Browse Questions

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

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

Toolbox:
• 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,
$\;\;\;=\begin{bmatrix}\frac{a}{2}\sqrt{a^2-a^2}+\frac{a^2}{2}\sin^{-1}\big(\frac{a}{a}\big)-\frac{a}{2\sqrt2}\sqrt{a^2-\big(\frac{a}{\sqrt 2}\big)^2}-\frac{a^2}{2}\sin^{-1}\big(\frac{a}{a\sqrt 2}\big)\end{bmatrix}$
$\Rightarrow A=\frac{a^2}{2}\big(\frac{\pi}{2}\big)-\frac{a}{2\sqrt 2}.\frac{a}{\sqrt 2}-\frac{a^2}{2}\big(\frac{\pi}{4}\big)$
$\;\;\;\;\;\;\;=\frac{\pi a^2}{4}-\frac{a^2}{4}-\frac{\pi a^2}{8}$
$\;\;\;\;\;\;\;=\frac{a^2}{4}\begin{bmatrix}\pi-1-\frac{\pi}{2}\end{bmatrix}$
$\;\;\;\;\;\;\;=\frac{a^2}{4}\begin{bmatrix}\frac{\pi}{2}-1\end{bmatrix}$ sq.units.
Hence the required area is $\frac{a^2}{2}\begin{bmatrix}\frac{\pi}{2}-1\end{bmatrix}$sq.units.