Community Answer

【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/2a)Given polar curves are\( \mathrm{{r}={1}} \)\( \mathrm{{r}^{{2}}={2}{\sin{{\left({2}\theta\right)}}}} \)To find the points of intersectionsubstitute one equation with another equationthen we get \( \mathrm{{1}={2}{\sin{{\left({2}\theta\right)}}}} \)\( \mathrm{{\sin{{\left({2}\theta\right)}}}=\frac{{1}}{{2}}} \)\( \mathrm{{\sin{{\left({2}\theta\right)}}}={\sin{{\left({2}{n}\pi+{30}^{{o}}\right)}}}} \)where \( \mathrm{{n}={0},{1},{2},{3},{4}\ldots.} \)then \( \mathrm{{2}\theta={2}{n}\pi+{30}^{{o}}} \)\( \mathrm{\theta={n}\pi+{15}^{{o}}} \) where \( \mathrm{{n}={0},{1},{2},{3},{4},\ldots.} \)Explanation:Please refer to solution in this step.Step2/2b)To find the area of the region that lies inside \( \mathrm{{r}^{{2}}={2}{\sin{{\left({2}\theta\right)}}}} \)but outside \( \mathrm{{r}={1}} \)then in 1st quadrantthe curves meet at pi/8 thenA_between\( \ ... See the full answer