Question

Find the area, lying above x-axis and included between the circle x² + y² = 8x and the parabola y² = 4x.

Answer 1

The given equations are $x^2+y^2=8x\cdots \left(1\right)$ and $y^2=4x\cdots \left(2\right)$
Clearly the equation $x^2+y^2=8x$ is a circle with centre $\left(4,0\right)$ and has a radius 4.Also $y^2=4x$ is a parabola with vertex at origin and the axis along the x-axis opening in the positive direction .
To find the intersecting points of the curves ,we solve both the equation.(2
$\therefore$ $x^2+4x=8x$
$\Rightarrow$ $x^2-4x=0$
$\Rightarrow$$x\left(x-4\right)=0$
$\Rightarrow$$x=0\mathrm{and}x=4$
When $x=0,y=0$
When $x=4,y=\pm 4$

To approximate the area of the shaded region the length $=\left|y_2-y_1\right|$ and the width = dx
$A={\int }_0^4\left|y_2-y_1\right|dx$
$={\int }_0^4\left(y_2-y_1\right)dx$ $\left[\because y_2>y_1\therefore \left|y_2-y_1\right|=y_2-y_1\right]$
$={\int }_0^4\left[\sqrt{\left(16-{\left(x-4\right)}^2\right)}-\sqrt{4x}\right]\mathit{d}x\left\{\therefore y_2=\sqrt{16-{\left(x-4\right)}^2}\mathrm{and}{y}_1=2\sqrt{x}\right\}$

$={\int }_0^4\sqrt{16-{\left(x-4\right)}^2}dx-{\int }_0^4\sqrt{4x}dx$
$={{\left[\frac{\left(x-4\right)}{2}\sqrt{16-{\left(x-4\right)}^2}+\frac{16}{2}{\sin }^{-1}\left(\frac{x-4}{4}\right)\right]}^4}_0-{{\left[\frac{4x^{{\displaystyle \frac{3}{2}}}}{3}\right]}^4}_0$
$=\left[0+0-0-8{\sin }^{-1}\left(\frac{-4}{4}\right)\right]-\frac{4}{3}\times 4^{\frac{3}{2}}$
$$\begin{gathered} =\frac{8\mathrm{\pi }}{2}-\frac{32}{3} \\ =4\mathrm{\pi }-\frac{32}{3} \end{gathered}$$
Hence the required area is $4\mathrm{\pi }-\frac{32}{3}$ square units.