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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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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} $

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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 $y^2=2x$ 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 $x^2-2x=0$
$x(x-2)=0$
(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,
$\;\;=2\begin{bmatrix}\sqrt{4-(x-2)^2}+\large\frac{4}{2}\normalsize \sin^{-1}\big(\large\frac{x-2}{2}\big)\end{bmatrix}_0^2$
$\;\;=-\begin{bmatrix}\sqrt{2}\big(\large\frac{x^{\large\frac{3}{2}}}{\Large\frac{3}{2}}\big)\end{bmatrix}_0^2$
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 $\sin^{-1}(0)=0$
$\;\;=2[2\times\large\frac{\pi}{2}-\frac{8}{3}]$
$\;\;=2[\pi-\large\frac{8}{3}]$sq.units.
Hence the required area is =$2[\pi-\large\frac{8}{3}]$sq.units.
answered May 8, 2013 by sreemathi.v
 
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