Question

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

• A. $\frac{{\pi }^3-8}{4}$ sq. units
• B. $\frac{{\pi }^3+8}{4}$ sq. units
• C. $4({\pi }^3-8)$ sq. units
• D. $\frac{\pi -8}{4}$ sq. units

Answer 1

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

Area of the circle with radius $\pi$ is ${\pi }^3$

Then, area of the arc of circle in the $firstquadrantis$\dfrac{\pi^3 }{4}sq\:units$ ....... $\left(i\right)$
Also, the area of the curve $y=\sin x$ is $\stackrel{\pi }{\underset{0}{\int }}\sin xdx=2squnits$ ....... $\left(ii\right)$
So, the required area is obtained by taking out $\left(ii\right)$ from $\left(i\right)$ i.e. $\left(i\right)-\left(ii\right)$

Thus, the reuqired area
$=\frac{{\pi }^3}{4}-2=\frac{{\pi }^3-8}{4}squnits.$