The exercise is to derive the formula for the rotation of a polar curve of the form $ r = r ( \theta ) $ about the $ x $ – axis. When I asked my lecturer, he suggested that I consider a series of infinitely thin cylinders with circumference $ 2 \pi y $, where $ y = r \sin \theta $ denotes the vertical distance from the $ x $ axis and an area given by $ \text{d} A = r \; \text{d} r \; \text{d} \theta $ to obtain $ V = \int r^3 \sin \theta \; \text{d} \theta $, which ended up giving the correct numerical answer. I instead tried converting to Cartesian as below which instead resulted in the following $$ \begin{aligned}
V &= \pi \int y^2 \; \text{d} x \\
x &= r \cos \theta \\
\text{d} x &= – r \sin \theta \; \text{d} \theta \\
y^2 &= r^2 \sin ^2 \theta \\
V &= \pi \int r^3 \sin ^3 \theta \; \text{d} \theta \\
\end{aligned} $$. Why does my method not give the correct formula?
[Math] Volume of revolution of polar curve
calculuspolar coordinatessolid of revolutionvolume
Related Solutions
An asymptote is going to mean that $r\rightarrow\infty$, which happens when the denominator approaches zero. Vertical or horizontal means that $\theta$ is a multiple of $\pi/2$ (vertical for even, horizontal for odd multiples). The only asymptote would therefore be when $\theta\rightarrow\pi$, which would be a horizontal one above (and parallel to) the negative $x$ axis.
Now as $\theta\rightarrow\pi$, $$ y = r\sin\theta = \frac{\theta\sin\theta}{\pi-\theta} = \frac{\theta\sin(\pi-\theta)}{\pi-\theta} \rightarrow \theta \rightarrow \pi $$ so the horizontal asymptote is $y=\pi$.
For visualization, we can reason that as $\theta$ increases from $0$ to $\pi$, $r$ increases from $0$ to $\infty$, crossing $1$ at the positive $y$ axis. Here are some plots for $\theta\in[0,.6\pi]$ and $[0,.99\pi]$ made with sage (online):
x_coords = [t/(pi-t)*cos(t) for t in srange(0, 0.6*pi, 0.02)]
y_coords = [t/(pi-t)*sin(t) for t in srange(0, 0.6*pi, 0.02)]
list_plot(zip(x_coords, y_coords))
x_coords = [t/(pi-t)*cos(t) for t in srange(0, 0.99*pi, 0.02)]
y_coords = [t/(pi-t)*sin(t) for t in srange(0, 0.99*pi, 0.02)]
list_plot(zip(x_coords, y_coords))
Horizontal and vertical tangents can be found by finding the roots of $\frac{dy}{d\theta}$ and $\frac{dx}{d\theta}$ respectively (assuming that they are never simultaneously zero, in which case we'd have to resort to higher order derivatives). A simple way to find these roots is to observe that $$r = \left(\frac{\pi}{\theta}-1\right)^{-1} \implies \frac{dr}{d\theta} = -\left(\frac{\pi}{\theta}-1\right)^{-2} \cdot \left(-\pi\theta^{-2}\right) = \frac{\pi}{(\theta-\pi)^2} $$ so that $$ 0 = \frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta-r\sin\theta \iff \tan\theta = \frac{1}{r} \frac{dr}{d\theta} = \frac{\pi}{\theta(\pi-\theta)} $$ and $$ 0 = \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta+r\cos\theta \iff -\cot\theta = \frac{1}{r} \frac{dr}{d\theta} = \frac{\pi}{\theta(\pi-\theta)} $$ which show us that they never vanish simultaneously. From the shape of $\frac{\pi}{\theta(\pi-\theta)}$, which is symmetric and positive on $(0,\pi)$ with global minimum at $\frac{\pi}{2}$ and vertical asymptotes at the endpoints, we see that it will meet $\tan\theta$ at exactly one point $\theta_0\in(0,\frac{\pi}{2})$; thus, there is one unique vertical tangent to our curve. From the graph above, we might estimate $$ \tan\theta_0\approx\frac{.4}{.25}=1.6 \implies \theta_0 \approx \frac{\pi}{2}-\sqrt{\frac{\pi}{2}\left(\frac{\pi}{2}-\frac{5}{4}\right)} \approx .8609 $$ where we used the quadratic equation to deduce $\theta$ from $\tan\theta$. However, the actual solution seems to be closer to $0.97803904765198235$:
t=var('t'); find_root(t*(pi-t)*tan(t)==pi, 0, pi/2)
For horizontal tangents, the same function $\frac{\pi}{\theta(\pi-\theta)}$ must now meet $-\cot\theta$, which is equivalent to the equation $$\tan\theta=\theta\left(1-\frac{\pi}{\theta}\right) $$ which on $[0,\pi]$ has only solutions at the two endpoints, i.e., on our curve, at the origin and at the horizontal asymptote found above.
To generate the right-hand loop of the lemniscate $r^2=\cos(2\theta)$, you need to let $\theta$ vary from $-\frac{\pi}{4}$ to $\frac{\pi}{4}$; i.e. $\theta \in [-\frac{\pi}{4},\frac{\pi}{4}] $. To see this, we have
$$ r = \bar{+} \sqrt{\cos(2 \theta) } \implies \cos(2\theta)\geq 0 \implies -\frac{\pi}{2}\leq2\theta \leq \frac{\pi}{2}\implies -\frac{\pi}{4}\leq \theta \leq \frac{\pi}{4}. $$
Here is the plot
Here is the full graph
Best Answer
Your equation is not quite correct. I approached it this way...
$$V=2\pi\int\!\!\!\int y~dy~dx=2\pi\int\!\!\!\int r\sin\theta\cdot r~dr~d\theta=\frac{2\pi}{3}\int r^3\sin\theta~d\theta$$
You can see how this would work for a sphere by integrating over $\theta\in[0,\pi]$ to get $V=\frac{4\pi r^3}{3}$.