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 both curves.
\({\rm{r = sin2\theta ,r = cos2\theta }}\)

The resulting area of the zone between the two curves\(\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ - 1}}\).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Polar region.

As shown, there are nine intersection locations, one at the pole and the other eight in each quadrant. By setting the two equations equal to each other, we can determine one of the remaining eight (at least its\({\rm{\theta }}\)-value).

\(\begin{aligned}{c}{\rm{sin2\theta = cos2\theta }}\\\frac{{{\rm{sin2\theta }}}}{{{\rm{cos2\theta }}}}{\rm{ = 1}}\\{\rm{tan2\theta = 1}}\\{\rm{2\theta = ta}}{{\rm{n}}^{{\rm{ - 1}}}}{\rm{1}}\\{\rm{2\theta = }}\frac{{\rm{\pi }}}{{\rm{4}}}\\{\rm{\theta = }}\frac{{\rm{\pi }}}{{\rm{8}}}\end{aligned}\)

The sought-after territory is divided into eight divisions. Each quadrant has two such portions, each of which resembles an elongated and "pointed" ellipse.

Because each portion has the same area, we may find the area of this region by multiplying the area of one section by eight. To calculate the area of a single section, take the area of the polar region from the blue curve from 0 to\({\rm{\pi /8}}\) and multiply it by the area of the polar region from the purple curve from to. However, the area of the polar region from the purple curve\({\rm{\pi /8}}\) to\({\rm{\pi /4}}\) is identical to the area of the polar region from the blue curve\({\rm{\pi /8}}\) to \({\rm{\pi /4}}\).

Step 2: Resulting.

Polar region

\(\begin{aligned}{c}{\rm{A = }}\frac{{\rm{1}}}{{\rm{2}}}\int_{\rm{0}}^{{\rm{\pi /8}}} {{{{\rm{(sin2\theta )}}}^{\rm{2}}}} {\rm{d\theta + }}\frac{{\rm{1}}}{{\rm{2}}}\int_{{\rm{\pi /8}}}^{{\rm{\pi /4}}} {{{{\rm{(cos2\theta )}}}^{\rm{2}}}} {\rm{d\theta }}\\{\rm{ = 2 \times }}\frac{{\rm{1}}}{{\rm{2}}}\int_{\rm{0}}^{{\rm{\pi /8}}} {{{{\rm{(sin2\theta )}}}^{\rm{2}}}} {\rm{d\theta }}\\{\rm{ = }}\int_{\rm{0}}^{{\rm{\pi /8}}} {{\rm{si}}{{\rm{n}}^{\rm{2}}}} {\rm{2\theta d\theta }}\\{\rm{ = }}\int_{\rm{0}}^{{\rm{\pi /8}}} {\frac{{\rm{1}}}{{\rm{2}}}} {\rm{(1 - cos4\theta )d\theta }}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {{\rm{\theta - }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{sin4\theta }}} \right)_{\rm{0}}^{{\rm{\pi /8}}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {\frac{{\rm{\pi }}}{{\rm{8}}}{\rm{ - }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{(1)}}} \right)\end{aligned}\)

\({\rm{ = }}\frac{{\rm{\pi }}}{{{\rm{16}}}}{\rm{ - }}\frac{{\rm{1}}}{{\rm{8}}}\)

Therefore the resulting region of the area is

\(\begin{aligned}{c}{\rm{A = 8}}\left( {\frac{{\rm{\pi }}}{{{\rm{16}}}}{\rm{ - }}\frac{{\rm{1}}}{{\rm{8}}}} \right)\\{\rm{ = }}\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ - 1}}\end{aligned}\)

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Essential Calculus: Early Transcendentals

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Short Answer

Find the area of the region that lies inside both curves.
\({\rm{r = sin2\theta ,r = cos2\theta }}\)

Step by Step Solution

Step 1: Polar region.

Step 2: Resulting.

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Find the area of the region that lies inside both curves.\({\rm{r = sin2\theta ,r = cos2\theta }}\)

Step by Step Solution

Step 1: Polar region.

Step 2: Resulting.

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Find the area of the region that lies inside both curves.
\({\rm{r = sin2\theta ,r = cos2\theta }}\)