Question

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

\({\rm{r = 2cos\theta - sec\theta }}\)

Short answer

To double the result, use the polar area formula.\({\rm{2 - }}\frac{{\rm{\pi }}}{{\rm{2}}}\)

Step-by-step solution

Because it is symmetrical, use the integration limits and twice the result to find the area.

\({\rm{r = 2cos\theta - sec\theta }}\)

Half of the figure is made up of \({\rm{0}} \le {\rm{\theta < }}\frac{{\rm{\pi }}}{{\rm{2}}}\)

The second part of \(\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ < \theta }} \le {\rm{\pi }}\)

The top part of the loop is \({\rm{0}} \le {\rm{\theta }} \le \frac{{\rm{\pi }}}{{\rm{4}}}\),and the bottom half is \({\rm{0}} \le {\rm{\theta }} \le \frac{{\rm{\pi }}}{{\rm{4}}}\).

Because the area is symmetrical, use those as integration limits and twice the result to find it.

Use the following formula to reduce the amount of power you use.

\(\begin{aligned}{c}{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{\theta = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{(1 + cos(2\theta ))}}\\{\rm{r = 2cos\theta - sec\theta }}\\{{\rm{r}}^{\rm{2}}}{\rm{ = 4co}}{{\rm{s}}^{\rm{2}}}{\rm{\theta - 4cos\theta sec\theta + se}}{{\rm{c}}^{\rm{2}}}{\rm{\theta }}\\{\rm{ = 4co}}{{\rm{s}}^{\rm{2}}}{\rm{\theta - 4 + se}}{{\rm{c}}^{\rm{2}}}{\rm{\theta }}\end{aligned}\)

\(\begin{aligned}{c}{\rm{ = 4}}\left( {\frac{{\rm{1}}}{{\rm{2}}}{\rm{(1 + cos(2\theta ))}}} \right){\rm{ - 4 +

se}}{{\rm{c}}^{\rm{2}}}{\rm{\theta }}\\{\rm{ = 2 + 2cos(2\theta ) - 4 + se}}{{\rm{c}}^{\rm{2}}}{\rm{\theta }}\\{\rm{ = 2cos(2\theta ) - 2 + se}}{{\rm{c}}^{\rm{2}}}{\rm{\theta }}\end{aligned}\)

To double the result, add \({\rm{2}}\) to the polar area formula.

\(\begin{aligned}{l}{\rm{A = 2 \times }}{\frac{{\rm{1}}}{{\rm{2}}}_{\rm{a}}}^{\rm{b}}{{\rm{r}}^{\rm{2}}}{\rm{d\theta }}\\{\rm{ = }}_{\rm{0}}^{{\rm{\pi /4}}}\left( {{\rm{2cos(2\theta ) - 2 + se}}{{\rm{c}}^{\rm{2}}}{\rm{\theta }}} \right){\rm{d\theta }}\\{\rm{ = (sin(2\theta ) - 2\theta + tan\theta )}}_{\rm{0}}^{{\rm{\pi /4}}}\\{\rm{ = }}\left( {{\rm{sin}}\left( {\frac{{\rm{\pi }}}{{\rm{2}}}} \right){\rm{ - }}\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ + tan}}\frac{{\rm{\pi }}}{{\rm{4}}}{\rm{ - (0)}}} \right)\\{\rm{ = 1 - }}\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ + 1}}\\{\rm{ = 2 - }}\frac{{\rm{\pi }}}{{\rm{2}}}\end{aligned}\)

Twice the result using the polar area formula\({\rm{2 - }}\frac{{\rm{\pi }}}{{\rm{2}}}\)