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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the area of the region $\{(x,4):y^2\leq 4x,4x^2+4y^2\leq9\}$ using method of integration.

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$4x^2+4y^2=9$
$x^2+y^2=(\frac{3}{2})^2$
This represents a circle with centre at the origin and radius $\frac{3}{2}$ and the parabola $y^2=4x$
The required area bounded by the parabola $y^2=4x$ and the circle $x^2+y^2=(\frac{3}{2})^2$ is shown by the shaded portion.
The point of intersection are
$y^2=4x$
$x^2+y^2=(\frac{3}{2})^2$
$x^2+4x=\frac{9}{4}\Rightarrow 4x^2+16x-9=0$
$\Rightarrow (2x+9)(2x-1)=0$
Therefore x=$\frac{1}{2}$ or $x=\frac{9}{2}$
If x=$\frac{1}{2}\Rightarrow y=\pm \sqrt 2$ and $x=\frac{9}{2}$,y is imaginary.
The points of intersections are ($\frac{1}{2}.\sqrt 2)$ and $\frac{1}{2},-\sqrt 2)$
Therefore Area=2[area of OAMO+area of AMPA]
$\qquad\qquad=2\begin{bmatrix}\int_0^{\frac{1}{2}}2\sqrt xdx+2\int_{\frac{1}{2}}^{\frac{3}{4}}\sqrt{\frac{9}{4}-x^2}dx\end{bmatrix}$
$\qquad\qquad=4.\frac{2}{3}x^{\frac{3}{2}}]^{\frac{1}{2}}+2\begin{bmatrix}\frac{1}{2}x+\sqrt{\frac{9}{4}-x^2}+\frac{1}{2}\frac{9}{4}sin^{-1}\frac{2x}{3}\end{bmatrix}_{\frac{1}{2}}^{\frac{3}{2}}$
$A=\frac{8}{3}[\frac{1}{2\sqrt{2}}-0]+\frac{9}{4}sin^{-1}(1)-\frac{1}{\sqrt2}+\frac{9}{4}sin^{-1}(\frac{1}{3})$
$\;\;=\frac{2\sqrt 2}{3}+\frac{9\pi}{8}-\frac{1}{\sqrt 2}-\frac{9}{4}sin^{-1}\frac{1}{3}$
Therefore $A=\frac{\sqrt 2}{6}+\frac{9\pi}{8}-\frac{9}{4}sin^{-1}\big(\frac{1}{3}\big)$ sq.units
answered Mar 25, 2013 by sreemathi.v
 

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