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Essential Calculus: Early Transcendentals
Found in: Page 528
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

Find the area of the region that lies inside the first curve and outside the second curve.

\({\rm{r = 2cos\theta ,}}\;\;\;{\rm{r = 1}}\).

The area is \(\frac{{\rm{\pi }}}{{\rm{3}}}{\rm{ + }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}\).

See the step by step solution

Step by Step Solution

Step 1: Explanation of solution.

Inside \({\rm{r = 2cos\theta }}\) (Green graph)

Outside \({\rm{r = 1}}\) (Bink graph)

Find for intersection

\(\begin{aligned}{c}{\rm{2cos\theta = 1}}\\{\rm{cos\theta = }}\frac{{\rm{1}}}{{\rm{2}}}\\{\rm{\theta = }}\frac{{\rm{\pi }}}{{\rm{3}}}{\rm{, - }}\frac{{\rm{\pi }}}{{\rm{3}}}\end{aligned}\)

Double the integral from \({\rm{0}}\) to \(\frac{{\rm{\pi }}}{{\rm{3}}}\) and symmetry.

Step 2: Calculation.

Take the inner curves \({{\rm{r}}^{\rm{2}}}\) and subtract them from the outer curves \({{\rm{r}}^{\rm{2}}}\).

\(\begin{aligned}{c}{\rm{A = 2 \times }}\frac{{\rm{1}}}{{\rm{2}}}\int_{\rm{a}}^{\rm{b}} {\left( {{\rm{r}}_{\rm{1}}^{\rm{2}}{\rm{ - r}}_{\rm{2}}^{\rm{2}}} \right)} {\rm{d\theta }}\\{\rm{ = }}\int_{\rm{0}}^{{\rm{\pi /3}}} {\left( {{{{\rm{(2cos\theta )}}}^{\rm{2}}}{\rm{ - }}{{\rm{1}}^{\rm{2}}}} \right)} {\rm{d\theta }}\\{\rm{ = }}\int_{\rm{0}}^{{\rm{\pi /3}}} {\left( {{\rm{4co}}{{\rm{s}}^{\rm{2}}}{\rm{\theta - 1}}} \right)} {\rm{d\theta }}\end{aligned}\)

\({\rm{ = }}\int_{\rm{0}}^{{\rm{\pi /3}}} {\left( {{\rm{4}}\left( {\frac{{\rm{1}}}{{\rm{2}}}{\rm{(1 + cos2\theta )}}} \right){\rm{ - 1}}} \right)} {\rm{d\theta }}\) Power reduction formula

\({\rm{ = }}\int_{\rm{0}}^{{\rm{\pi /3}}} {{\rm{(2 + 2cos2\theta - 1)}}} {\rm{d\theta }}\)

\(\begin{aligned}{l}{\rm{ = }}\int_{\rm{0}}^{{\rm{\pi /3}}} {{\rm{(1 + 2cos2\theta )}}} {\rm{d\theta }}\\{\rm{ = }}\left( {{\rm{\theta + 2 \times }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{sin2\theta }}} \right)_{\rm{0}}^{{\rm{\pi /3}}}\\{\rm{ = }}\left( {\frac{{\rm{\pi }}}{{\rm{3}}}{\rm{ + }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}{\rm{ - (0 + 0)}}} \right)\\{\rm{ = }}\frac{{\rm{\pi }}}{{\rm{3}}}{\rm{ + }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}.\end{aligned}\)

Therefore the area is \(\frac{{\rm{\pi }}}{{\rm{3}}}{\rm{ + }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}\).

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