The portion of the area enclosed by the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\)=1, that lies in the first quadrant is

\(\frac{πab}{4}\)

Ellipse us symmetrical about both x-axis & y-axis.

∴ Area of ellipse = 4 × Area of OAB.

$= 4 × \int_0^aydx$

given, \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

$\frac{y^2}{b^2}=1-\frac{x^2}{a^2}⇒\frac{y^2}{b^2}=\frac{a^2-x^2}{a^2}⇒y^2=\frac{b^2}{a^2}(a^2-x^2)$

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

Since OAB is in 1st quadrant, so the value of y is positive

$∴y=\frac{b}{a}\sqrt{a^2-x^2}$

Area of ellipse = $= 4 × \int_0^ay.dx⇒4 \int_0^a\frac{b}{a}\sqrt{a^2-x^2}dx$

$⇒\frac{4b}{a} \int_0^a\sqrt{a^2-x^2}dx$

∴ According to the formula → $\sqrt{a^2-x^2}dx=\frac{1}{2}x\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}\frac{x}{a}+C$

$∴ \frac{4b}{a}[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}\frac{x}{a}]_0^a$

$\frac{4b}{a}[(\frac{a}{2}\sqrt{a^2-a^2}+\frac{a^2}{2}sin^{-1}\frac{a}{a})-(\frac{0}{2}\sqrt{a^2-0}+\frac{0^2}{2}sin^{-1}(0))]$

$=\frac{4b}{a}[0+\frac{a^2}{2}sin^{-1}(1)-0-0]⇒\frac{4b}{a}×\frac{a^2}{2}sin^{-1}(1)$

$=2ab×sin^{-1}(1)⇒2ab×\frac{\pi}{2}⇒\pi ab$

Required area = $\pi ab$ square units.

So, option 1 is correct.