Chapter 13: Q 52. (page 1039)

Let $S = \{(x, y) | x^{2} + y^{2} \leq 1, x \geq 0\}$
Find the centroid of S.

Short Answer

Expert verified

Answer is $(\frac{4}{3} π, 0)$

Step by step solution

Step 1. Given information

$S = \{(x, y) | x^{2} + y^{2} \leq 1, x \geq 0\}$

Step 2. Explanation

$S = \{(x, y) | x^{2} + y^{2} \leq 1, x \geq 0\}$

$\begin{array}{rcl} \bar{x} & = & \frac{\int_{0}^{1} \int_{- \sqrt{1 - x^{2}}}^{\sqrt{1 - x^{2}}} x d y d x}{\int_{0}^{1} \int_{- \sqrt{1 - x^{2}}}^{\sqrt{1 - x^{2}}} d y d x} \\ \bar{x} & = & \frac{\int_{0}^{1} x [y]_{- \sqrt{1 - x^{2}}}^{\sqrt{1 - x^{2}}} d x}{\int_{0}^{1} [y]_{- \sqrt{1 - x^{2}}}^{\sqrt{1 - x^{2}}} d x} \\ = & \frac{2 \int_{0}^{1} x \sqrt{1 - x^{2}} d x}{2 \int_{0}^{1} \sqrt{1 - x^{2}} d x} \\ = & \frac{\int_{0}^{1} x \sqrt{1 - x^{2}}}{\int_{0}^{1} \sqrt{1 - x^{2}} d x} \\ = & \frac{{[- \frac{1}{3} {(1 - x^{2})}^{3 / 2}]}_{0}^{1}}{{[\frac{1}{2} \{x \sqrt{1 - x^{2}} + \sin^{- 1} x\}]}_{0}^{1}} \\ = & \frac{(\frac{1}{3})}{(\frac{π}{4})} \\ = & \frac{4}{3} π \end{array}$

$\begin{array}{rcl} \bar{y} & = & \frac{\int_{0}^{1} \int_{- \sqrt{1 - x^{2}}}^{\sqrt{1 - x^{2}}} y d y d x}{\int_{0}^{1} \int_{- \sqrt{1 - x^{2}}}^{\sqrt{1 - x^{2}}} d y d x} \\ = & \frac{\int_{0}^{1} {[\frac{y^{2}}{2}]}_{- \sqrt{1 - x^{2}}}^{\sqrt{1 - x^{2}}} d x}{\int_{0}^{1} [y]_{- \sqrt{1 - x^{2}}}^{\sqrt{1 - x^{2}}} d x} \\ = & \frac{\int_{0}^{1} [0] d x}{\int_{0}^{1} [0] d x} \\ = & 0 \end{array}$

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Let $S = \{(x, y) | x^{2} + y^{2} \leq 1, x \geq 0\}$
Find the centroid of S.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Explanation

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Let S=x,y|x2+y2≤1,x≥0Find the centroid of S.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Explanation

One App. One Place for Learning.

Most popular questions from this chapter

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Theoretical and Mathematical Physics

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Let $S = \{(x, y) | x^{2} + y^{2} \leq 1, x \geq 0\}$
Find the centroid of S.