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 \[x=\pi \]You need to login to perform this action.
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