The smaller area between the ellipse $\frac{x^2}{4}+\frac{y^2}{36}=1$ and the line $\frac{x}{2}+\frac{y}{6}=1$ is given by:

$3 \int\limits_0^2 \sqrt{4-x^2} d x-\int\limits_0^2(6-3 x) d x$

The correct answer is Option (1) → $3 \int\limits_0^2 \sqrt{4-x^2} d x-\int\limits_0^2(6-3 x) d x$

$\frac{x^2}{2^2}+\frac{y^2}{6^6}=1$, $\frac{x}{2}+\frac{y}{6}=1$

we know that area of ellipse is $πab$ if $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

area of quarter ellipse = $\frac{π×2×6}{4}=3π$

area of triangle = $\frac{1}{2}×2×6=6$

required area = $3π-6$

Now using equations

$y=3\sqrt{4-x^2}$ → eq. of ellipse

$y=6-3x$ → eq. of line

so by integration required area = $3 \int\limits_0^2 \sqrt{4-x^2} d x-\int\limits_0^2(6-3 x) d x$