Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
0 votes

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

Can you answer this question?

1 Answer

0 votes
  • 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.
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,
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
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App