Ask Questions, Get Answers

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

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

Can you answer this question?

1 Answer

0 votes
  • 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}$$\sin^{-1}\big(\large\frac{x}{a}\big)$.
Here $x=x$ and $a=2$
Step 2:
Hence on integrating we get,
On applying limits we get,
$\;\;=4\times 2[\sin^{-1}(1)-\sin^{-1}(0)].$
But $\sin^{-1}(0)=0$ and $\sin^{-1}(1)=\large\frac{\pi}{2}$
$\;\;=8.\large\frac{\pi}{2}$$=4\pi$ sq.units.
Hence the required area is $4\pi$ sq. units.
answered Apr 26, 2013 by sreemathi.v
edited Apr 28, 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