Question

Sketch the curve and find the area that it encloses.

\({\rm{r = 4 + 3sin\theta }}\)

Short answer

The area of the curve is \({\rm{A = }}\frac{{{\rm{82\pi }}}}{{\rm{4}}}.\)

Step-by-step solution

the curve for \({\rm{r = 4 + 3sin\theta }}\) is,

Area formula is \({\rm{A = }}\frac{{\rm{1}}}{{\rm{2}}}\int_{\rm{a}}^{\rm{b}} {{{\rm{r}}^{\rm{2}}}} {\rm{d\theta ,}}\) 

\(\begin{aligned}{c}{\rm{A = }}\frac{{\rm{1}}}{{\rm{2}}}\int_{\rm{a}}^{\rm{b}} {{{\rm{r}}^{\rm{2}}}} {\rm{d\theta }}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\int_{{\rm{\pi /2}}}^{{\rm{3\pi /2}}} {\rm{2}} {{\rm{(4 + 3sin\theta )}}^{\rm{2}}}{\rm{d\theta }}\\{\rm{ = }}\int_{{\rm{\pi /2}}}^{{\rm{3\pi /2}}} {\left( {{\rm{16 + 21sin\theta + 9si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}} \right)} {\rm{d\theta }}\end{aligned}\)

\(\begin{aligned}{c}{\rm{ = }}\int_{{\rm{\pi /2}}}^{{\rm{3\pi /2}}} {\left( {{\rm{16 + 24sin\theta + }}\frac{{\rm{9}}}{{\rm{2}}}{\rm{(1 - cos2\theta )}}} \right)} {\rm{d\theta }}\\\left. {{\rm{ = }}\int_{{\rm{\pi /2}}}^{{\rm{3\pi /2}}} {\left( {\frac{{{\rm{41}}}}{{\rm{2}}}{\rm{ + 24sin\theta - }}\frac{{\rm{9}}}{{\rm{2}}}{\rm{cos2\theta }}} \right)} } \right){\rm{d\theta }}\end{aligned}\)

Integrating and getting,

Know that,

\({\rm{cos2\theta = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{sin2\theta }}\)

\(\begin{aligned}{c}{\rm{ = }}\left( {\frac{{{\rm{41}}}}{{\rm{2}}}{\rm{\theta - 24cos\theta - }}\frac{{\rm{9}}}{{\rm{4}}}{\rm{sin2\theta }}} \right)_{{\rm{\pi /2}}}^{{\rm{3\pi /2}}}\\{\rm{ = }}\frac{{{\rm{123\pi }}}}{{\rm{4}}}{\rm{ - 0 - 0 - }}\left( {\frac{{{\rm{41\pi }}}}{{\rm{4}}}{\rm{ - 0 - 0}}} \right)\\{\rm{A = }}\frac{{{\rm{82\pi }}}}{{\rm{4}}}.\end{aligned}\)

Therefore, the area of the curve is \({\rm{A = }}\frac{{{\rm{82\pi }}}}{{\rm{4}}}\)