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

Answer: (A)

$x^2+y^2=8$
$x^2=2y$
$\Rightarrow y^2+2y-8=0\Rightarrow (y+4)(y-2)=0$
$\Rightarrow y=-4\text{ (rejected), }y=2\Rightarrow x=\pm 2$
$A=\underset{-2}{\overset{2}{\int }}\sqrt{8-x^2}dx-\underset{-2}{\overset{0}{\int }}\frac{x^2}{2}dx-\underset{0}{\overset{2}{\int }}xdx$
$=2\underset{0}{\overset{2}{\int }}\sqrt{8-x^2}dx-\frac{1}{6}{\left(x^3\right)}_{-2}^0-{\left(\frac{x^2}{2}\right)}_0^2=2{\left[\frac{x}{2}\sqrt{8-x^2}+\frac{8}{2}{\sin }^{-1}\left(\frac{x}{2\sqrt{2}}\right)\right]}_0^2-\frac{(0+8)}{6}-\frac{(4-0)}{2}$
$=2\left[2+4{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)-0\right]-\frac{4}{3}-2=4+\frac{8\pi }{4}-\frac{10}{3}=2\pi +\frac{2}{3}$