Question

Find the area of the region that lies inside both curves.

\({{\rm{r}}^{\rm{2}}}{\rm{ = 2sin2\theta ,}}\;\;\;{\rm{r = 1}}\).

Short answer

The resulting area of the zone between the two curves\({\rm{2 - }}\sqrt {\rm{3}} {\rm{ + }}\frac{{\rm{\pi }}}{{\rm{3}}}\).

Step-by-step solution

Quadrant cross.

The number\({\rm{\theta }}\) at which the two graphs in Quadrant one cross

\(\begin{aligned}{l}{\rm{2sin2\theta = }}{{\rm{1}}^{\rm{2}}}\\{\rm{2sin2\theta = 1}}\end{aligned}\)

Divide each half by two

\(\begin{aligned}{l}{\rm{sin2\theta = }}\frac{{\rm{1}}}{{\rm{2}}}\\{\rm{2\theta = }}\frac{{\rm{\pi }}}{{\rm{6}}}{\rm{,}}\frac{{{\rm{5\pi }}}}{{\rm{6}}}\\{\rm{\theta = }}\frac{{\rm{\pi }}}{{{\rm{12}}}}{\rm{,}}\frac{{{\rm{5\pi }}}}{{{\rm{12}}}}\end{aligned}\)

Resulting.

Area\({\rm{ = }}\)Area inside the ribbon\({\rm{ - }}\)Area outside the circle but inside the ribbon

\(\begin{aligned}{l}\int_{\rm{0}}^{{\rm{\pi /2}}} {\rm{2}} {\rm{sin2\theta d\theta - }}\int_{{\rm{\pi /12}}}^{{\rm{5\pi /12}}} {\rm{2}} {\rm{sin2\theta - 1d\theta }}\\{\rm{ = ( - cos2\theta )}}_{\rm{0}}^{{\rm{\pi /2}}}{\rm{ - ( - cos2\theta - \theta )}}_{{\rm{\pi /12}}}^{{\rm{5\pi /12}}}\\{\rm{ = 2 - }}\left( {{\rm{ - cos}}\frac{{{\rm{5\pi }}}}{{\rm{6}}}{\rm{ - }}\frac{{{\rm{5\pi }}}}{{{\rm{12}}}}} \right){\rm{ + }}\left( {{\rm{ - cos}}\frac{{\rm{\pi }}}{{\rm{6}}}{\rm{ - }}\frac{{\rm{\pi }}}{{{\rm{12}}}}} \right)\\{\rm{ = 2 - }}\left( {\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}{\rm{ - }}\frac{{{\rm{5\pi }}}}{{{\rm{12}}}}} \right){\rm{ + }}\left( {{\rm{ - }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}{\rm{ - }}\frac{{\rm{\pi }}}{{{\rm{12}}}}} \right)\\{\rm{ = 2 - }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}{\rm{ + }}\frac{{{\rm{5\pi }}}}{{{\rm{12}}}}{\rm{ - }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}{\rm{ - }}\frac{{\rm{\pi }}}{{{\rm{12}}}}\\{\rm{ = 2 - }}\sqrt {\rm{3}} {\rm{ + }}\frac{{\rm{\pi }}}{{\rm{3}}}\end{aligned}\)

Therefore, the area of the region that lies inside both curves is\({\rm{2 - }}\sqrt {\rm{3}} {\rm{ + }}\frac{{\rm{\pi }}}{{\rm{3}}}\).