Oriented surface element of an ellipsoid of revolution ($d\mathbf{S}=(ab{\sin }^2\theta \cos \varphi ,ab{\sin }^2\theta \sin \varphi ,b^2\sin \theta \cos \theta )d\theta d\varphi$)

Parametrize the ellipsoid by $\mathbf{x}=(b\sin \theta \cos \varphi ,b\sin \theta \sin \varphi ,a\cos \theta )$, with positive semiaxes $a,b$, $0\le \theta \le \pi$ and $0\le \varphi <2\pi$. The cross product ${\mathbf{x}}_{\theta }\times {\mathbf{x}}_{\varphi }$ gives the displayed outward vector area element. For $0<\theta <\pi$, it is the positive multiple $a{b}^2\sin \theta$ of the outward normal direction $(x/b^2,y/b^2,z/a^2)$, fixing its orientation. A flux integral can therefore be evaluated directly as $\int \int \mathbf{F}(\mathbf{x})\cdot ({\mathbf{x}}_{\theta }\times {\mathbf{x}}_{\varphi })d\theta d\varphi$, without computing the length of the normal separately. The scalar surface area element is the Euclidean norm of this vector element.