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Essential Calculus: Early Transcendentals
Found in: Page 713
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Evaluate the given integral by changing to polar coordinates.\(\iint_{\text{R}} {\left( {{\text{2x - y}}} \right)}{\text{ dA}}\)

Where \({\rm{R}}\)is the region in the first quadrant enclosed by the circle\({{\rm{x}}^2}{\rm{ + }}{{\rm{y}}^2}{\rm{ = }}4\)and the lines\({\rm{x = 0 and y = x}}\).

Thus, the value is \(\int_R {(2x - }y){\text{ }}dA = \left( {\frac{{16}}{3} - 4\sqrt 2 } \right)\)

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given information.

It is given that,

Where \(R\)is the region in the first quadrant enclosed by the circle\({x^2} + {\rm{ }}{y^2} = {\rm{ }}4\)and the lines\(x{\rm{ }} = {\rm{ }}0{\rm{ }}and{\rm{ }}y{\rm{ }} = {\rm{ }}x\).

Step 2: Evaluate by changing to polar coordinates.

Substitute\(x = r\cos \theta \),\(y = r\sin \theta \), and \(dA = rdrd\theta \).

The equation of circle will be,

\(\begin{array}{l}{x^2} + {\rm{ }}{y^2} = {\rm{ }}4\\ \Rightarrow {\left( {r\cos \theta } \right)^2} + {\left( {r\sin \theta } \right)^2} = 4\\ \Rightarrow {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta = 4\\ \Rightarrow {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) = 4\\{\cos ^2}\theta + {\sin ^2}\theta = 1\\ \Rightarrow {r^2} = 4\\ \Rightarrow r = 2\end{array}\)

So, \(0 \le r \le 2\).

The equation of lines will be,

\(\begin{array}{l}x{\rm{ }} = {\rm{ }}0\\ \Rightarrow r\cos \theta = 0\\ \Rightarrow \cos \theta = 0\\ \Rightarrow \theta = \frac{\pi }{2}\end{array}\)

And

\(\begin{array}{l}y{\rm{ }} = {\rm{ }}x\\ \Rightarrow r\sin \theta = r\cos \theta \\ \Rightarrow \frac{{r\sin \theta }}{{r\cos \theta }} = 1\\ \Rightarrow \tan \theta = 1\\ \Rightarrow \theta = \frac{\pi }{4}\end{array}\)

So, \(\frac{\pi }{4} \le \theta \le \frac{\pi }{2}\).

The value of the integral will be,

\(I = \int\limits_{\pi /4}^{\pi /2} {\int\limits_0^2 {\left( {2r\cos \theta - r\sin \theta } \right) \cdot rdrd\theta } } \)

Integrating on \(r\) we get,

\(\begin{array}{l}I = \int\limits_0^\pi {\left( {2\cos \theta - \sin \theta } \right) \cdot \left( {\frac{1}{3}{r^3}} \right)_0^2d\theta } \\ = \int\limits_0^\pi {\left( {2\cos \theta - \sin \theta } \right) \cdot \left( {\frac{1}{3}\left( {{2^3} - {0^3}} \right)} \right)d\theta } \\ = \frac{8}{3}\int\limits_0^\pi {\left( {2\cos \theta - \sin \theta } \right)d\theta } \end{array}\)

Integrating on \(\theta \) we get,

\(\begin{aligned} I &= \frac{{16}}{3}\left( {\sin \theta } \right)_{\pi /4}^{\pi /2} - \frac{8}{3}\left( { - \cos \theta } \right)_{\pi /4}^{\pi /2}\\ &= \frac{{16}}{3}\left( {\sin \frac{\pi }{2} - \sin \frac{\pi }{4}} \right) + \frac{8}{3}\left( {\cos \frac{\pi }{2} - \cos \frac{\pi }{4}} \right)\\ &= \frac{{16}}{3}\left( {1 - \frac{1}{{\sqrt 2 }}} \right) + \frac{8}{3}\left( {0 - \frac{1}{{\sqrt 2 }}} \right)\\ &= \frac{{16}}{3} - \frac{{16}}{{3\sqrt 2 }} - \frac{8}{{3\sqrt 2 }}\\ &= \left( {\frac{{16}}{3} - \frac{{24}}{{3\sqrt 2 }}} \right)\\ &= \left( {\frac{{16}}{3} - 4\sqrt 2 } \right)\end{aligned}\)

Therefore, \(\iint_R {(2x - }y){\text{ }}dA = \left( {\frac{{16}}{3} - 4\sqrt 2 } \right)\)

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