Ask Questions, Get Answers

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

Find the area of the region bounded by the curve $y^2=2x$ and $x^2+y^2=4x.$

$\begin{array}{1 1}(A) \pi-\large\frac{8}{3}sq.units. \\ (B) \pi+\large\frac{8}{3}sq.units. \\(C) 2[\pi-\large\frac{8}{3}]sq.units. \\ (D) 2[\pi+\large\frac{8}{3}]sq.units \end{array} $

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 $y^2=2x$ and $x^2+y^2=4x$
Consider the curve $y^2=2x$
Clearly this curve represents a parabola open rightward with vertex at (0,0) and along $x$-axis.
Next consider the curve $x^2+y^2=4x$
$\Rightarrow x^2-4x+y^2=0$
(i.e)$ (x-2)^2-4+y^2=0$
$\Rightarrow (x-2)^2+y^2=4$
Clearly this represents the equation of a circle whose centre is (2,0) and radius 2.
The required area is the shaded region shown in the fig.
Step 2:
Let us find the points of intersection.
$2x=4x-x^2$ or $x^2-2x=0$
(i.e) $x=0$ or $x=2$ and $y=0$ and 4.
Hence the point of intersection are (0,0) and (2,2).
Since the area enclosed is in the I and Iv quadrants,
The required area is $A=2\int_0^2(y_2-y_1)dx.$
Where $y_2=\sqrt{(4)-(x-2)^2}$ and $y_1=\sqrt{2x}$
$A=2\int_0^2[\sqrt{4-(x-2)^2}-\sqrt {2x}]dx.$
$\;\;=2\int_0^2\sqrt{4-(x-2)^2}dx-\sqrt 2\int_0^2\sqrt x dx.$
Step 3:
On integrating we get,
$\;\;=2\begin{bmatrix}\sqrt{4-(x-2)^2}+\large\frac{4}{2}\normalsize \sin^{-1}\big(\large\frac{x-2}{2}\big)\end{bmatrix}_0^2$
On applying limits we get,
$2\begin{bmatrix}\large\frac{2-2}{2}\normalsize (\sqrt{4-(2-2)^2}+2\sin^{-1}\big(\large\frac{2-2}{2}\big)\end{bmatrix}-\frac{0-2}{2}\sqrt{4-(0-2)^2}-2\sin^{-1}\big(\frac{-2}{2}\big)-\begin{bmatrix}\sqrt{2}\times \frac{2}{3}(2)^{\large\frac{3}{2}}-0\end{bmatrix}$
$\;\;=2[0+2\sin^{-1}0)-(0-2\sin^{-1}(-1))-\sqrt 2\times \large\frac{2}{3}\normalsize \times 2\sqrt{2}$
$\sin^{-1}(-1)=\large-\frac{\pi}{2}$ and $\sin^{-1}(0)=0$
Hence the required area is =$2[\pi-\large\frac{8}{3}]$sq.units.
answered May 8, 2013 by sreemathi.v
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App