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

The correct answer is $\pi a^2$

Given equation of circle is:
$x^2+y^2=a^2$

⇒ $y^2=a^2-x^2$

⇒ $y=\sqrt{a^2-x^2}$​



Area of circle = 4×Area of first quadrant

=$4 \int_{0}^{a} y$ $dx$

=$4 \int_{0}^{a} \sqrt{a^2-x^2}$ $dx$

=4 $\left[ \dfrac{x}{2} \sqrt{a^2-x^2} + \dfrac{a^2}{2} sin^{-1} \dfrac{x}{a}\right]_{0}^{a}$

=4 $\left[ 0 + \dfrac{a^2}{2} sin^{-1} \dfrac{a}{a} - 0 - \dfrac{a^2}{2} sin^{-1} \dfrac{0}{a}\right]$

=4 $\left[ \dfrac{a^2}{2} sin^{-1} 1 - \dfrac{a^2}{2} sin^{-1} 0\right]$

=4 $\left[ \dfrac{a^2}{2} \dfrac{\pi}{2} \right]$

= $\pi a^2$ sq. units