 Solution

The base is the disk $D:(x-a{)}^2+y^2\le a^2$, on which $x\ge 0$. The vertical thickness is $(p-q)x$. Using the symmetry of the disk about $x=a$, or integrating with $x=a+r\cos \varphi$, givesFor the curved cylindrical surface use $x=a(1+\cos \varphi )$, $y=a\sin \varphi$, with $0\le \varphi \le 2\pi$ and $qx\le z\le px$. The area element is $ad\varphi dz$. Its surface area is thereforeThe sloping planar faces are excluded from this curved surface area.