Volume of revolution of cardioid The parametric equations of a cardioid are $x=\cos\theta (1-\cos\theta)$ and $y=\sin\theta (1-\cos\theta)$, $0\le\theta\le 2\pi$. Diagram here. The region enclosed by the cardioid is rotated about the x-axis, find the volume of the solid. I am not allowed to use polar form, or double integrals, due to the limitations of the NSW mathematics syllabus. Basically I'm stuck with using a disk approximation. What I've got so far using the disk approximation is:
$\lim_{\delta x \to 0} \sum_{x=-2}^0 \pi y^2\delta x =\pi\int_{-2}^0 y^2 dx = \pi\int_{\pi}^{\frac\pi 2} \sin^2\theta(1-\cos\theta)^2 (2\cos\theta\sin\theta - \sin\theta) d\theta $ which gives the volume of the portion from x= -2 to 0, but I have no idea how to calculate the volume from x = 0 to 1/4. My gut feeling says to just extend the bounds of the previous integral to $\pi$ and 0, but because there exists two y-values for each x value in that domain, shouldn't I have to subtract the volume of the larger disk from the smaller disk i.e. form an annulus?
 A: (after some wrangling with trying to find a way to eliminate the parameter):
I think that, for the region $ \ x = 0 \ \text{to} \ x = \frac{1}{4} \ , \ $ you do want a subtractive approach, but it won't be through constructing an annulus. In order to manage that, you'd have to find the two angles which give the same $ \ x \ $ value and then calculate the values of $ \ y \ $ corresponding to those angles in order to get the radii for each annulus.
Instead, use the fact that the "switch-over" at $ \ x =  \frac{1}{4} \ $ from the "lower" portion of the cardioid  to the "upper" portion occurs at $ \ \theta =  \frac{\pi}{3} \ . $  You will want to find the volume $ \ \pi \ \int \ y^2 \ dx \ $ running from $ \ \theta =  \frac{\pi}{3} \ \ \text{to} \ \ \theta =  \frac{\pi}{2} \ $ , and then subtract off the volume $ \ \pi \ \int \ y^2 \ dx \ $ running from $ \ \theta =  0 \ \ \text{to} \ \ \theta =  \frac{\pi}{3} \ . $
EDIT (3/15) --  [Now that I've had a chance to come back to this one]
Here's a graph of the situation:

Since you are producing disks centered on the $ \ x-$ axis, your basic integral $ \ \pi \ \int \  y^2 \ dx \ $ is correct. The complication in interpreting the curve is that the cardioid is being expressed in terms of the so-called "angle parameter", which is not the same as the use of angle in polar coordinates.  So it is possible to have multiple values of a coordinate variable in terms of $ \ \theta \ $ , a common situation with parametric curves.  
On the interval $ \ \frac{\pi}{2} \ \le \theta \ \le \ \pi \ , $ corresponding to $ \ x = 0 \ $ to $ \ x = -2 \ , $ there is no difficulty, so that portion of the volume of revolution [the section in orange] can be covered by $ \ \pi \ \int_{-2}^0 \  y^2 \ dx \ $  $ = \  \ \pi \ \int_{\pi/2}^{\pi} \  [y(\theta)]^2 \ \cdot  \frac{dx}{d\theta} \  d\theta \ \ , $ as you'd already done.  
What I am describing is a way to deal with the difficulty of the value of $ \ y  \ $ not being unique on the interval $ \ x = 0 \ $ to $ \ x = \frac{1}{4} \ . $  The maximum value of $ \ x \ $ does occur at $ \ x = \frac{1}{4} \ , $ so we need to break the integration at the corresponding value of the angle-parameter, $ \ \theta = \frac{\pi}{3} \ . $  
Since the disk radii run from the curve down to the $ \ x-$ axis, we can start with summing the volumes of those disks over the interval in $ \ x \ $ by using $ \ \pi \ \int_{0}^{1/4} \  y^2 \ dx \ $  $ = \  \ \pi \ \int_{\pi/3}^{\pi/2} \  [y(\theta)]^2 \ \cdot  \frac{dx}{d\theta} \  d\theta \ \ . $ [error -- see below]  That covers the sections in both green and blue.  But the blue section is not part of the interior of the solid of revolution, so we must now subtract the disks over the same interval in $ \ x \ , $ which are represented by the interval in angle-parameter $ \ 0 \ \le \ \theta \ \le \ \frac{\pi}{3} \ . $
Edit by robtob (3/16)
The integral should be $$  \pi \ \left[ \ \int_{\pi}^{\pi/2} \  [y(\theta)]^2 \ \cdot  \frac{dx}{d\theta} \  d\theta \ + \ \int_{\pi/2}^{\pi/3} \  [y(\theta)]^2 \ \cdot  \frac{dx}{d\theta} \  d\theta \ - \ \int_{0}^{\pi/3} \  [y(\theta)]^2 \ \cdot  \frac{dx}{d\theta} \  d\theta \ \right] \ \ $$ $=$ $$  \pi \ \left[ \ \int_{\pi}^{\pi/2} \  [y(\theta)]^2 \ \cdot  \frac{dx}{d\theta} \  d\theta \ + \ \int_{\pi/2}^{\pi/3} \  [y(\theta)]^2 \ \cdot  \frac{dx}{d\theta} \  d\theta \ + \ \int_{\pi/3}^0 \  [y(\theta)]^2 \ \cdot  \frac{dx}{d\theta} \  d\theta \ \right] \ \ $$  $=$ $$\pi \ \left[ \ \int_{\pi}^0 \  [y(\theta)]^2 \ \cdot  \frac{dx}{d\theta} \  d\theta \right]$$
as the bounds run from left to right, so $\pi$ to $\pi/2$ and $\pi/2$ to $\pi/3$ are the appropriate bounds when converted to parametric form.
