The area (in sq. units) of the bigger portion of region enclosed by the curves $4x^2+9y^2 = 36$ and $2x + 3y = 6$ is

$\frac{3}{2} (3π+2)$

The correct answer is Option (2) → $\frac{3}{2} (3π+2)$

Ellipse: $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$; line: $2x+3y=6\Rightarrow y=\frac{6-2x}{3}$.

Intersection points: $(0,2)$ and $(3,0)$.

Smaller segment area (between ellipse and line):

$A_s=\displaystyle\int_{0}^{3}\left(\frac{2}{3}\sqrt{9-x^{2}}-\frac{6-2x}{3}\right)\,dx =\frac{1}{3}\left[2\int_{0}^{3}\sqrt{9-x^{2}}\,dx-\int_{0}^{3}(6-2x)\,dx\right]$

$\int_{0}^{3}\sqrt{9-x^{2}}\,dx=\frac{\pi\cdot 3^{2}}{4}=\frac{9\pi}{4},\quad \int_{0}^{3}(6-2x)\,dx=9$

$A_s=\frac{1}{3}\left(2\cdot\frac{9\pi}{4}-9\right)=\frac{1}{3}\left(\frac{9\pi}{2}-9\right)=\frac{3\pi}{2}-3$

Total ellipse area $=\pi ab=6\pi$.

Bigger portion area $=6\pi-\left(\frac{3\pi}{2}-3\right)=\frac{9\pi}{2}+3$

Final Area = $\frac{9\pi}{2}+3$