Cardioid

Consider a 2d cardioid in the xz-plane rotating about the z-axis. When $\vec C$ describes the surface, we get in a spherical coordinate system:

(Note: $r_{\mathrm v}$ denotes the variable from our coordinate system, not $r=r(\theta)$ . (We don't need $r_{\mathrm v}$ for the surface)

\begin{gather*} r(\theta)=1-\cos(\theta)\\ \vec C=r(\theta)\begin{pmatrix} \sin(\theta)\cos(\varphi)\\\sin(\theta)\sin(\varphi)\\\cos(\theta) \end{pmatrix} \end{gather*}

Lets calculate the surface $A$ :

$\begin{aligned} \mathrm dA&=|C_{\theta}\times C_{\varphi}|\mathrm d\theta\mathrm d\varphi\\ \vec C_{\theta}&=r_{\theta}\begin{pmatrix} \sin(\theta)\cos(\varphi)\\\sin(\theta)\sin(\varphi)\\\cos(\theta) \end{pmatrix}+r\begin{pmatrix} \cos(\theta)\cos(\varphi)\\\cos(\theta)\sin(\varphi)\\-\sin(\theta) \end{pmatrix}~~,~~ C_{\varphi}=r\begin{pmatrix} -\sin(\theta)\sin(\varphi)\\\sin(\theta)\cos(\varphi)\\0 \end{pmatrix}\\ r_\theta&=\sin(\theta)\\ P&=C_{\theta}\times C_{\varphi}=rr_\theta\begin{pmatrix} 0-\cos(\theta)\sin(\theta)\cos(\varphi)\\-\cos(\theta)\sin(\theta)\sin(\varphi)-0\\\sin^2(\theta)\cos^2(\varphi)+\sin^2(\theta)\sin^2(\varphi) \end{pmatrix}+r^2\sin(\theta)\begin{pmatrix} \sin(\theta)\cos(\varphi)\\\sin(\theta)\sin(\varphi)\\\cos(\theta) \end{pmatrix}\\ &=r\begin{pmatrix} \sin(\theta)cos(\varphi)\left(-\sin(\theta)\cos(\theta)+\sin(\theta)-\cos(\theta)\sin(\theta)\right)\\ \sin(\theta)\sin(\varphi)\left(-\sin(\theta)\cos(\theta)+\sin(\theta)-\cos(\theta)\sin(\theta)\right)\\ \sin^3(\theta)+\sin(\theta)\cos(\theta)-\sin(\theta)\cos^2(\theta) \end{pmatrix}\\ \frac{P^2}{r^2}&=\left(-2\sin(\theta)\cos(\theta)+\sin(\theta)\right)^2\sin^2(\theta)+\sin^2(\theta)\left(\sin^2(\theta)+\cos(\theta)-\cos^2(\theta)\right)^2\\ &=\sin^2(\theta)\left[4\sin^2(\theta)\cos^2(\theta)+\sin^2(\theta)-4\sin^2(\theta)\cos(\theta)+\sin^4(\theta)+\cos^2(\theta)\right.\\ &~~~~~~~~~\left.~~~~~~+2\sin^2(\theta)\cos(\theta)+\cos^4(\theta)-2\sin^2(\theta)\cos^2(\theta)-2\cos^3(\theta)\right]\\ &=\sin^2(\theta)\left[2\sin^2(\theta)\cos^2(\theta)+\sin^2(\theta)-2\sin^2(\theta)\cos(\theta)+\sin^4(\theta)+\cos^2(\theta)\right.\\ &~~~~~~~~~\left.~~~~~~+\cos^4(\theta)-2\cos^3(\theta)\right]\\ &=\sin^2(\theta)\left[1^2+1-2\sin^2(\theta)\cos(\theta)-2\cos^3(\theta)\right]\\ &=2\sin^2(\theta)\left[1-\cos(\theta)\right]=4\sin^2(\theta)\sin^2\left(\frac{\theta}{2}\right)\\ \mathrm dA&=|P|\mathrm d\theta\mathrm d\varphi=2\left(\sin(\theta)\sin\left(\frac{\theta}{2}\right)-\cos(\theta)\sin(\theta)\sin\left(\frac{\theta}{2}\right)\right)\mathrm d\theta\mathrm d\varphi \end{aligned}$

\begin{gather*} \int\sin(x)\sin(x/2)\mathrm dx=\frac 43\sin^3(x/2)\\ \int\cos(x)\sin(x)\sin(x/2)\mathrm dx=\frac 4{15}\sin^3(x/2)\left(3\cos(x)+2\right)\\ \Rightarrow A=2\pi\int_0^\pi|P|\mathrm d\theta=4\pi\left[\frac 43-\frac 4{15}(-3+2)\right]=\pi\frac{2^5}{5} \end{gather*}

$\\$

Lets calculate the volume $V$ too:

Note: $r_{\mathrm v}$ denotes a variable from our spherical coordinate system, and $r=r(\theta)$ denotes the bound of integration.

$\begin{aligned} \mathrm dV&=r_{\mathrm v}^2\sin(\theta)\mathrm dr_v\mathrm d\theta\mathrm d\varphi\\ V&=\int\limits_0^\pi\int\limits_0^{r(\theta)}\int\limits_0^{2\pi}r_{\mathrm v}^2\sin(\theta)\mathrm d\varphi\mathrm dr_{\mathrm v}\mathrm d\theta=\frac{2\pi}{3}\int\limits_0^\pi r^3(\theta)\sin(\theta)\mathrm d\theta\\ r^3(\theta)&=8\sin^6\left(\frac{\theta}{2}\right)=\frac{1}{4}\left(10-15\cos(\theta)+6\cos(2\theta)-\cos(3\theta)\right)\\ V&=\frac{\pi}{6}\int\limits_0^\pi \left(10\sin(\theta)-15\cos(\theta)\sin(\theta)+6\cos(2\theta)\sin(\theta)-\cos(3\theta)\sin(\theta)\right)\mathrm d\theta \end{aligned}$

\begin{gather*} \int\cos(x)\sin(x)\mathrm dx=-\frac 12\cos^2(x)\\ \int\cos(2x)\sin(x)\mathrm dx=\frac 1{6}\left(3\cos(x)-\cos(3x)\right)\\ \int\cos(3x)\sin(x)\mathrm dx=\frac 1{8}\left(4\cos^2(x)-\cos(4x)\right)\\ \Rightarrow V=\frac{\pi}{6}\left[-10\cdot(-2)-15\cdot(0)+(-2-2)-\frac 18\cdot(0)\right]=\pi\frac{2^3}{3} \end{gather*}

So: $A+B=\boxed{8}$