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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Using integration, find the area of the region bounded by the parabola \( y^2=4x\) and the circle \( 4x^2+4y^2=9\)

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  • If 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.
Step 1:
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=\large\frac{9}{4}$$\Rightarrow y^2=\large\frac{9}{4}-x^2$
$y=\sqrt {\large\frac{9}{4}-x^2}$
Let $R_2$ be the region lying inside the circle.
Hence the area of the required region is bounded by the parabola $x^2=4y$ and the circle $x^2+y^2=\large\frac{9}{4}.$
Step 2:
To find the point of intersection let us substitute $x^2=4y$ in the equation of the circle.
$4y+y^2=\large\frac{9}{4}$$\Rightarrow 4y^2+16y-9=0$
On factorising we get,
(2y+9)(2y-1)=0
$\Rightarrow x=\large\frac{-9}{2}$
and $y=\large\frac{1}{2}$.
Hence the points of intersection are ($\sqrt 2,\frac{1}{2})(\sqrt 2,\frac{-1}{2}).$
Step 3:
The required area is $A=\int_0^{\Large\frac{1}{\sqrt 2}}(R_2-R_1)dx$
$\Rightarrow\int_0^{\Large\frac{1}{\sqrt 2}}\large\frac{\sqrt {9-4x^2}}{2}-\int_0^\frac{1}{\sqrt 2}\frac{x^2}{4}$$dx$.
$A=\frac{1}{2}\int_0^{\Large\frac{1}{\sqrt 2}}\sqrt {9-4x^2}-\frac{1}{4}\int_0^\sqrt 2x^2dx.$
On integrating we get,
$\Rightarrow\large\frac{1}{2}.\frac{1}{2}\begin{bmatrix}\frac{2x}{2}\sqrt{9-4x^2}+\frac{9}{2}\sin^{-1}\frac{2x}{3}\end{bmatrix}_0^\sqrt 2-\frac{1}{4}\begin{bmatrix}\frac{x^3}{3}\end{bmatrix}_0^\sqrt 2$
On applying the limits we get,
$\Rightarrow\large\frac{\sqrt 2}{12}+\frac{9}{8}\sin^{-1}\big(\frac{2\sqrt 2}{3}\big)$.
Step 4:
But the curves are symmetrical about x-axis the required area is
$A=2\bigg[\large\frac{\sqrt 2}{12}+\frac{9}{8}$$\sin^{-1}\big(\large\frac{2\sqrt 2}{3}\big)\bigg]$
$\;\;\;=\large\frac{\sqrt 2}{6}+\frac{9}{4}$$\sin^{-1}\big(\frac{2\sqrt 2}{3}\big)$.
Hence the required area is $\large\frac{\sqrt 2}{6}+\frac{9}{4}$$\sin^{-1}\big(\large\frac{2\sqrt 2}{3}\big)$.
answered Sep 27, 2013 by sreemathi.v
 

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