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 \[=\int_{0}^{\pi }{({{y}_{2}}-{{y}_{1}})}\,dx\] \[=\int_{0}^{\pi }{(\sqrt{{{\pi }^{2}}-{{x}^{2}}}-\sin x)\,dx}\] \[=\left[ \frac{x}{2}\sqrt{{{\pi }^{2}}-{{x}^{2}}}+\frac{{{\pi }^{2}}}{2}{{\sin }^{-1}}\left( \frac{x}{\pi } \right)+\cos x \right]_{0}^{\pi }\] \[=0+\frac{{{\pi }^{3}}}{4}-1-(0+1)\] \[=\frac{{{\pi }^{3}}-8}{4}\,\text{sq}\,\text{unit}\]You need to login to perform this action.
You will be redirected in
3 sec