Using integration the area enclosed by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is equal to :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $πab$
By symetry
$I=II=III=IV$
Required area = $4×I$
from eq.
$y=\frac{b}{a}\sqrt{a^2-x^2}$
so area = $4\frac{b}{a}\int\limits_0^a\sqrt{a^2-x^2}dx$
$=4\frac{b}{a}\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_0^a$
$=4\frac{b}{a}×\frac{πa^2}{2×2}=πab$