Question

Using integration find the area of the region bounded by the curves $y=\sqrt{4-x^2},x^2+y^2-4x=0$ and the x-axis.

Answer 1

The given curves are $y=\sqrt{4-x^2}$ and $x^2+y^2-4x=0$.
$y=\sqrt{4-x^2}\Rightarrow x^2+y^2=4$ .....(1)
This represents a circle with centre O(0, 0) and radius = 2 units.
Also,
$x^2+y^2-4x=0\Rightarrow {\left(x-2\right)}^2+y^2=4$ .....(2)
This represents a circle with centre B(2, 0) and radius = 2 units.
Solving (1) and (2), we get
$$\begin{gathered} {\left(x-2\right)}^2=x^2 \\ \Rightarrow x^2-4x+4=x^2 \\ \Rightarrow x=1 \\ \therefore y^2=3\Rightarrow y=\pm \sqrt{3} \end{gathered}$$
Thus, the given circles intersect at $\mathrm{A}\left(1,\sqrt{3}\right)$ and $\mathrm{C}\left(1,-\sqrt{3}\right)$.


∴ Required area
= Area of the shaded region OABO
$$\begin{gathered} ={\int }_0^1\sqrt{4-{\left(x-2\right)}^2}dx+{\int }_1^2\sqrt{4-x^2}dx \\ ={\left[\frac{1}{2}\left(x-2\right)\sqrt{4-{\left(x-2\right)}^2}+\frac{4}{2}{\sin }^{-1}\left(\frac{x-2}{2}\right)\right]}_0^1 \\ +{\left[\frac{1}{2}x\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\left(\frac{x}{2}\right)\right]}_1^2 \\ =\left[-\frac{\sqrt{3}}{2}+2{\sin }^{-1}\left(-\frac{1}{2}\right)\right]-\left[0+2{\sin }^{-1}\left(-1\right)\right] \\ +\left(0-\frac{1}{2}\sqrt{3}\right)+2\left[{\sin }^{-1}\left(1\right)-{\sin }^{-1}\left(\frac{1}{2}\right)\right] \end{gathered}$$
$$\begin{gathered} =-\frac{\sqrt{3}}{2}-2\times \frac{\mathrm{\pi }}{6}+2\times \frac{\mathrm{\pi }}{2}-\frac{\sqrt{3}}{2}+2\times \frac{\mathrm{\pi }}{2}-2\times \frac{\mathrm{\pi }}{6} \\ =-\sqrt{3}+2\mathrm{\pi }-\frac{2\mathrm{\pi }}{3} \\ =\left(\frac{4\mathrm{\pi }}{3}-\sqrt{3}\right)\mathrm{square}\mathrm{units} \end{gathered}$$