= Solution
For positive semiaxes $a,b$, the <cross product> of the two parametric <tangent vectors> gives the outward <oriented surface element of an ellipsoid of revolution>:
$$
\mathbf x_\theta\times\mathbf x_\phi=(ab\sin^2\theta\cos\phi,\ ab\sin^2\theta\sin\phi,\ b^2\sin\theta\cos\theta).
$$
Substituting the parametrization into the <vector field> and taking its <dot product> with this element gives
$$
\mathbf G\cdot(\mathbf x_\theta\times\mathbf x_\phi)=a^3b^2\sin^3\theta\cos^2\theta\cos^2\phi+ab^4\sin^5\theta\cos\phi\sin^3\phi+a^3b^2\sin\theta\cos^4\theta.
$$
The middle term has zero <integral> over $0\leq\phi\leq2\pi$. In the other terms, $\int_0^{2\pi}\cos^2\phi\,d\phi=\pi$, $\int_0^\pi\sin^3\theta\cos^2\theta\,d\theta=4/15$, and $\int_0^\pi\sin\theta\cos^4\theta\,d\theta=2/5$. The directly evaluated <flux> is therefore
$$
\boxed{\int_S\mathbf G\cdot d\mathbf S=a^3b^2\left(\frac{4\pi}{15}+\frac{4\pi}{5}\right)=\frac{16\pi a^3b^2}{15}.}
$$
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