The area of the smaller region bounded by the ellipse $\frac{x^2}{16}+\frac{y^2}{9}= 1$ and the straight line $3x + 4y = 12$ is:

$(3\pi-6)$ sq units

The correct answer is Option (4) → $(3\pi-6)$ sq units

$\text{Ellipse: }\frac{x^{2}}{16}+\frac{y^{2}}{9}=1,\;\text{ line: }3x+4y=12\Rightarrow y=3-\frac{3}{4}x.$

$\text{Intersection points: }(4,0),(0,3).$

$\text{Smaller area between arc and chord }= \displaystyle\int_{0}^{4}\Big(3\sqrt{1-\frac{x^{2}}{16}}-\big(3-\frac{3}{4}x\big)\Big)\,dx.$

$\displaystyle\int_{0}^{4}3\sqrt{1-\frac{x^{2}}{16}}\,dx:\; x=4\sin t,\;dx=4\cos t\,dt\Rightarrow 3\!\int_{0}^{\pi/2}\!4\cos^{2}t\,dt=12\cdot\frac{\pi}{4}=3\pi.$

$\displaystyle\int_{0}^{4}\!\left(3-\frac{3}{4}x\right)\!dx=\left[3x-\frac{3}{8}x^{2}\right]_{0}^{4}=12-6=6.$

$\text{Required area}=3\pi-6=3(\pi-2).$