logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
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
Toolbox:
  • 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.
answered Apr 26, 2013 by sreemathi.v
edited Apr 28, 2013 by sreemathi.v
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...