Acknowledgment of error and additional material (3/17)
I thank robtob for the correction.  For some reason, I had led myself to believe that there was a minus sign in the integration that would reverse the direction in which the cardioid would be covered by the integration by the angle parameter.  The blue region about which I was concerned is correctly canceled by the integration straight through with $ \ \theta \ $ decreasing from $ \ \pi \ $ to 0 .  (Had I graphed the integrand at the time -- see below -- I would have spotted my mistake then...)
The integrand proves to be (after some time spent with product-to-sum formulas)
$$ -\frac{11}{8} \sin \theta \ + \ \frac{19}{16} \sin 2 \theta \ - \ \frac{1}{16} \sin 3 \theta \ - \ \frac{1}{2} \sin 4 \theta \ + \ \frac{5}{16} \sin 5 \theta \ - \ \frac{1}{16} \sin 6 \theta \ \ .  $$

The result of the definite integration is then
$$ \frac{\pi}{96} \ [ \ 132 \cos \theta \ - \ 57 \cos 2 \theta \ + \ 2 \cos 3 \theta \ + \ 12 \cos 4 \theta \ - \ 6 \cos 5 \theta \ + \  \cos 6 \theta \ ] \ \vert_{\pi}^0  $$
$$ \frac{\pi}{96} \ \cdot \ 2 \ \cdot ( 132 + 2 - 6 ) \ = \ \frac{128}{48} \pi \ = \ \frac{8}{3} \pi \ , $$
the terms with odd multiples of $ \ \theta \ $ canceling out.
A: I think if you  do this
$$\pi\int_{0}^{\pi} \sin^2\theta(1-\cos\theta)^2 (2\cos\theta\sin\theta - \sin\theta) d\theta$$
then thats the complete volume.
A: The intense machinations in the proffered solutions can be mitigated by solution in the complex plane with Pappus's $2^{nd}$ Centroid Theorem: the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid $R$, i.e., $2πR$. The volume is simply $V=2\pi RA$.
Consider a general cardioid in the complex plane, say
$$z=2a(1+\cos t)e^{it}$$
where $t\in[0,\pi]$ is in the upper-half plane in preparation for rotation about the $x$-axis.
Now, the area and centroid are given as follows and the complex plane,
$$
A=\frac{1}{2}\int \Im\{ z^*\dot z\}~dt\\
R=\frac{1}{3A}\int z~\Im\{ z^*\dot z\}~dt\\
R_y=\frac{1}{3A}\int \Im\{z\}~\Im\{ z^*\dot z\}~dt\\
$$
where $R_y$ is the centroid for rotation about the $x$-axis.
Thus we develop the solution as follows,
$$
z=2a(1+\cos t)e^{it}\\
z^*=2a(1+\cos t)e^{-it}\\
\dot z=2a\big( (1+\cos t)i-\sin t\big)e^{it}\\
z^*\dot z=2a(1+\cos t) 2a\big( (1+\cos t)i-\sin t\big)\\
\Im\{z^*\dot z\}=4a^2(1+\cos t)^2\\
\Im\{z\}~\Im\{ z^*\dot z\}=8a^3\sin t (1+\cos t)^3\\
$$
The volume is then give by
$$
\begin{align}
V
&=2\pi RA\\
&=2\pi \frac{1}{3}\int_0^{\pi}8a^3\sin t(1+\cos t)^3 ~dt\\
&=\frac{16\pi a^3}{3}\int_0^{\pi}\sin t(1+\cos t)^3 ~dt\\
&=\frac{64\pi a^3}{3}
\end{align}
$$
For your case, where $a=1/2$, we have $V=\frac{8\pi}{3}$, as shown by @colormegone. I have verified this solution numerically.
