Volume of Common Region

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Let's work in 2D with the cross section and then revolve it to get the solid volume of intersection.

Center distance = $\sqrt{(10-2)^2+(12-4)^2+(18-8)^2} = \sqrt{64+64+100} = \sqrt{228}$

$\cos \alpha = \frac{8^2+8^2-10^2}{2(8)(\sqrt{228}}$ and $10 \sin \beta = 8 \sin \alpha$

Giving angles: $\alpha = 37.37$ and $\beta = 29.05$

Areas: $A_1 = 8^2 (\frac{\alpha}{2}-\frac{\sin 2\alpha}{4}) = 5.436$ and $A_2 = 10^2 (\frac{\beta}{2}-\frac{\sin 2\beta}{4})=4.1269$

Centroids: $h_1 = \frac{2R}{3}\frac{2-2\cos^3\alpha-3\cos\alpha \sin^2 \alpha}{2 \alpha \sin 2\alpha} = 1.8487$ $h_2 = \frac{2R}{3}\frac{2-2\cos^3\beta-3\cos\beta \sin^2 \beta}{2 \beta \sin 2\beta} = 1.8372$

Combined Centroid = $h = \frac{A_1 h_1 + A_2 h_2}{A_1 + A_2} = 1.8438$

Volume = $2 \pi h (A_1 + A_2) = 110.78$