The area enclosed by the curve $x=8 \cos t, y=4 \sin t$ between the lines $x=0$ and $x=4 \sqrt{3}$ is:

$8 \sqrt{3}+\frac{32 \pi}{3}$ units

The correct answer is Option (3) - $8 \sqrt{3}+\frac{32 \pi}{3}$ units

$x=8 \cos t$, $y=4 \sin t$

$\frac{x}{2}=4\cos t$, $y=4\sin t$

so $\frac{x^2}{4}+y^2=4^2(\cos^2t+\sin^2t)$

$⇒\frac{x^2}{2^2}+\frac{y^2}{12}=4^2$

or $\frac{x^2}{8^2}+\frac{y^2}{4^2}=1$, $y=\frac{1}{2}\sqrt{8^2-x^2}$

area I = area II by symmetry

so area = $2×\int\limits_0^{4\sqrt{3}}\frac{1}{2}\sqrt{8^2-x^2}dx$

$=\int\limits_0^{4\sqrt{3}}\sqrt{8^2-x^2}dx$

$=\left[\frac{x}{2}\sqrt{8^2-x^2}+\frac{8^2}{2}\sin^{-1}\frac{x}{8}\right]_0^{4\sqrt{3}}$

$=8\sqrt{3}+\frac{32 \pi}{3}$