question_answer

The area bounded by the curves \[{{x}^{2}}+{{y}^{2}}=25,\] \[4y=|4-{{x}^{2}}|\] and \[x=0\] above the x- axis is

A)  \[24\,\,{{\sin }^{-1}}\left( \frac{4}{5} \right)\]

B)  \[25\,\,{{\sin }^{-1}}\left( \frac{4}{5} \right)\]

C)  \[2+\frac{25}{2}{{\sin }^{-1}}\left( \frac{4}{5} \right)\]

D)  None of these

Correct Answer: C

Solution :

Required area \[=\int_{0}^{4}{\sqrt{(25-{{x}^{2}})}}dx-\int_{0}^{2}{\frac{4-{{x}^{2}}}{4}}dx-\int_{2}^{4}{\frac{{{x}^{2}}-4}{4}}dx\] \[=\left[ \frac{x}{2}\sqrt{25-{{x}^{2}}}+\frac{25}{2}\,{{\sin }^{-1}}\frac{x}{5} \right]_{0}^{4}\] \[-\frac{1}{4}\left[ 4x-\frac{{{x}^{3}}}{3} \right]_{0}^{2}-\frac{1}{4}\left[ \frac{{{x}^{3}}}{3}-4x \right]_{2}^{4}\] \[=\left( 2\times 3+\frac{25}{2}\,{{\sin }^{-1}}\left( \frac{4}{5} \right) \right)-\frac{1}{4}\left( 8-\frac{8}{3} \right)\] \[-\frac{1}{4}\left( \frac{64}{3}-16 \right)+\frac{1}{4}\left( \frac{8}{3}-8 \right)\] \[=2+\frac{25}{2}si{{n}^{-1}}\left( \frac{4}{5} \right)\]