Question

Find the area of the circle $x^2+y^2=a^2$ by integration.

Answer 1

Given equation of circle is:
$x^2+y^2=a^2$
$\Rightarrow y^2=a^2-x^2$
$\Rightarrow y=\sqrt{a^2-x^2}$
$\therefore$ Area of circle $=4\times$Area of first quadrant
$=4{\int }_0^{a}ydx$
$=4{\int }_0^a\sqrt{a^2-x^2}dx$
$=4{\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}\frac{x}{a}\right]}_0^a$
$=4\left[0+\frac{a^2}{2}{\sin }^{-1}\frac{a}{a}-\left(0+\frac{a^2}{2}{\sin }^{-1}\frac{0}{a}\right)\right]$
$=4\left[\frac{a^2}{2}{\sin }^{-1}1-0\right]$
$=2a^2\cdot \frac{\pi }{2}$
$=\pi a^2$ square units.