The centers of the spheres are 10 units apart.
$$R_1 = 8 \qquad R_2 = 4$$
I've come up with
$$64 = a^2 + r^2 \text{ and } 16 = b^2 + r^2 \Longrightarrow 10 = a + b$$
Therefore $a = 10 - b$, this leads to
$$64 = 100 - 20b + b^2 + r^2 \Longrightarrow 16 = b^2 + r^2$$
Leading to
$$b = 2.6 \quad a = 7.4 \quad r^2 = 9.24$$
From there $h_1 = 0.6$ and $h_2 = 1.4$
Following the formula $V = \dfrac{\pi h}{6} (3r^2 + h^2$) this gives the total volume
$V = \dfrac{\pi h_1}{6}(3r^2 + h_1^2) + \dfrac{\pi h_2}{6}(3r^2 + h_2^2)$
$V = \dfrac{\pi \cdot0.6}{6}(3\cdot9.24 + 0.6^2) + \dfrac{\pi \cdot 1.4}{6}(3\cdot9.24 + 1.4^2)$
Which results in $V \approx 30.58$
which is half of what the result should be.
