# Determine the area under the curve $y=\sqrt{a^2-x^2}$ included between the lines $x=0$ and $x=a$.

Toolbox:
• The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
• $\int\sqrt{a^2-x^2}=\large\frac{x}{2}$$\sqrt{a^2-x^2}+\large\frac{a^2}{2}$$\sin^{-1}\big(\large\frac{x}{a}\big)$$+c. Step 1: Given y=\sqrt{a^2-x^2} We are asked to find the area of the curve bounded between the lines x=0 and x=a. Consider y=\sqrt{a^2-x^2} Squaring on both sides we get, y^2=a^2-x^2\Rightarrow x^2+y^2=a^2 Clearly this is an equation of the circle with radius a,whose centre is the origin (0,0). The area of the required region is the shaded region. Step 2: A=2\times \int_0^a ydx. Since the curve is symmetrical about x-axis and it is bounded between the lines x=0 and x=a (i.e) the y-axis and the line x=a,it includes the I and IV quadrant. Hence we take it as twice the area of the I quadrant. A=2\int_0^a\sqrt{a^2-x^2}dx. On integrating we get, A=2\begin{bmatrix}\large\frac{x}{2}\normalsize \sqrt{a^2-x^2}+\large\frac{a^2}{2}\normalsize \sin^{-1}\big(\large\frac{x}{a}\big)\end{bmatrix}_0^a Step 3: On applying limits we get, \;\;=2\begin{bmatrix}\large\frac{a}{2}\normalsize\sqrt{a^2-x^2}+\large\frac{a^2}{2}\normalsize\sin^{-1}\big(\large\frac{a}{a}\big)\end{bmatrix} \;\;=2[0+\large\frac{a^2}{2}$$\sin^{-1}(1)]$
But $\sin^{-1}=\large\frac{\pi}{2}$
$A=2\big(\large\frac{a^2}{2}.\frac{\pi}{2}\big)=\frac{\pi a^2}{2}$
Hence the required area is $\large\frac{\pi a^2}{2}$