• # question_answer The area in the first quadrant between ${{x}^{2}}+{{y}^{2}}={{\pi }^{2}}$ and $y=\sin x$ is                                      [MP PET 1997] A)            $\frac{({{\pi }^{3}}-8)}{4}$   B)            $\frac{{{\pi }^{3}}}{4}$ C)            $\frac{({{\pi }^{3}}-16)}{4}$ D)            $\frac{({{\pi }^{3}}-8)}{2}$

Area of the circle in first quadrant is $\frac{\pi ({{\pi }^{2}})}{4}$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 $\frac{{{\pi }^{3}}}{4}-2=\frac{{{\pi }^{3}}-8}{4}$.