MCQ
The area in the first quadrant between ${x^2} + {y^2} = {\pi ^2}$ and $y = \sin x$ is
- ✓$\frac{{({\pi ^3} - 8)}}{4}$
- B$\frac{{{\pi ^3}}}{4}$
- C$\frac{{({\pi ^3} - 16)}}{4}$
- D$\frac{{({\pi ^3} - 8)}}{2}$
$i.e.,$ $\frac{{{\pi ^3}}}{4}$.
Also area bounded by curve $y = \sin x$ and $x -$ axis is $2$ sq. unit.
Hence required area is $\frac{{{\pi ^3}}}{4} - 2 = \frac{{{\pi ^3} - 8}}{4}$.
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