# Sketch the region $\{(x,0):y=\sqrt {4-x^2}$} and $x$-axis.Find the area of the region using integration.

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• The area enclosed by a curve $y=f(x)$,the $x$-axis and the ordinate $x=a$ and $y=b$ is given by $\int_a^b ydx.$
• $\int\sqrt{a^2-x^2}dx=\large\frac{x}{2}$$\sqrt{a^2-x^2}+\large\frac{a^2}{2}$$\sin^{-1}\big(\large\frac{x}{a}\big)$$+c. Step 1: Given \{(x,0):y=\sqrt{4-x^2}\} y=\sqrt{4-x^2}.Squaring on both sides we get y^2=4-x^2\Rightarrow x^2+y^2=4 Clearly this equation represents a circle of radius 2. The required area is between x=0 and x=\pm 2 The area of the shaded region can be found by A=4\times \int_0^2\sqrt{4-x^2}dx. We are now taking this as 4 times the region because the curve the curve is symmetrical about x-axis and it includes all the 4 quadrants. A=4\times \int_0^2\sqrt{2^2-x^2}dx. Clearly this is of the form \int\sqrt{a^2-x^2}dx=\large\frac{x}{2}$$\sqrt{a^2-x^2}+\large\frac{a^2}{2}$$\sin^{-1}\big(\large\frac{x}{a}\big). Here x=x and a=2 Step 2: Hence on integrating we get, A=4\begin{bmatrix}\large\frac{x}{2}\normalsize\sqrt{4-x^2}+\large\frac{4}{2}\normalsize\sin^{-1}\big(\large\frac{x}{2}\big)\end{bmatrix}_0^2 On applying limits we get, A=4\begin{bmatrix}\large\frac{2}{2}\normalsize\sqrt{4-4}+2\sin^{-1}\big(\large\frac{2}{2}\big)\end{bmatrix} \;\;=4\times 2[\sin^{-1}(1)-\sin^{-1}(0)]. But \sin^{-1}(0)=0 and \sin^{-1}(1)=\large\frac{\pi}{2} \;\;=8.\large\frac{\pi}{2}$$=4\pi$ sq.units.
Hence the required area is $4\pi$ sq. units.
edited Apr 28, 2013