= Oriented surface element of an ellipsoid of revolution
{title2=$d\mathbf S=(ab\sin^2\theta\cos\phi,ab\sin^2\theta\sin\phi,b^2\sin\theta\cos\theta)d\theta d\phi$}
Parametrize the <ellipsoid> by $\mathbf x=(b\sin\theta\cos\phi,b\sin\theta\sin\phi,a\cos\theta)$, with positive semiaxes $a,b$, $0\leq\theta\leq\pi$ and $0\leq\phi<2\pi$. The <cross product> $\mathbf x_\theta\times\mathbf x_\phi$ gives the displayed outward vector <area element>. For $0<\theta<\pi$, it is the positive multiple $ab^2\sin\theta$ 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 $\int\int\mathbf F(\mathbf x)\cdot(\mathbf x_\theta\times\mathbf x_\phi)d\theta d\phi$, without computing the length of the normal separately. The scalar <surface area> element is the <Euclidean norm> of this vector element.
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