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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area of the circle \(4x^2 + 4y^2 = 9\) which is interior to the parabola \(x^2 = 4y.\)

$\begin{array}{1 1}\frac{\sqrt 2}{6}+\frac{9}{4}\sin^{-1}\big( \frac{2\sqrt 2}{3}\big) \\ \frac{\sqrt 2}{3}+\frac{5}{4}\sin^{-1}\big( \frac{\sqrt 2}{3}\big) \\ \frac{\sqrt 2}{3}-\frac{9}{4}\sin^{-1}\big( \frac{2\sqrt 2}{3}\big)\\ \frac{\sqrt 2}{6}+\frac{9}{4}\cos^{-1}\big( \frac{2\sqrt 2}{3}\big)\end{array} $

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  • If we are given two curves represented by y=f(x),y=g(x) where $f(x)\geq g(x)$ in [a,b],the point of intersection of two curves are given by x=a and x=b by taking common values of y from the equation of the two curves.
The curve $x^2=4y$ is the equation of the parabola with vertex at the origin and axis along y-axis and open upwards.
Let $R_1$ be the region lying inside the parabola $x^2=4y\Rightarrow y=\frac{x^2}{4}$
We have $4x^2+4y^2=9$
$x^2+y^2=\frac{9}{4}\Rightarrow x^2+y^2=\frac{9}{4}\Rightarrow y^2=\frac{9}{4}-x^2$
$y=\sqrt {\frac{9}{4}-x^2}$
Clearly this represents the equation of a circle with the centre at the origin and radius $\frac{3}{2}$.
Let $R_2$ be the region lying inside the circle.
Hence the area of the required region is bounded by the parabola $x^2=4y$ and the circle $x^2+y^2=\frac{9}{4}.$
The area of the required region is the shaded portion shown in the fig.
To find the point of intersection let us substitute $x^2=4y$ in the equation of the circle.
$4y+y^2=\frac{9}{4}\Rightarrow 4y^2+16y-9=0$
On factorising we get,
$\Rightarrow x=\frac{-9}{2}$ or $y=\frac{1}{2}$.
If $y=\frac{1}{2},x=\pm\sqrt 2$ and if $y=\frac{-9}{2},$x is imaginary.
Hence the points of intersection are ($\sqrt 2,\frac{1}{2})(\sqrt 2,\frac{-1}{2}).$
Clearly the curves are symmetrical about the x-axis.
Hence the required area is
$A=\int_0^\sqrt 2(R_2-R_1)dx$
$\;\;\;=\int_0^\sqrt 2\frac{\sqrt {9-4x^2}}{2}-\int_0^\sqrt 2\frac{x^2}{4}dx$.
$A=\frac{1}{2}\int_0^\sqrt 2\sqrt {9-4x^2}-\frac{1}{4}\int_0^\sqrt 2x^2dx.$
On integrating we get,
$\;\;\;\;=\frac{1}{2}.\frac{1}{2}\begin{bmatrix}\frac{2x}{2}\sqrt{9-4x^2}+\frac{9}{2}\sin^{-1}\frac{2x}{3}\end{bmatrix}_0^\sqrt 2-\frac{1}{4}\begin{bmatrix}\frac{x^3}{3}\end{bmatrix}_0^\sqrt 2$
On applying the limits we get,
$\;\;\;=\frac{1}{4}\begin{bmatrix}\sqrt 2\sqrt{9-8}+\frac{9}{2}\sin^{-1}\big(\frac{2\sqrt 2}{3}\big)\end{bmatrix}-\frac{1}{12}(\sqrt 2)^3$
$\;\;\;\;=\frac{\sqrt 2}{4}+\frac{9}{8}\sin^{-1}\big(\frac{2\sqrt 2}{3}\big)-\frac{\sqrt 2}{6}$.
$\;\;\;=\frac{\sqrt 2}{12}+\frac{9}{8}\sin^{-1}\big(\frac{2\sqrt 2}{3}\big)$.
But since the curves are symmetrical about x-axis the required area is
$A=2\begin{bmatrix}\frac{\sqrt 2}{12}+\frac{9}{8}\sin^{-1}\big(\frac{2\sqrt 2}{3}\big)\end{bmatrix}$.
$\;\;\;=\frac{\sqrt 2}{6}+\frac{9}{4}\sin^{-1}\big(\frac{2\sqrt 2}{3}\big)$.
Hence the required area is $\frac{\sqrt 2}{6}+\frac{9}{4}\sin^{-1}\big(\frac{2\sqrt 2}{3}\big)$.
answered Dec 21, 2013 by yamini.v
edited Sep 16, 2014 by sreemathi.v

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