question_answer

The area in the first quadrant between \[{{x}^{2}}+{{y}^{2}}={{\pi }^{2}}\]and \[y=\sin x\]is:

A) \[\frac{{{\pi }^{3}}}{4}sq\]unit           

B)         \[\frac{{{\pi }^{3}}-16}{4}\text{sq}\,\text{unit}\]

C)         \[\frac{{{\pi }^{3}}-8}{2}\text{sq}\,\text{unit}\]

D)         \[\frac{{{\pi }^{3}}-8}{4}sq\,unit\]

Correct Answer: D

Solution :

The point of intersection between the curves \[{{x}^{2}}+{{y}^{2}}={{\pi }^{2}}\]and \[y=\sin x\]in the first quadrant is at \[x=\pi \]                 \[\therefore \]Required area \[=\int_{0}^{\pi }{({{y}_{2}}-{{y}_{1}})}\,dx\]                                 \[=\int_{0}^{\pi }{(\sqrt{{{\pi }^{2}}-{{x}^{2}}}-\sin x)\,dx}\]                                 \[=\left[ \frac{x}{2}\sqrt{{{\pi }^{2}}-{{x}^{2}}}+\frac{{{\pi }^{2}}}{2}{{\sin }^{-1}}\left( \frac{x}{\pi } \right)+\cos x \right]_{0}^{\pi }\]                 \[=0+\frac{{{\pi }^{3}}}{4}-1-(0+1)\]                 \[=\frac{{{\pi }^{3}}-8}{4}\,\text{sq}\,\text{unit}\]