Question

Find the area of the region enclosed by one loop of the curve.

\({\rm{r = 1 + 2sin\theta }}\)

Short answer

Within the inner loop, there is an area.\({\rm{A = \pi - }}\frac{{\rm{3}}}{{\rm{2}}}\sqrt {\rm{3}} .\)

Step-by-step solution

The provided formula.

\({\rm{r = 1 + 2sin\theta }}\)

The formula for computing area is as follows

\({\rm{A}}{{\rm{ = }}_{\rm{a}}}\frac{{\rm{1}}}{{\rm{2}}}{{\rm{r}}^{\rm{2}}}{\rm{d\theta }}\)

Step 2:  To find two different \({\rm{\theta }}\)for which \({\rm{r = 0,}}\)you must first find two different \({\rm{\theta }}{\rm{.}}\)

\(\begin{aligned}{c}{\rm{r = 0}}\\{\rm{1 + 2sin\theta = 0}}\\{\rm{sin\theta = - }}\frac{{\rm{1}}}{{\rm{2}}}\\{\rm{\theta = }}\frac{{{\rm{7\pi }}}}{{\rm{6}}}\\{\rm{\theta = }}\frac{{{\rm{11\pi }}}}{{\rm{6}}}\end{aligned}\)

The integral's inner loop limitations are\({\rm{\theta = }}\frac{{{\rm{7\pi }}}}{{\rm{6}}},{\rm{\theta = }}\frac{{{\rm{11\pi }}}}{{\rm{6}}}\)

One loop's area is. 

\(\begin{aligned}{c}{\rm{A = }}_{\frac{{{\rm{7\pi }}}}{{\rm{6}}}}^{\frac{{{\rm{11\pi }}}}{{\rm{6}}}}\frac{{\rm{1}}}{{\rm{2}}}{{\rm{r}}^{\rm{2}}}{\rm{d\theta }}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\;\;\;_{\frac{{{\rm{7\pi }}}}{{\rm{6}}}}^{\frac{{{\rm{11\pi }}}}{{\rm{6}}}}\left( {{\rm{1 + 4sin\theta + 4si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}} \right){\rm{d\theta }}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\;\;\;_{\frac{{{\rm{7\pi }}}}{{\rm{6}}}}^{\frac{{{\rm{11\pi }}}}{{\rm{6}}}}{\rm{d\theta + }}\frac{{\rm{4}}}{{\rm{2}}}\int_{\frac{{{\rm{7\pi }}}}{{\rm{6}}}}^{\frac{{{\rm{11\pi }}}}{{\rm{6}}}} {{\rm{sin}}} {\rm{\theta d\theta + }}\frac{{\rm{4}}}{{\rm{2}}}\int_{\frac{{{\rm{7\pi }}}}{{\rm{6}}}}^{\frac{{{\rm{11\pi }}}}{{\rm{6}}}} {{\rm{si}}{{\rm{n}}^{\rm{2}}}} {\rm{\theta d\theta }}\end{aligned}\)

\(\begin{aligned}{c}{\rm{ = }}\left. {\frac{{\rm{1}}}{{\rm{2}}}{\rm{\theta }}} \right|_{\frac{{{\rm{7\pi }}}}{{\rm{6}}}}^{\frac{{{\rm{11\pi }}}}{{\rm{6}}}}{\rm{ - }}\left. {{\rm{2cos\theta }}} \right|_{\frac{{{\rm{7\pi }}}}{{\rm{6}}}}^{\frac{{{\rm{11\pi }}}}{{\rm{6}}}}{\rm{ + 2}}\sum\limits_{\frac{{{\rm{7\pi }}}}{{\rm{6}}}}^{\frac{{{\rm{11\pi }}}}{{\rm{6}}}} {\frac{{{\rm{1 - cos2\theta }}}}{{\rm{2}}}} {\rm{d\theta }}\\{\rm{ = }}\left. {\frac{{\rm{1}}}{{\rm{4}}}{\rm{\theta }}} \right|_{\rm{0}}^{\frac{{\rm{\pi }}}{{\rm{2}}}}{\rm{ - }}\left. {\frac{{\rm{1}}}{{\rm{4}}}{\rm{sin4\theta }}} \right|_{\rm{0}}^{\mid {\rm{frac\pi 2}}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times }}\frac{{{\rm{4\pi }}}}{{\rm{6}}}{\rm{ - 2}}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}{\rm{ - - }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}{\rm{ + }}\left. {\rm{\theta }} \right|_{\frac{{{\rm{7\pi }}}}{{\rm{6}}}}^{\frac{{{\rm{11\pi }}}}{{\rm{6}}}}{\rm{ - }}\left. {\frac{{\rm{1}}}{{\rm{2}}}{\rm{sin2\theta }}} \right|_{\frac{{{\rm{7\pi }}}}{{\rm{6}}}}^{\frac{{{\rm{11\pi }}}}{{\rm{6}}}}\\{\rm{ = }}\frac{{\rm{\pi }}}{{\rm{3}}}{\rm{ - 2}}\sqrt {\rm{3}} {\rm{ + }}\frac{{{\rm{2\pi }}}}{{\rm{3}}}{\rm{ - }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{sin}}\frac{{{\rm{11\pi }}}}{{\rm{3}}}{\rm{ - sin}}\frac{{{\rm{7\pi }}}}{{\rm{3}}}\\{\rm{ = \pi - 2}}\sqrt {\rm{3}} {\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}\sqrt {\rm{3}} \\{\rm{ = \pi - }}\frac{{\rm{3}}}{{\rm{2}}}\sqrt {\rm{3}} \end{aligned}\)

Inside the inner circle, there is a \({\rm{A = \pi - }}\frac{{\rm{3}}}{{\rm{2}}}\sqrt {\rm{3}} .\)