Question

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

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

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

Required area

= Area of circle (1st quadrant) $-{\int }_0^{\pi }sinxdx$

$=\frac{\pi .{\pi }^2}{4}-[-cosx{]}_0^{\pi }=\frac{{\pi }^3}{4}+(cos\pi -cos0)$

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

Hence, option A is correct.