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 
\[\left( x - 2 \right)^2 = x^2 \]
\[ \Rightarrow x^2 - 4x + 4 = x^2 \]
\[ \Rightarrow x = 1 \]
\[ \therefore y^2 = 3 \Rightarrow y = \pm \sqrt{3}\]
Thus, the given circles intersect at \[A\left( 1, \sqrt{3} \right)\] and \[C\left( 1, - \sqrt{3} \right)\]

∴ Required area
= Area of the shaded region OABO
\[= \int_0^1 \sqrt{4 - \left( x - 2 \right)^2} dx + \int_1^2 \sqrt{4 - x^2} dx\]
\[ = \left[ \frac{1}{2}\left( x - 2 \right)\sqrt{4 - \left( x - 2 \right)^2} + \frac{4}{2} \sin^{- 1} \left( \frac{x - 2}{2} \right) \right]_0^1 \]
\[ + \left[ \frac{1}{2}x\sqrt{4 - x^2} + \frac{4}{2} \sin^{- 1} \left( \frac{x}{2} \right) \right]_1^2 \]
\[ = \left[ - \frac{\sqrt{3}}{2} + 2 \sin^{- 1} \left( - \frac{1}{2} \right) \right] - \left[ 0 + 2 \sin^{- 1} \left( - 1 \right) \right]\]
\[ + \left( 0 - \frac{1}{2}\sqrt{3} \right) + 2\left[ \sin^{- 1} \left( 1 \right) - \sin^{- 1} \left( \frac{1}{2} \right) \right]\]
\[= - \frac{\sqrt{3}}{2} - 2 \times \frac{\pi}{6} + 2 \times \frac{\pi}{2} - \frac{\sqrt{3}}{2} + 2 \times \frac{\pi}{2} - 2 \times \frac{\pi}{6}\]
\[ = - \sqrt{3} + 2\pi - \frac{2\pi}{3}\]
\[ = \left( \frac{4\pi}{3} - \sqrt{3} \right)\text{ square units }\]