# 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?

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(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.

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At each angle from 0 to $\pi/2$, the radius is defined by the parametric form of y. I don't think that we are forming a ring or anything. I think it suffices if she runs the integral from $\pi$ to 0. Am I wrong? –  satish ramanathan Mar 9 '14 at 3:38
The cardioid has two $\ y-$ values for $\ \theta = 0 \$ to $\ \theta = \frac{\pi}{2} \ .$ (Take a look at the curve in the linked diagram.) OP is limited as to what means they may use to find volume of a solid of revolution. And the issue is exactly that there is no simple way to calculate the volume of a "slice" on the "positive side" of the $\ y-$ axis. –  RecklessReckoner Mar 9 '14 at 3:45
I completely understand and had a look at the diagram but I still think that she has to run the integrand from 0 to $\pi$. May be you are right!! –  satish ramanathan Mar 9 '14 at 3:48
You need a way to remove the portion below the "lower" part of the revolved cardioid -- simply integrating isn't going to manage that. [The problem is that the "upper semi-cardioid" is not a function of $\ x \$ over $\ 0 \ \le \ x \ \le \ \frac{1}{4} \ .$] The calculation needs to be "broken" into two separate integrals for $\ x \ \ge \ 0 \ .$ –  RecklessReckoner Mar 9 '14 at 3:51
Your solution is slightly incorrect as the integral should be from $\pi$ to $\pi/2$ for the first one, then $\pi/2$ to $\pi/3$ If you sum the boundaries: $\pi$ to $\pi /2$, $\pi/2$ to $\pi/3$, and flipping the last boundary to $\pi/3$ to $0$ and changing the sign to a $+$ you get $\pi \int_{\pi}^0 y(\theta)^2 \cdot \frac{dx}{d\theta}$, which gives the correct answer $\frac{8\pi}{3}$ –  robtob Mar 17 '14 at 10:52

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.

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How do you deal with the two $\ y-$ values for $\ \theta = 0 \$ to $\ \theta = \frac{\pi}{2} \$ ? That is what the concern is. –  RecklessReckoner Mar 9 '14 at 3:49