Question

Find the area inside the larger loop and outside the smaller loop of the limaçon\({\rm{r = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + cos\theta }}\).

Short answer

The limacon's area inside the larger loop and outside the smaller loop is\({\rm{\pi + 3}}\sqrt {\rm{3}} \).

Step-by-step solution

Doubling.

Given \({\rm{r = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + cos\theta }}\)

Finished the figures \({\rm{0}} \le {\rm{\theta < 2\pi }}\)

The interval top half of the loop is \(\left( {{\rm{0,}}\frac{{{\rm{2\pi }}}}{{\rm{3}}}} \right)\)

The interval bottom half of the loop is \(\left( {\frac{{{\rm{2\pi }}}}{{\rm{3}}}{\rm{,\pi }}} \right)\)

The areas of the big and small loops are obtained by doubling the integrals over these intervals.

Create the boundaries.

First, the integral

\(\begin{aligned}{l}{\rm{A = 2 \times }}\frac{{\rm{1}}}{{\rm{2}}}\int_{\rm{a}}^{\rm{b}} {{{\rm{r}}^{\rm{2}}}} {\rm{d\theta }}\\{\rm{ = }}\int {{{\left( {\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + cos\theta }}} \right)}^{\rm{2}}}} {\rm{d\theta }}\\{\rm{ = }}\int {\left( {\frac{{\rm{1}}}{{\rm{4}}}{\rm{ + cos\theta + co}}{{\rm{s}}^{\rm{2}}}{\rm{\theta }}} \right)} {\rm{d\theta }}\\{\rm{ = }}\int {\left( {\frac{{\rm{1}}}{{\rm{4}}}{\rm{ + cos\theta + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{(1 + cos2\theta )}}} \right)} {\rm{d\theta }}\\{\rm{ = }}\int {\left( {\frac{{\rm{3}}}{{\rm{4}}}{\rm{ + cos\theta + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{cos2\theta }}} \right)} {\rm{d\theta }}\\{\rm{ = }}\frac{{\rm{3}}}{{\rm{4}}}{\rm{\theta + sin\theta + }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{sin2\theta }}\\\end{aligned}\)

Subtract the little loop from the huge loop to create the boundaries.

\(\begin{aligned}{l}{\rm{A = }}\left( {\frac{{\rm{3}}}{{\rm{4}}}{\rm{\theta + sin\theta + }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{sin2\theta }}} \right)_{\rm{0}}^{{\rm{2\pi /3}}}{\rm{ - }}\left( {\frac{{\rm{3}}}{{\rm{4}}}{\rm{\theta + sin\theta + }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{sin2\theta }}} \right)_{{\rm{2\pi /3}}}^{\rm{\pi }}\\{\rm{ = }}\left( {\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ + }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}{\rm{ - }}\frac{{\sqrt {\rm{3}} }}{{\rm{8}}}{\rm{ - 0}}} \right){\rm{ - }}\left( {\frac{{{\rm{3\pi }}}}{{\rm{4}}}{\rm{ + 0 + 0 - }}\left( {\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ + }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}{\rm{ - }}\frac{{\sqrt {\rm{3}} }}{{\rm{8}}}} \right)} \right)\\{\rm{ = }}\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ + }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}{\rm{ - }}\frac{{\sqrt {\rm{3}} }}{{\rm{8}}}{\rm{ - }}\frac{{{\rm{3\pi }}}}{{\rm{4}}}{\rm{ + }}\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ + }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}{\rm{ - }}\frac{{\sqrt {\rm{3}} }}{{\rm{8}}}\\{\rm{ = }}\frac{{{\rm{2\pi + 2}}\sqrt {\rm{3}} }}{{\rm{2}}}{\rm{ - }}\frac{{{\rm{2}}\sqrt {\rm{3}} }}{{\rm{8}}}{\rm{ - }}\frac{{{\rm{3\pi }}}}{{\rm{4}}}\\{\rm{ = }}\frac{{{\rm{4\pi + 4}}\sqrt {\rm{3}} {\rm{ - }}\sqrt {\rm{3}} {\rm{ - 3\pi }}}}{{\rm{4}}}\\{\rm{ = }}\frac{{{\rm{\pi + 3}}\sqrt {\rm{3}} }}{{\rm{4}}}\\\end{aligned}\)

The area inside the larger loop and outside the smaller loop of the limaçon is\({\rm{\pi + 3}}\sqrt {\rm{3}} \).