Lateral surface area of an oblique elliptical cone

Let the apex be at point $A$ . And the base is represented parametrically as $b(t) = v_0 + v_1 \cos t + v_2 \sin t$ . In this problem $A = (0, 0, 12) , v_0 = (-4, 0, 0) , v_1 = (4, 0, 0), v_2 = (0, 4, 0)$ .

Then points on the lateral surface of the cone, are given by

$p(t, s) = A + s (b(t) - A) = A + s ( v_0 - A + v_1 \cos t + v_2 \sin t )$ where $t \in [0, 2\pi)$ and $s \in [0, 1]$ .

The surface area integral for such a parametrically defined surface is given by,

$S = \displaystyle \iint_{t, s} | p_s \times p_t | dt ds$

where $p_s = \dfrac{\partial p }{\partial s }$ , and $p_t = \dfrac{\partial p}{\partial t}$ . Hence, we have,

$p_s = v_0 - A + v_1 \cos t + v_2 \sin t$ and $p_t = s ( - v_1 \sin t + v_2 \cos t )$ . Their cross product is,

$p_s \times p_t = s ( - ((v_0 - A) \times v_1 ) \sin t + ( (v_0 - A) \times v_2 ) \cos t + v_1 \times v_2 )$

Integrating with respect to $s$ introduces a factor of $\frac{1}{2}$ , and the integral reduces to the one-dimensional integral,

$S = \frac{1}{2} \displaystyle \int_{0}^{2 \pi} | - ((v_0 - A) \times v_1 ) \sin t + ( (v_0 - A) \times v_2 ) \cos t + v_1 \times v_2 |\hspace{8pt} dt$

This integral can be evaluated numerically, and the final answer is $S \approx 162.5$