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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area of the region in the first quadrant enclosed by \(x\) - axis, line \(x = \sqrt {3}\: y\) and the circle \(x^2 + y^2 = 4.\)

$\begin{array}{1 1} \large \frac{\pi}{3} sq. units. \\ \large \frac{\pi}{6} sq. units. \\\large \frac{\pi}{9} sq. units. \\ \large \frac{\pi}{4} sq. units \end{array} $

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Toolbox:
  • Area of the region bounded between a curve y=f(x) and a line is given by \[A=\int_a^by\;dx=\int_a^bf(x)\;dx \]
  • where a and b are the point of intersection of the line and curve.
  • To find the point of intersection,we can solve the two equations.
To find the area of the region bounded by the circle $x^2+y^2=4$ and $x=\sqrt 3y$ and the x-axis.
The required area of the region is given by the shaded portion in the fig.
To find the point of intersection ,let us solve the equation;
$x^2+y^2=4$-----(1) and $x=\sqrt 3y\Rightarrow x^2=3y^2$
$\Rightarrow \frac{x^2}{3}=y^2$--------(2)
substituting for $y^2$ in equ(1) we get
$x^2+\frac{x^2}{3}=4 \Rightarrow 3x^2+x^2=12$
$\Rightarrow 4x^2=12$
$\Rightarrow x^2=3$
$\Rightarrow x=\pm \sqrt 3$
If x=$\sqrt 3$ then y=1.
Hence we can take the limit as $\sqrt 3$ to 2 .since the radius of the circle is 2.
The required area $A=\int_\sqrt 3^2(y_1+y_2)dx$
where $y_1=\sqrt {4-x^2}$
$\;\;\;\;\;\;\;\;\;\;y_2$=area of the triangle.
$A=\int_\sqrt 3^2\sqrt {4-x^2}$+area of the triangle
on integrating $y_1$ we get,
$A=\begin{bmatrix}\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}(\frac{x}{2})\end{bmatrix}_\sqrt 3^2$
on applying the limits we get,
$A=\begin{bmatrix}\frac{2}{2}\sqrt{4-4}+\frac{4}{2}\sin^{-1}(\frac{2}{2})\end{bmatrix}-\begin{bmatrix}\frac{\sqrt 3}{2}\sqrt{4-3}+\frac{4}{2}\sin^{-1}(\frac{\sqrt 3}{2})\end{bmatrix}$
$A=\begin{bmatrix}0+2(\frac{\pi}{2})-\frac{\sqrt 3}{2}-2(\frac{\pi}{3})\end{bmatrix}$
$A\;\;\;=\begin{bmatrix}\pi-\frac{\sqrt 3}{2}-\frac{2\pi}{3}\end{bmatrix}=\begin{bmatrix}\frac{\pi}{3}-\frac{\sqrt 3}{2}\end{bmatrix}$------(3)
$y_2$=area of the triangle bounded between x-axis and the line $x=\sqrt 3y$
$\;\;\;=\frac{1}{2}\times \sqrt 3\times 1=\frac{\sqrt 3}{2}$-------(4)
Required area can be obtained by combining (3) and (4)
$A=\frac{\pi}{3}-\frac{\sqrt 3}{2}+\frac{\sqrt 3}{2}=\frac{\pi}{3}$ sq.units.
Hence the required area=$\Large \frac{\pi}{3}$ sq. units.
answered Dec 20, 2013 by yamini.v
 

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