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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Determine the area under the curve $y=\sqrt{a^2-x^2}$ included between the lines $x=0$ and $x=a$.

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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 $\large\frac{\pi a^2}{2}$
answered May 8, 2013 by sreemathi.v
 
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