Ask Questions, Get Answers

Home  >>  CBSE XII  >>  Math  >>  Application of Integrals

Find the area of the smaller part of the circle \(x^2 + y^2 = a^2\) cut off by the line \( x =\Large { \frac{a}{\sqrt 2}}.\)

$\begin{array}{1 1} \large \frac{a^2}{2}\begin{bmatrix}\frac{\pi}{2}-1\end{bmatrix}sq.units. \\\large \frac{a^3}{3}\begin{bmatrix}\frac{\pi}{2}-1\end{bmatrix}sq.units . \\ \large \frac{a^2}{2}\begin{bmatrix}\frac{\pi}{4}-1\end{bmatrix}sq.units. \\ \large \frac{a^2}{2}\begin{bmatrix}\frac{\pi}{\sqrt 2}-1\end{bmatrix}sq.units. \end{array} $

Download clay6 mobile app

1 Answer

  • Area of the region bounded between a curve and a line is given by \[A=\int_a^by\;dx=\int_a^bf(x)\;dx\]
  • where a and b are the point of intersection of the line and the curve.
  • The point of intersection can be found by solving the two equations.
The area of the required region is the smaller part of the circle$x^2+y^2=a^2$,cut off by the line $x=\frac{a}{\sqrt 2}$
This is the shaded portion shown in the fig.
clearly the point of intersection is $(\frac{a}{\sqrt 2},\pm\frac{a}{\sqrt 2})$
Hence the required area is\[A=\int_\frac{a}{\sqrt 2}^ay \;dx=\int_\frac{a}{\sqrt 2}^a\sqrt {a^2-x^2} \;dx\]\[(x^2+y^2=a^2\Rightarrow y=\sqrt{a^2-x^2})\]
on integrating we get,
$\;\;\;=\begin{bmatrix}\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\big(\frac{x}{a}\big)\end{bmatrix}_\frac{a}{\sqrt 2}^a$
on applying limits we get,
$\;\;\;=\begin{bmatrix}\frac{a}{2}\sqrt{a^2-a^2}+\frac{a^2}{2}\sin^{-1}\big(\frac{a}{a}\big)-\frac{a}{2\sqrt2}\sqrt{a^2-\big(\frac{a}{\sqrt 2}\big)^2}-\frac{a^2}{2}\sin^{-1}\big(\frac{a}{a\sqrt 2}\big)\end{bmatrix}$
$\Rightarrow A=\frac{a^2}{2}\big(\frac{\pi}{2}\big)-\frac{a}{2\sqrt 2}.\frac{a}{\sqrt 2}-\frac{a^2}{2}\big(\frac{\pi}{4}\big)$
$\;\;\;\;\;\;\;=\frac{\pi a^2}{4}-\frac{a^2}{4}-\frac{\pi a^2}{8}$
$\;\;\;\;\;\;\;=\frac{a^2}{4}\begin{bmatrix}\frac{\pi}{2}-1\end{bmatrix}$ sq.units.
Hence the required area is $\frac{a^2}{2}\begin{bmatrix}\frac{\pi}{2}-1\end{bmatrix}$sq.units.
answered Dec 20, 2013 by yamini.v

Related questions

Ask Question