Question

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

The correct option is A $\frac{{\pi }^3-8}{4}$

$A={\int }_0^{\pi }(\sqrt{{\pi }^2-x^2}-\sin x)dx$

$=\frac{x}{2}\sqrt{{\pi }^2-x^2}+\frac{{\pi }^2}{2}\sin +\frac{x}{\pi }+\cos x{\int }_0^{\pi }$

$=\frac{{\pi }^2}{2}\left(\frac{\pi }{2}\right)-2=\frac{{\pi }^3}{4}-2$ $Sq.unit$

$=\frac{{\pi }^3-8}{4}$ $Sq.unit$