Question

The area bounded by the circle $x^2+y^2=8$, the parabola $x^2=2y$ and the line $y=x$ in $y\ge 0$ is

• A. $\frac{2}{3}+2\pi$
• B. $\frac{2}{3}-2\pi$
• C. $\frac{2}{3}+\pi$
• D. $\frac{2}{3}-\pi$

Answer 1

The correct option is A $\frac{2}{3}+2\pi$

The required area

$={\int }_{-2}^2\sqrt{8-x^2}dx-{\int }_{-2}^0\frac{1}{2}{x}^{2}dx-{\int }_0^{2}xdx$

$=2{\int }_0^2\sqrt{8-x^2}dx-{\left[\frac{x^3}{6}\right]}_{-2}^0-{\left[\frac{x^2}{2}\right]}_0^2$

$=2{[\frac{x}{2}\sqrt{8-x^2}+\frac{8}{2}{\sin }^{-1}\frac{x}{2\sqrt{2}}]}_0^2-\frac{4}{3}-2$

$=2[2+4.\frac{\pi }{4}]-\frac{10}{3}=\frac{2}{3}+2\pi$