{2}$ .

Short Answer

Expert verified

The centroid is $\bar{x} = \frac{147}{85}, \bar{y} = 0$ .

Step by step solution

Given Information

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

$ρ (x, y) = k x^{2}$

Finding Center of Mass

The required formula is

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

Use $ρ (x, y) = k x^{2}$

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

$\bar{x} = \frac{\int_{1}^{2} \int_{- x + 1}^{x - 1} k x^{3} d y d x}{\int_{1}^{2} \int_{- x + 1}^{x - 1} k x^{2} d y d x}$

Solving inner integral first

$\bar{x} = \frac{\int_{1}^{2} k x^{3} [y]_{- x + 1}^{x - 1} d x}{\int_{1}^{2} k x^{2} [y]_{- x + 1}^{x - 1} d x}$

$\bar{x} = \frac{\int_{1}^{2} k x^{3} [(x - 1) - (- x + 1)] d x}{\int_{1}^{2} k x^{2} [(x - 1) - (- x + 1)] d x} = \frac{\int_{1}^{2} k x^{3} [2 x - 2] d x}{\int_{1}^{2} k x^{2} [2 x - 2] d x} = \frac{\int_{1}^{2} k (x^{4} - x^{3}) d x}{\int_{1}^{2} k (x^{3} - x^{2}) d x}$

Simplifying further

$\bar{x} = \frac{k {[\frac{x^{5}}{5} - \frac{x^{4}}{4}]}_{1}^{2}}{k {[\frac{x^{4}}{4} - \frac{x^{3}}{3}]}_{1}^{2}} = \frac{k [\frac{49}{20}]}{k [\frac{17}{12}]} = \frac{147}{85}$

Simplification

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

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

$\bar{y} = \frac{\int_{1}^{2} k x^{2} {[\frac{y^{2}}{2}]}_{- x + 1}^{x - 1} d x}{\int_{1}^{2} k x^{2} [y]_{- x + 1}^{x - 1} d x} = \frac{\int_{1}^{2} k x^{2} [\frac{(x - 1)^{2} - (- x + 1)^{2}}{2}] d x}{\int_{1}^{2} 2 k (x^{3} - x^{2}) d x} = \frac{\int_{1}^{2} k x^{2} [0] d x}{\int_{1}^{2} 2 k (x^{3} - x^{2}) d x}$

$\bar{x} = \frac{147}{85}, \bar{y} = 0$

Hence, Center of Mass is $\bar{x} = \frac{147}{85}, \bar{y} = 0$

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

Short Answer

Step by step solution

Given Information

Finding Center of Mass

Simplification

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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 square of the point’s distance from the y-axis, find the center of mass of T2.

Short Answer

Step by step solution

Given Information

Finding Center of Mass

Simplification

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Theoretical and Mathematical Physics

Logic and Functions

Statistics

Discrete Mathematics

Pure Maths

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