Discussion Forum



1)

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


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

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

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

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

Answer:

Option A

Explanation:

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

 6102021584_12.PNG

 

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}$