Question

The parabola $y=\frac{x^2}{2}$ divides the circle $x^2+y^2=8$ into two parts. Find the area of both parts.

• A. $6\pi +\frac{4}{3}$ , $2\pi -\frac{4}{3}$
• B. $6\pi -\frac{4}{3}$ , $2\pi +\frac{4}{3}$
• C. $6\pi +\frac{2}{3}$ , $2\pi -\frac{2}{3}$
• D. $6\pi -\frac{2}{3}$ , $2\pi +\frac{2}{3}$

Answer 1

Answer: (A)

$6\pi +\frac{4}{3}$ , $2\pi -\frac{4}{3}$

Let us first find the x-values of the points of intersection using the given equations of parabola $y=\frac{x^2}{2}$ and circle $x^2+y^2=8$ as follows:

$x^2+y^2=8\Rightarrow x^2+{\left(\frac{x^2}{2}\right)}^2=8(\because y=\frac{x^2}{2})\Rightarrow x^2+\frac{x^4}{4}=8\Rightarrow 4x^2+x^4=32$

$\Rightarrow x^4+4x^2-32=0\Rightarrow x^4+8x^2-4x^2-32=0\Rightarrow x^2(x^2+8)-4(x^2+8)=0\Rightarrow (x^2-4)(x^2+8)=0\Rightarrow (x^2-4)=0,(x^2+8)=0\Rightarrow x^2=4,x^2=-8\Rightarrow x=\pm \sqrt{4}\Rightarrow x=\pm 2$

Let $A_1$ be the area of the region inside the circle and above the parabola and $A_2$ be the area of the region inside the circle and below the parabola. Then we have,

$A_1={\int }_{-2}^2(\sqrt{8-x^2}-\frac{1}{2}{x}^2)dx=2{\int }_0^2(\sqrt{8-x^2}-\frac{1}{2}{x}^2)dx=2[\frac{1}{2}\times 8{\sin }^{-1}\left(\frac{2}{\sqrt{8}}\right)+\frac{1}{2}\times 2\sqrt{8-2^2}-\frac{1}{2}{\left[\frac{1}{3}{x}^3\right]}_0^2]=8{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)+2\sqrt{4}-\frac{8}{3}$

$=8\times \frac{\pi }{4}+4-\frac{8}{3}=2\pi +\frac{4}{3}$

We know that the area of the circle is $\pi r^2$, therefore the area of the circle $x^2+y^2=8$ with radius $r=\sqrt{8}$ is:

$\pi {\left(\sqrt{8}\right)}^2=8\pi$

Thus, we have

$A_2=8\pi -(2\pi +\frac{4}{3})=6\pi -\frac{4}{3}$

Hence, the area of parabola and circle is $2\pi +\frac{4}{3}$ and $6\pi -\frac{4}{3}$ respectively.