question_answer

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

A) \[2+\frac{25}{2}{{\sin }^{-1}}\frac{4}{5}\]

B) \[2+\frac{25}{4}{{\sin }^{-1}}\frac{4}{5}\]

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

D) None of these

Correct Answer: A

Solution :

[a] The required area \[=\int_{0}^{4}{\sqrt{25-{{x}^{2}}}dx-\int_{0}^{2}{\frac{4-{{x}^{2}}}{4}dx-\int_{2}^{4}{\frac{{{x}^{2}}-4}{4}dx}}}\] \[=\left[ \frac{x}{2}\sqrt{25-{{x}^{2}}}+\frac{25}{2}{{\sin }^{-1}}\frac{x}{5} \right]_{0}^{4}\] \[=-\frac{1}{4}\left[ 4x-\frac{{{x}^{3}}}{3} \right]_{0}^{2}-\frac{1}{4}\left[ \frac{{{x}^{3}}}{3}-4x \right]_{2}^{4}=2+\frac{25}{2}{{\sin }^{-1}}\frac{4}{5}.\]