Let $\Omega_1, \Omega_2, \Omega_3$ denote three bodies in $R^3$ and assume that we know both the volumes of the intersections $$ |\Omega_1 \cap \Omega_2| = V_{12}, \; |\Omega_1 \cap \Omega_3| = V_{13}, \;|\Omega_2 \cap \Omega_3| = V_{23}, \;|\Omega_1 \cap \Omega_2 \cap \Omega_3| = V_{123} $$ and the corresponding surface areas $$ |\partial \Omega_{12}| = S_{12}, \; |\partial \Omega_{13} | = S_{13}, \;|\partial \Omega_{23}| = S_{23}, \; |\partial \Omega_{123} |=S_{123}. $$
What is the volume and surface area of the union of the three bodies?
Using the analogy with a Venn diagram of three intersecting disks, the volume and surface area seem to be given by $$ |\Omega_1 \cup \Omega_2 \cup \Omega_3| = |\Omega_1| + |\Omega_2| + |\Omega_3| - V_{12} - V_{13} - V_{23} + V_{123}$$ and $$ |\partial(\Omega_1 \cup \Omega_2 \cup \Omega_3)| = |\partial \Omega_1| + |\partial \Omega_2| + |\partial \Omega_3| - S_{12} - S_{13} - S_{23} + S_{123}.$$ Is this statement true?