Find the area of the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ by integration.

The correct answer is $\pi ab$

Given equation of ellipse is:
$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

⇒ $\dfrac{y^2}{b^2}=1 - \dfrac{x^2}{a^2}$

⇒ $\dfrac{y^2}{b^2}=\dfrac{a^2 - x^2}{a^2}$

⇒ $y^2=\dfrac{b^2}{a^2}(a^2 - x^2)$

⇒ $y= \pm \dfrac{b}{a} \sqrt{(a^2 - x^2)}$



As AOB is in 1st quadrant, value if $y$ is positive.
∴ $y= \dfrac{b}{a} \sqrt{(a^2 - x^2)}$


Area of ellipse = 4×Area of first quadrant

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

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

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

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

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

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

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

= $\pi ab$ sq. units