= Solution
The base is the disk $D:(x-a)^2+y^2\leq a^2$, on which $x\geq0$. The vertical thickness is $(p-q)x$. Using the symmetry of the disk about $x=a$, or integrating with $x=a+r\cos\phi$, gives
$$
V=(p-q)\iint_Dx\,dA=(p-q)a\,\pi a^2,
\qquad \boxed{V=\pi a^3(p-q).}
$$
For the curved cylindrical surface use $x=a(1+\cos\phi)$, $y=a\sin\phi$, with $0\leq\phi\leq2\pi$ and $qx\leq z\leq px$. The <area element> is $a\,d\phi\,dz$. Its <surface area> is therefore
$$
A=\int_0^{2\pi}a(p-q)a(1+\cos\phi)\,d\phi,
\qquad \boxed{A=2\pi a^2(p-q).}
$$
The sloping planar faces are excluded from this curved <surface area>.
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