Step 1

Consider the following curves:

\(\displaystyle{r}^{{{2}}}={50}{\sin{{\left({2}\theta\right)}}},\ {r}={5}\)

The objective is to find the area of the region that lies inside both the curves.

Sketch the region as follows:

Find the intersection points as follows:

From the curves,

\(\displaystyle{50}{\sin{{2}}}\theta={5}^{{{2}}}\)

\(\displaystyle{\sin{{2}}}\theta={\frac{{{25}}}{{{50}}}}\)

\(\displaystyle{\sin{{2}}}\theta={\frac{{{1}}}{{{2}}}}\)

\(\displaystyle{2}\theta={\frac{{\pi}}{{{6}}}},\ {\frac{{{5}\pi}}{{{6}}}},\ {\frac{{{13}\pi}}{{{6}}}},\ {\frac{{{17}\pi}}{{{6}}}}\)

\(\displaystyle\theta={\frac{{\pi}}{{{12}}}},\ {\frac{{{5}\pi}}{{{12}}}},\ {\frac{{{13}\pi}}{{{12}}}},\ {\frac{{{17}\pi}}{{{12}}}}\)

The formula for the area A of the polar region \(\displaystyle{\mathbb{{{R}}}}\) is,

\(\displaystyle{A}={\int_{{\alpha}}^{{{b}}}}{\frac{{{1}}}{{{2}}}}{\left[{f{{\left(\theta\right]}}}^{{{2}}}{d}\theta\right.}\)

Step 2

The desired area can be found by adding the area inside the cardioid between \(\displaystyle\theta={0}\) and \(\displaystyle{\frac{{\pi}}{{{4}}}}\) from the area inside the circle from 0 to \(\displaystyle{\frac{{\pi}}{{{4}}}}\)

Since the region is symmetric about \(\displaystyle\theta={\frac{{\pi}}{{{4}}}}\), you can write the area is,

\(\displaystyle{A}={2}{\left\lbrace{2}\times{\frac{{{1}}}{{{2}}}}{\int_{{{0}}}^{{{\frac{{\pi}}{{{12}}}}}}}{\left(\sqrt{{{50}{\sin{{\left({2}\theta\right)}}}}}\right)}^{{{2}}}{d}\theta+{2}\times{\frac{{{1}}}{{{2}}}}{\int_{{{\frac{{\pi}}{{{12}}}}}}^{{{\frac{{\pi}}{{{4}}}}}}}{5}^{{{2}}}{d}\theta\right\rbrace}\)

\(\displaystyle={2}{\left\lbrace{\int_{{{0}}}^{{{\frac{{\pi}}{{{12}}}}}}}{50}{\sin{{\left({2}\theta\right)}}}{d}\theta+{\int_{{{\frac{{\pi}}{{{12}}}}}}^{{{\frac{{\pi}}{{{4}}}}}}}{25}{d}\theta\right\rbrace}\)

\(\displaystyle={2}{\left\lbrace{{\left[-{\frac{{{50}}}{{{2}}}}{\cos{{2}}}\theta\right]}_{{{0}}}^{{{\frac{{\pi}}{{{12}}}}}}}+{25}{{\left[\theta\right]}_{{{\frac{{\pi}}{{{12}}}}}}^{{{\frac{{\pi}}{{{4}}}}}}}\right\rbrace}\)

\(\displaystyle={2}{\left[-{25}{\left({\cos{{\frac{{\pi}}{{{6}}}}}}-{\cos{{0}}}\right)}+{25}{\left({\frac{{\pi}}{{{4}}}}-{\frac{{\pi}}{{{12}}}}\right)}\right]}\)

\(\displaystyle={2}{\left[-{25}{\left({\frac{{\sqrt{{{3}}}}}{{{2}}}}-{1}\right)}+{25}{\left({\frac{{{2}\pi}}{{{12}}}}\right)}\right]}\)

\(\displaystyle={2}{\left(-{\frac{{{25}\sqrt{{{3}}}}}{{{2}}}}+{25}+{\frac{{{25}\pi}}{{{6}}}}\right)}\)

\(\displaystyle=-{25}\sqrt{{{3}}}+{50}+{\frac{{{25}\pi}}{{{3}}}}\)

Thus, the area of the region that lies inside both the curves is

\(\displaystyle-{25}\sqrt{{{3}}}+{50}+{\frac{{{25}\pi}}{{{3}}}}\)

Consider the following curves:

\(\displaystyle{r}^{{{2}}}={50}{\sin{{\left({2}\theta\right)}}},\ {r}={5}\)

The objective is to find the area of the region that lies inside both the curves.

