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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area of the region \({(x, y) : y^2\: \leq \: 4x, 4x^2 + 4y^2\: \leq\: 9}\)

This question has appeared in model paper 2012.

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Toolbox:
  • Suppose we are given two curves represented by f(x),y=g(x) where $f(x)\geq g(x)$ in [a,b] the points of intersection of these two curves are given by x=a and x=b by taking common values of y from the given equation of the two curves.
  • Hence the required area is given by $A=\int_a^b[f(x)-g(x)dx.$
Step1:
The area bounded by the curves is $\left\{(x,y):y^2\leq 4x,4x^2+4y^2\leq 9\right \}$
Here we understand that since the area bounded between the two curves is between the circle $4x^2+4y^2=9$ and the parabola $y^2=4x$, the point of intersection can be found by solving the two equations.
$4x^2+4y^2=9$ --------(1)
$\;\;\;\;\;\;\;\;\;4x=y^2$ ------(2)
substituting for $y^2$ in equ(1) we get
$4x^2+4(4x)=9$
$4x^2+16x-9=0$
on factorising we get,
(2x-1)(2x+9)=0
$\Rightarrow x=\frac{1}{2}$ and $x=\frac{-9}{2}$
when $x=\frac{1}{2};y=\pm\sqrt 2$ and when $x=\frac{-9}{2};$y=imaginary.
Hence $x=\frac{-9}{2}$ is admirable,
Hence let us take the points of intersection as ($\frac{1}{2},\sqrt 2)$ and $(\frac{1}{2},-\sqrt 2).$
The area of the region is the shaded portion as shown in the fig
Thus the required area is the region bounded by the parabola $y^2=4x$ and the circle $x^2+y^2=\bigg(\frac{3}{2}\bigg)^2$
clearly the curves are symmetrical about the x-axis.
Hence the required area=2(area of the shaded region lying above the x-axis).
Let $y_1$ be the area bounded by the parabolic curve and the x-axis.
As if moves from 0 to $\frac{1}{2}$
$a=2\int_0^\frac{1}{2}y_1dx$
$\;\;\;=2\times\int_0^\frac{1}{2}2\sqrt x dx$
on integrating we get,
$A=2\times \begin{bmatrix}2\times \frac{x^{\frac{3}{2}}}{\frac{3}{2}}\end{bmatrix}_0^\frac{1}{2}$
on applying limits we get,
$A=2\times \begin{bmatrix}2\times\frac{2}{3}\times \frac{1}{2\sqrt 2}\end{bmatrix}$
$\;\;\;=\frac{8}{3}\times\frac{1}{2\sqrt 2}=\frac{2\sqrt 2}{3}$---------(1)
Let $y_2$ be the area bounded by the circle and the x-axis.As it moves from $\frac{1}{2}$ to $\frac{3}{2}$.
$A=2\times\int_\frac{1}{2}^\frac{3}{2}\sqrt {\frac{9}{4}-x^2}dx$.
on integrating we get ,
$A=2\times\begin{bmatrix}\frac{1}{2}x\sqrt {\frac{9}{4}-x^2}+\frac{1}{2}.\frac{9}{4}\sin^{-1}\frac{2x}{3}\end{bmatrix}_\frac{1}{2}^\frac{3}{2}$
on applying limits we get,
$A=\begin{bmatrix}\frac{9}{4}\sin^{-1}(1)-\big(\frac{1}{\sqrt 2}+\frac{9}{4}\sin^{-1}(\frac{1}{3})\end{bmatrix}$
$\;\;\;=\frac{9}{4}.\frac{\pi}{2}-\frac{1}{\sqrt 2}-\frac{9}{4}\sin^{-1}\big(\frac{1}{3}\big)$----------(2)
on combining area in equ(1) and area in equ(2) we get,
$A=\frac{2\sqrt 3}{2}+\frac{9\pi}{8}-\frac{1}{\sqrt 2}-\frac{9}{4}\sin^{-1}\big(\frac{1}{3}\big)$
on simplifying we get.
$A=\frac{\sqrt 2}{6}+\frac{9\pi}{8}-\frac{9}{4}\sin^{-1}\big(\frac{1}{3}\big)$
Hence the required area is
$A=\frac{\sqrt 2}{6}+\frac{9\pi}{8}-\frac{9}{4}\sin^{-1}\big(\frac{1}{3}\big)$sq.units.
answered Jan 23, 2013 by sreemathi.v
edited Dec 22, 2013 by balaji.thirumalai
 

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