Mathematics | VITEEE 2014 | Discussion Page

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

$\frac{\pi^{3}-8}{4}$

$\frac{\pi^{3}}{4}$

$\frac{\pi^{3}-16}{4}$

$\frac{\pi^{3}-8}{2}$

Option A

$x^{2}+y^{2}=\pi^{2}$ is a circle of radius $\pi$ and centre at the origin

Required of area= Area of circle (1 st quadrant) - $\int_{0}^{\pi} \sin x dx$

= $\frac{\pi \pi^{2}}{4}-[-\cos x]_{0}^{\pi}=\frac{\pi^{3}}{4}+(\cos \pi-\cos 0)$

= $\frac{\pi^{3}}{4}+(-1-1)=\frac{\pi^{3}}{4}-2=\frac{\pi^{3}-8}{4}$

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