= Solution
The handle and the two end pieces have disjoint interiors; their common boundary <circles> contribute no <volume>. Hence the total <mass> is $M=M_h+2M_e$. With $\delta=\alpha-\gamma$, the results above give
$$
\boxed{M=2\pi\rho a^3\left[\gamma(\beta^2-\delta^2)+\frac{2\beta^3}{3}+\beta^2\delta-\frac{\delta^3}{3}\right].}
$$
Equivalently, since $\gamma+\delta=\alpha$,
$$
\boxed{M=2\pi\rho a^3\left[\alpha\beta^2+\frac{2\beta^3}{3}-\gamma(\alpha-\gamma)^2-\frac{(\alpha-\gamma)^3}{3}\right].}
$$
Back to article page
