Calculating the surface of revolution of a cardioid. I have the cardioid $r=1+cos(t)$ for $0\leq{t}\leq{2\pi}$ and I want to calculate the surface of revolution of said curve. How can I calculate it?
The parematrization of the cardioid is:
$$x(t)=(1+cos(t))cos(t)$$
$$y(t)=(1+cos(t))sin(t)$$
and 
$$\frac{dx}{dt}=\left(-2\cos\left(t\right)-1\right)\sin\left(t\right)$$
$$\frac{dy}{dt}=\cos\left(t\right)\left(\cos\left(t\right)+1\right)-\sin^2\left(t\right)
$$
To calculate the surface of revolution I know I can use the formula (since I want to revolve it around the x-axis) $$2\pi \int_{a}^{b} y(t)\sqrt{\bigg(\frac{dx}{dt}\bigg)^2+\bigg(\frac{dy}{dt}\bigg)^2}  dt$$
However, when I do that integral, the result is $0$ (using online calculators). Why is this wrong? Obviously the surface can't be 0, right? What's the correct answer? 
 A: First a tip: it is often easier to substitute back for $r$ in terms of $t$ after differentiating.  We have
$$\displaylines{
  x=r\cos t\ ,\quad y=r\sin t\cr
  \frac{dx}{dt}=\frac{dr}{dt}\cos t-r\sin t\ ,\quad
    \frac{dy}{dt}=\frac{dr}{dt}\sin t+r\cos t\cr
  \eqalign{\biggl(\frac{dx}{dt}\biggr)^2+\biggl(\frac{dy}{dt}\biggr)^2
    &=\biggl(\frac{dr}{dt}\biggr)^2+r^2\cr
    &=\sin^2t+(1+\cos t)^2\cr
    &=2+2\cos t\cr
    &=4\cos^2\Bigl(\frac t2\Bigr)\cr}\cr
  \sqrt{\biggl(\frac{dx}{dt}\biggr)^2+\biggl(\frac{dy}{dt}\biggr)^2}
    =2\Bigl|\cos\Bigl(\frac t2\Bigr)\Bigr|\ .\cr}$$
Note the absolute value - omitting this is often a reason why answers come out wrong.  To find the area of the surface we have to integrate from $0$ to $\pi$, because rotating just the top half of the cardioid gives the whole surface:
$$\eqalign{A
  &=2\pi \int_0^\pi y\sqrt{\bigg(\frac{dx}{dt}\bigg)^2
          +\bigg(\frac{dy}{dt}\bigg)^2}\ dt\cr
  &=4\pi \int_0^\pi (1+\cos t)\sin t\Biggl|\cos\Bigl(\frac t2\Bigr)\biggr|\ dt\cr
  &=8\pi\int_0^{\pi/2} (1+\cos2u)\sin2u\,|\cos u|\,du\cr
  &=32\pi\int_0^{\pi/2} \cos^4u\sin u\,du\cr
  &=\frac{32\pi}5\ .\cr}$$
A: Pappus's $(1^{st})$ Centroid Theorem states that the surface area $A$ of a surface of revolution generated by rotating a plane curve $C$ about an axis external to $C$and on the same plane is equal to the product of the arc length $s$ of $C$ and the distance d traveled by its geometric centroid. Simply put, $S=2\pi RL$, where $R$ is the normal distance of the centroid to the axis of revolution and $L$ is curve length. The centroid of a curve is given by
$$\mathbf{R}=\frac{\int \mathbf{r}ds}{\int ds}=\frac{1}{L} \int \mathbf{r}ds$$
Second, I'll demonstrate how to solve this in the complex plane. And finally, I'll carry out the solution for an arbitrary cardioid given by
$$z=2a(1+\cos t)e^{it}$$
In the complex plane, the surface area of a curve rotated about the $x$-axis is given by
$$S=2\pi\int \Im\{z\}|\dot z| du,\quad z=z(u)$$
Thus we have
$$
\begin{align}
&z=2a(1+\cos t)e^{it},\quad \theta\in[0,\pi]\\
&\Im\{z\}=2a(1+\cos t)\sin t\\
&\dot z=2a\big( (1+\cos t)i-\sin t\big)e^{it}\\
&|\dot z|=2\sqrt{2}a\sqrt{1+\cos t}\\
\end{align}
$$
So that finally
$$
\begin{align}
S
&=2\pi\int_0^{\pi}2a(1+\cos t)\sin t ~2\sqrt{2}a\sqrt{1+\cos t}~dt\\
&=8\sqrt{2}\pi a^2\int_0^{\pi}\sin t (1+\cos t)^{3/2}~dt\\
&=8\sqrt{2}\pi a^2 \frac{8\sqrt{2}}{5}\\
&=\frac{128\pi a^2}{5}\\
\end{align}
$$
This has been tested numerically for arbitrary $a$ and, of course, agrees with @David for $a=1/2$. See alos my solution for the volume of said cardioid of revolution here
