CircleHard

Question

The area of the region enclosed between the circles $x^{2} + y^{2} = 4$ and $x^{2} + (y - 2)^{2} = 4$ is :

Options

A.$\frac{2}{3}(2\pi - 3\sqrt{3})$
B.$\frac{4}{3}(2\pi - 3\sqrt{3})$
C.$\frac{4}{3}(2\pi - \sqrt{3})$
D.$\frac{2}{3}(4\pi - 3\sqrt{3})$

Solution

Sol.

$$\begin{matrix} & A = 2\int_{0}^{\sqrt{3}}\mspace{2mu}\mspace{2mu}\left\lbrack \sqrt{4 - x^{2}} - \left( 2 - \sqrt{4 - x^{2}} \right) \right\rbrack dx \\ & \ = 2\int_{0}^{\sqrt{3}}\mspace{2mu}\mspace{2mu}\left( 2\sqrt{4 - x^{2}} - 2 \right)dx \\ & \ = 4\int_{0}^{\sqrt{3}}\mspace{2mu}\mspace{2mu}\left( \sqrt{4 - x^{2}} - 1 \right)dx \\ & \ = \left\lbrack 4\left\lbrack \frac{1}{2}\left( x\sqrt{4 - x^{2}} + 4\sin^{- 1}\frac{x}{2} \right) - x \right\rbrack \right\rbrack_{0}^{\sqrt{3}} \\ & \ = 4\left\lbrack \frac{1}{2}\left( \sqrt{3} + 4 \times \frac{\pi}{3} \right) - \sqrt{3} \right\rbrack = 4\left\lbrack \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \right\rbrack \\ & \ = \frac{8\pi}{3} - 2\sqrt{3}\text{~}\text{(Sq. units)}\text{~} \end{matrix}$$

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