Question

Graph the curve and find the area that it encloses.

\({\rm{r = 3 - 2cos4\theta }}\)

Short answer

The radius of the curve and the territory it encompasses\({\rm{A = 11\pi }}{\rm{.}}\;\;\)

Step-by-step solution

The integration interval must be found.

The Square being enlarged.

\(\begin{aligned}{c}{\rm{A = }}\frac{{\rm{1}}}{{\rm{2}}}\;\;\;{{\rm{r}}^{\rm{b}}}{\rm{d\theta }}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}_{\rm{0}}^{{\rm{2\pi }}}{{\rm{(3 - 2cos4\theta )}}^{\rm{2}}}{\rm{d\theta }}\\\left. {{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}_{\rm{0}}^{{\rm{2\pi }}}{\rm{9 - 12cos4\theta + 4co}}{{\rm{s}}^{\rm{2}}}{\rm{4\theta }}} \right)\;\;\;{\rm{d\theta }}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\int_{\rm{0}}^{{\rm{2\pi }}} {\;\;\;} {\rm{9 - 12cos4\theta + 4}}\;\;\;\frac{{\rm{1}}}{{\rm{2}}}{\rm{(1 + cos8\theta )}}\;\;\;{\rm{d\theta }}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}_{\rm{0}}^{{\rm{2\pi }}}{\rm{(9 - 12cos4\theta + 2 + 2cos8\theta )d\theta }}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}_{\rm{0}}^{{\rm{2\pi }}}{\rm{(11 - 12cos4\theta + 2cos8\theta )d\theta }}\\\left. {{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\;\;\;{\rm{11\theta - 3sin4\theta + }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{sin8\theta }}} \right)_{\rm{0}}^{{\rm{2\pi }}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{(22\pi - 0 + 0 - (0 - 0 + 0))}}_{\rm{0}}^{{\rm{2\pi }}}\\{\rm{ = 11\pi }}\;\;\\\;{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{\theta = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{(1 + cos2\theta )}}\end{aligned}\)

The curve and the space it encompasses\({\rm{A = 11\pi }}{\rm{.}}\;\;\)