Let
\[B\]
be a uniform density region entirely above pr below the \[x\]
axis. If \[B\]
is rotated about the \[x\]
axis then the volume of the solid generated is \[V= 2 \pi S y_c\]
where \[S\]
is the area of \[B\]
and \[(x_c , y_c )\]
is the centroid of \[B\]
Proof The volume of the solid generated by revolving
\[B\]
about the \[x\]
axis is\[\begin{equation} \begin{aligned} V &= \int^b_a \pi y^2_2 \: dx - \int^b_a \pi y^2_1 \: dx \\ &= - \pi (\int^b_a y^2_1 \: dx + \int^a_b y^2_2 \: dx) \\ &= - \pi ( \int_{C_1} y^2 \: dx + \int_{C_2} y^2 \: dx ) \\ &= - \pi \oint_C y^2 \: dx\end{aligned} \end{equation}\]
Apply Green's Theorem with
\[P= \pi y^2, \: Q=0\]
to give\[V = - \pi \oint_C y^2 \: dx =2 \pi \int_B y \: dx \; dy \]
The centre of mass of the centroid is
\[(x_c , y_c ) =(\frac{ \int_B x \rho \: dx \: dy}{M} , \frac{ \int_B y \rho \: dx \: dy}{M} ) =(\frac{\int_B x \: dx \:dy}{S} , \frac{\int_B y \: dx \:dy}{S} ) \]
if the lamina has uniform density.
Then
\[V= 2 \pi S y_c\]