First Quadrant Portion of an Ellipse

The ellipse can be represented in parametric form as

$r(t) = r_c + v_1 \cos(t) + v_2 \sin(t)$

where $r_c = [3, 4 ,0 ]^T, v_1 = R [3, 0,0]^T, v_2 = R [0, 6,0]^T$

with $R$ being the rotation matrix by $30^{\circ}$ or $\pi / 6$

That is,

$R = \begin{bmatrix} \cos \frac{\pi}{6} && -\sin \frac{\pi}{6} && 0 \\ \sin \frac{\pi}{6} && \cos \frac{\pi}{6} && 0 \\ 0 && 0 && 1 \end{bmatrix}$

Based on Green's Theorem, one can derive the following formula for the area,

$A = \frac{1}{2} \| \oint_C r(t) \times \frac{dr}{dt} dt \|$

The rest is calculating the intervals of the parameter $t$ where $r(t)$ lies in the first quadrant, i.e. has positive $x(t)$ , and $y(t)$ , and performing the above integral. There are two intervals in this case which satisfy this condition, one corresponds to the upper and right part of the ellipse, and the other to the lower part. The other segments of the boundary are along the x-axis and y-axis, and their contribution to the above integral is zero.