Solution
= Solution
Use <cylindrical coordinates> $(r,\phi,z)$ about the dumbbell axis. Their <Jacobian determinant> contributes the volume factor $r$, so the handle's <mass> is the <volume integral>
$$
M_h=\rho\int_{-\gamma a}^{\gamma a}\int_0^{2\pi}\int_0^H r\,dr\,d\phi\,dz=\rho(2\gamma a)(2\pi)\frac{H^2}{2}.
$$
Inserting the <radius> from part (a) yields
$$
\boxed{M_h=2\pi\rho\gamma a^3\left[\beta^2-(\alpha-\gamma)^2\right].}
$$