= Solution
Apply the same dimensionless transformation independently to the two cups, giving $X,Y\in[-\pi/2,\pi/2]$. The total-volume condition becomes $X+Y\geq\pi/2$. Since neither variable can exceed $\pi/2$, this region has $0\leq X\leq\pi/2$ and $\pi/2-X\leq Y\leq\pi/2$, the triangle shown above.
By <independence of random variables>, the <joint probability density> is $\tfrac14\cos x\cos y$. The desired <double integral> is
$$
\begin{aligned}
P&=\frac14\int_0^{\pi/2}\cos x\int_{\pi/2-x}^{\pi/2}\cos y\,dy\,dx\\
&=\frac14\int_0^{\pi/2}\cos x(1-\cos x)\,dx\\
&=\frac14\left[1-\int_0^{\pi/2}\frac{1+\cos2x}{2}\,dx\right]
=\frac14-\frac\pi{16}.
\end{aligned}
$$
Thus
$$
\boxed{\mathbb P(v_1+v_2\geq630\,\mathrm{mL})=\frac{4-\pi}{16}.}
$$
Boundary equality has <probability> zero for these continuous random variables.
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