find integral dyda..ून--yodystu...2Rndyda..9.

To evaluate the following integrals: A=∫02R∫0R2-x-R2dydx x=∫02R∫0R2-x-R2ydydx y=∫02R∫0R2-x-R2xdydx

Answered: 2文 A- dyda. adyda...

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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find integral

Expert Solution

Step 1

To evaluate the following integrals:

$A = \int_{0}^{2 R} \int_{0}^{\sqrt{R^{2} - {(x - R)}^{2}}} d y d x$
$x = \int_{0}^{2 R} \int_{0}^{\sqrt{R^{2} - {(x - R)}^{2}}} y d y d x$
$y = \int_{0}^{2 R} \int_{0}^{\sqrt{R^{2} - {(x - R)}^{2}}} x d y d x$

Step 2

i. To evaluate $A = \int_{0}^{2 R} \int_{0}^{\sqrt{R^{2} - {(x - R)}^{2}}} d y d x$ :

$A = \int_{0}^{2 R} \int_{0}^{\sqrt{R^{2} - {(x - R)}^{2}}} d y d x = \int_{0}^{2 R} {[y]}_{0}^{\sqrt{R^{2} - {(x - R)}^{2}}} d x = \int_{0}^{2 R} \sqrt{R^{2} - {(x - R)}^{2}} d x S u b s t i t u t e u = x - R \Rightarrow d u = d x & x : 0 \to 2 R \Rightarrow u : - R \to R = \int_{- R}^{R} \sqrt{R^{2} - u^{2}} d u (W e h a v e \int \sqrt{a^{2} - x^{2}} d x = \frac{x \sqrt{a^{2} - x^{2}}}{2} + \frac{a^{2}}{2} \sin^{- 1} (\frac{x}{a})) = {[\frac{u \sqrt{R^{2} - u^{2}}}{2} + \frac{R^{2}}{2} \sin^{- 1} (\frac{u}{R})]}_{- R}^{R} = [\frac{R^{2}}{2} \sin^{- 1} (1) - \frac{R^{2}}{2} \sin^{- 1} (- 1)] = [\frac{R^{2}}{2} \frac{π}{2} - \frac{R^{2}}{2} \times - \frac{π}{2}] = \frac{π R^{2}}{2}$

Thus, $A = \int_{0}^{2 R} \int_{0}^{\sqrt{R^{2} - {(x - R)}^{2}}} d y d x = \frac{π R^{2}}{2}$

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