Question

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 1

Answer: (A)

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

Required area
= Area of circle (1st quadrant ) - $\underset{0}{\overset{\pi }{\int }}\sin xdx$
$=\frac{\pi .{\pi }^2}{4}-{\left[-\cos x\right]}_0^{\pi }$
$=\frac{{\pi }^2}{4}+\left(\cos \pi -\cos 0\right)$
$=\frac{{\pi }^3}{4}+\left(-1-1\right)=\frac{{\pi }^3}{4}-2$
$=\frac{{\pi }^3-8}{4}$