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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area bounded by curves \(x^2 + (y-1)^2 = 1\) and \(x^2 + y^2 = 1.\)

$\begin{array}{1 1}\big(\large \frac{2\pi}{3}-\frac{\sqrt 3}{2}\big)\; sq. units. \\\big(\large \frac{\pi}{3}-\frac{\sqrt 3}{2}\big)\; sq. units. \\ \big(\large \frac{3\pi}{2}+\frac{\sqrt 3}{2}\big)\; sq. units. \\ \big(\large \frac{2\pi}{3}-\frac{\sqrt 2}{3}\big)\; sq. units\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 points 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.
Given the equation of the curves are
$x^2+y^2=1$ and $x^2+(y-1)^2=1.$
Clearly $x^2+y^2=1$ represents a circle with centre at (0,0) and radius=1.
$x^2+(y-1)^2=1$ represents a circle with centre at (0,1) and radius=1.
To find the points of the intersection of the given curves ,let us solve the given equations.\[x^2+y^2=1\Rightarrow x^2=1-y^2.\]
Substituting this in the second equation we get,
$(1-y^2)+(y-1)^2=1$
$\Rightarrow 1-y^2+y^2+1-2y=1$
$\Rightarrow\:y=\large\frac{1}{2}$
If $y=\large\frac{1}{2},x=\pm\frac{\sqrt 3}{2}.$
Hence the points of intersection are $\big(\frac{\sqrt 3}{2},\frac{1}{2}\big),\big(\frac{-\sqrt 3}{2},\frac{1}{2}\big)$
The required area is the shaded portion shown in the fig.
<< Enter Text >>
The curve $x=\sqrt {1-(y-1)^2}$ moves from y=0 to $ y=\frac{1}{2}.$
The curve $y=\sqrt{1-x^2}$ moves from $\frac{1}{2}$ to 1.
But the shaded region is symmetric about the line $y=\large\frac{1}{2}$ and
also about $y\:axis$
$\Rightarrow\:$ The required area is given by
$A=4\times\begin{bmatrix}\int_\frac{1}{2}^1\sqrt{1-y^2}dy\end{bmatrix}$
On integrating we get,
$A=4\begin{bmatrix}\frac{y}{2}\sqrt {1-y^2}+\frac{1}{2}\sin^{-1}(y)\end{bmatrix}_\frac{1}{2}^1$
on applying limits we get,
$A=4\begin{bmatrix}\large\frac{1}{2}sin^{-1}1-\large\frac{1}{4}\large\frac{\sqrt 3}{2}-\large\frac{1}{2}sin^{-1}(\large\frac{1}{2})\end {bmatrix}$
$\;\;\;=4\begin{bmatrix}\frac{\pi}{4}-\frac{\sqrt 3}{8}-\frac{1}{2}\frac{\pi}{6}\end{bmatrix}$
$\;\;\;=\big(\frac{2\pi}{3}-\frac{\sqrt 3}{2}\big)$ sq.units.
Hence the required area is $\big(\frac{2\pi}{3}-\frac{\sqrt 3}{2}\big)$ sq. units.
answered Feb 4, 2014 by rvidyagovindarajan_1
edited Feb 7, 2014 by balaji.thirumalai
 
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