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Find the area bounded by curves \((x - 1)^2 + y^2 = 1\) and \(x^2 + y^2 = 1.\)

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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 $(x-1)^2+(y-0)^2=1.$
Clearly $x^2+y^2=1$ represents a circle with centre at (0,0) and radius=1.
$x^2+y^2=1$ represents a circle with centre at (1,0) and radius=1.
To find the points of the intersection of the given curves ,let us solve the given equations.\[x^2+y^2=1\Rightarrow y^2=1-x^2.\]
Substituting this in the second equation we get,
$\Rightarrow x^2-2x+1+1-x^2=1$
$\quad\;\;-2x=-1 \Rightarrow x=\frac{1}{2}$.
If x=$\frac{1}{2},y=\pm\frac{\sqrt 3}{2}.$
Hence the points of intersection are $\big(\frac{1}{2},\frac{\sqrt 3}{2}\big),\big(\frac{1}{2},\frac{-\sqrt 3}{2}\big)$
The required area is the shaded portion shown in the fig.
The curve $y=\sqrt {1-(x-1)^2}$ moves from x=0 to $ x=\frac{1}{2}.$
The curve $y=\sqrt{1-x^2}$ moves from $\frac{1}{2}$ to 1.
Hence the required area is the sum of the above areas.
since it is symmetrical about x-axis.
$A=2\times\begin{bmatrix}\int_0^\frac{1}{\ 2}\sqrt{1-(x-1)^2}dx+\int_\frac{1}{2}^1\sqrt{1-(x-1)^2}dx\end{bmatrix}$
On integrating we get,
$A=2\begin{bmatrix}\frac{1}{2}.(x-1)\sqrt{1-(x-1)^2}+\frac{1}{2}\sin^{-1}\big(\frac{x-1}{1}\big)\end{bmatrix}_0^\frac{1}{2}+\begin{bmatrix}\frac{x}{2}\sqrt {1-x^2}+\frac{1}{2}\sin^{-1}(x)\end{bmatrix}_\frac{1}{2}^1$
on applying limits we get,
$A=\begin{bmatrix}\frac{-\sqrt 3}{4}+\sin^{-1}\big(\frac{-1}{2}\big)-\sin^{-1}(-1)\end{bmatrix}+\begin{bmatrix}\sin^{-1}(1)-\frac{\sqrt 3}{4}-\sin^{-1}\big(\frac{1}{2}\big)\end{bmatrix}$
$\;\;\;=\begin{bmatrix}\frac{-\sqrt 3}{4}-\frac{\pi}{6}+\frac{\pi}{2}+\frac{\pi}{2}-\frac{\sqrt 3}{4}-\frac{\pi}{6}\end{bmatrix}$
$\;\;\;=\big(\frac{2\pi}{3}-\frac{\sqrt 3}{2}\big)$ 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
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