Sketch the region as follows:

Find the intersection points as follows:

From the curves,

\(\displaystyle{50}{\sin{{2}}}\theta={5}^{{{2}}}\)

\(\displaystyle{\sin{{2}}}\theta={\frac{{{25}}}{{{50}}}}\)

\(\displaystyle{\sin{{2}}}\theta={\frac{{{1}}}{{{2}}}}\)

\(\displaystyle{2}\theta={\frac{{\pi}}{{{6}}}},\ {\frac{{{5}\pi}}{{{6}}}},\ {\frac{{{13}\pi}}{{{6}}}},\ {\frac{{{17}\pi}}{{{6}}}}\)

\(\displaystyle\theta={\frac{{\pi}}{{{12}}}},\ {\frac{{{5}\pi}}{{{12}}}},\ {\frac{{{13}\pi}}{{{12}}}},\ {\frac{{{17}\pi}}{{{12}}}}\)

The formula for the area A of the polar region \(\displaystyle{\mathbb{{{R}}}}\) is,

\(\displaystyle{A}={\int_{{\alpha}}^{{{b}}}}{\frac{{{1}}}{{{2}}}}{\left[{f{{\left(\theta\right]}}}^{{{2}}}{d}\theta\right.}\)

Step 2

The desired area can be found by adding the area inside the cardioid between \(\displaystyle\theta={0}\) and \(\displaystyle{\frac{{\pi}}{{{4}}}}\) from the area inside the circle from 0 to \(\displaystyle{\frac{{\pi}}{{{4}}}}\)

Since the region is symmetric about \(\displaystyle\theta={\frac{{\pi}}{{{4}}}}\), you can write the area is,

\(\displaystyle{A}={2}{\left\lbrace{2}\times{\frac{{{1}}}{{{2}}}}{\int_{{{0}}}^{{{\frac{{\pi}}{{{12}}}}}}}{\left(\sqrt{{{50}{\sin{{\left({2}\theta\right)}}}}}\right)}^{{{2}}}{d}\theta+{2}\times{\frac{{{1}}}{{{2}}}}{\int_{{{\frac{{\pi}}{{{12}}}}}}^{{{\frac{{\pi}}{{{4}}}}}}}{5}^{{{2}}}{d}\theta\right\rbrace}\)

\(\displaystyle={2}{\left\lbrace{\int_{{{0}}}^{{{\frac{{\pi}}{{{12}}}}}}}{50}{\sin{{\left({2}\theta\right)}}}{d}\theta+{\int_{{{\frac{{\pi}}{{{12}}}}}}^{{{\frac{{\pi}}{{{4}}}}}}}{25}{d}\theta\right\rbrace}\)

\(\displaystyle={2}{\left\lbrace{{\left[-{\frac{{{50}}}{{{2}}}}{\cos{{2}}}\theta\right]}_{{{0}}}^{{{\frac{{\pi}}{{{12}}}}}}}+{25}{{\left[\theta\right]}_{{{\frac{{\pi}}{{{12}}}}}}^{{{\frac{{\pi}}{{{4}}}}}}}\right\rbrace}\)

\(\displaystyle={2}{\left[-{25}{\left({\cos{{\frac{{\pi}}{{{6}}}}}}-{\cos{{0}}}\right)}+{25}{\left({\frac{{\pi}}{{{4}}}}-{\frac{{\pi}}{{{12}}}}\right)}\right]}\)

\(\displaystyle={2}{\left[-{25}{\left({\frac{{\sqrt{{{3}}}}}{{{2}}}}-{1}\right)}+{25}{\left({\frac{{{2}\pi}}{{{12}}}}\right)}\right]}\)

\(\displaystyle={2}{\left(-{\frac{{{25}\sqrt{{{3}}}}}{{{2}}}}+{25}+{\frac{{{25}\pi}}{{{6}}}}\right)}\)

\(\displaystyle=-{25}\sqrt{{{3}}}+{50}+{\frac{{{25}\pi}}{{{3}}}}\)

Thus, the area of the region that lies inside both the curves is

\(\displaystyle-{25}\sqrt{{{3}}}+{50}+{\frac{{{25}\pi}}{{{3}}}}\)