Question

Findthe area of the shaded region.

Short answer

The area of the shaded region is\(\frac{{{\rm{41\pi }}}}{{\rm{4}}}\).

Step-by-step solution

Find the area of the shaded region.

Rays rotate from\({\rm{\theta = - \pi /2to\theta = \pi /2}}\), covering the shaded region.

Fill up the blanks with the formula\({\rm{\# 4}}\) from this section.

\(\begin{aligned}{c}{\rm{A = }}\int_{{\rm{ - \pi /2}}}^{{\rm{\pi /2}}} {\frac{{\rm{1}}}{{\rm{2}}}} {{\rm{r}}^{\rm{2}}}{\rm{d\theta }}\\{\rm{ = }}\int_{{\rm{ - \pi /2}}}^{{\rm{\pi /2}}} {\frac{{\rm{1}}}{{\rm{2}}}} {{\rm{(4 + 3sin\theta )}}^{\rm{2}}}{\rm{d\theta }}\\{\rm{ = }}\int_{{\rm{ - \pi /2}}}^{{\rm{\pi /2}}} {\frac{{\rm{1}}}{{\rm{2}}}} \left( {{\rm{16 + 24sin\theta + 9si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}} \right){\rm{d\theta }}\\{\rm{ = }}\int_{{\rm{ - \pi /2}}}^{{\rm{\pi /2}}} {\rm{8}} {\rm{ + 12sin\theta + 4}}{\rm{.5}} \cdot {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta d\theta }}\\{\rm{Recall that:}}\left. {{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta = }}\frac{{{\rm{1 - cos2\theta }}}}{{\rm{2}}}} \right)\\{\rm{ = }}\int_{{\rm{ - \pi /2}}}^{{\rm{\pi /2}}} {\rm{8}} {\rm{ + 12sin\theta + 4}}{\rm{.5}} \cdot \left( {\frac{{{\rm{1 - cos2\theta }}}}{{\rm{2}}}} \right){\rm{d\theta }}\\{\rm{ = }}\int_{{\rm{ - \pi /2}}}^{{\rm{\pi /2}}} {\rm{8}} {\rm{ + 12sin\theta + 2}}{\rm{.25}} \cdot {\rm{(1 - cos2\theta )d\theta }}\\{\rm{ = }}\int_{{\rm{ - \pi /2}}}^{{\rm{\pi /2}}} {\rm{1}} {\rm{0}}{\rm{.25 + 12sin\theta - 2}}{\rm{.25}} \cdot {\rm{cos2\theta d\theta }}\\{\rm{ = }}\left( {{\rm{10}}{\rm{.25}} \cdot {\rm{\theta + 12( - cos\theta ) - 2}}{\rm{.25}} \cdot \frac{{\rm{1}}}{{\rm{2}}}{\rm{sin2\theta }}} \right)_{{\rm{ - \pi /2}}}^{{\rm{\pi /2}}}\\{\rm{ = (10}}{\rm{.25}} \cdot {\rm{\theta - 12cos\theta - 1}}{\rm{.125 \times sin2\theta )}}_{{\rm{ - \pi /2}}}^{{\rm{\pi /2}}}\\{\rm{ = 10}}{\rm{.25}} \cdot {\rm{((\pi /2) - ( - \pi /2))}}\\{\rm{ = 10}}{\rm{.25}} \cdot {\rm{\pi }}\\{\rm{ = }}\frac{{{\rm{41\pi }}}}{{\rm{4}}}\end{aligned}\)

Result.

\({\rm{A = }}\frac{{{\rm{41\pi }}}}{{\rm{4}}}\)

Therefore, the area of the shaded region is\(\frac{{{\rm{41\pi }}}}{{\rm{4}}}\).