Chapter 13: Q 31. (page 1039)

Let $T_{2}$ be triangular region with vertices $(1, 0), (2, 1), and (2, - 1)$
Find centroid of $T_{2}$

Short Answer

Expert verified

The centroid is $\bar{x} = \frac{17}{10}, \bar{y} = 0$

Step by step solution

Given Information

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

Finding x-

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

Density is uniform

$ρ (x, y)$ is proportional to point's distance from $y$ axis.

$ρ (x, y) = k x$

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

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

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

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

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

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

$\bar{x} = \frac{[(\frac{16}{4} - \frac{8}{3}) - (\frac{1}{4} - \frac{1}{3})]}{[(\frac{8}{3} - \frac{4}{2}) - (\frac{1}{3} - \frac{1}{2})]}$

$\bar{x} = \frac{17}{10}$

Finding y-

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

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

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

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

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

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

Hence

$\bar{x} = \frac{17}{10}, \bar{y} = 0$

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Let $T_{2}$ be triangular region with vertices $(1, 0), (2, 1), and (2, - 1)$
Find centroid of $T_{2}$

Short Answer

Step by step solution

Given Information

Finding x-

Finding y-

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Let T2be triangular region with vertices (1,0),(2,1),and(2,-1)Find centroid ofT2

Short Answer

Step by step solution

Given Information

Finding x-

Finding y-

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Theoretical and Mathematical Physics

Mechanics Maths

Decision Maths

Applied Mathematics

Probability and Statistics

Study anywhere. Anytime. Across all devices.

Let $T_{2}$ be triangular region with vertices $(1, 0), (2, 1), and (2, - 1)$
Find centroid of $T_{2}$