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Expert-verified Found in: Page 528 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # 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$$\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ - 1}}$$.

See the step by step solution

## 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} ### Want to see more solutions like these? 