Cardiod-oid

We are being asked for the surface of revolution of the cardioid $r = 1 + \cos\theta$ , which is $\begin{aligned} S & = \; 2\pi\int_0^\pi r\sin\theta \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta \; = \; 2\pi\int_0^\pi (1 + \cos\theta)\sin\theta \sqrt{2(1 + \cos\theta)}\,d\theta \\ & = \; 2\pi\sqrt{2}\int_0^\pi (1 + \cos\theta)^{\frac32}\sin\theta\,d\theta \; = \; 2\pi\sqrt{2}\Big[-\tfrac25(1 + \cos\theta)^{\frac{5}{2}}\Big]_0^\pi \; = \; \tfrac{32}{5}\pi \end{aligned}$ making the answer $32+5=\boxed{37}$ .