MHB Surface of revolution - Hypocycloid

[Dethrone]

Find the area of the surface of revolution generated by revolving about the x-axis the hypocycloid $x = a \cos^3\left({\theta}\right)$, $y= a \sin^3\left({\theta}\right)$.

I'm following the textbook example, and it says that "the required surface is generated by revolving the arc from $\theta = 0$ to $\theta$ to $\pi$ .

Then they set up the integral:
$$S=2\pi \int_{0}^{\pi} \sqrt{(\d{x}{\theta}+\d{y}{\theta}})\,d\theta$$
$$=(2)2\pi \int_{0}^{\pi /2} \sqrt{(\d{x}{\theta}+\d{y}{\theta}})\,d\theta$$

Now I have a couple of questions, especially since I haven't done a lot of parametric equation questions. First of all, why are they only revolving the arc from $\theta = 0$ to $\pi$ ? Why not the whole curve from $\theta = 0$ to $2\pi$ ? If this were a solids of revolution question, we would have rotated the whole curve, or would we? (I notice that $\pi$ to $2\pi$ is just the negative portion of the first half)

Also, if you were to do this question, would you graph this? If so, how would you go about doing so? Would you try out the important angles: $\theta =0$, $\frac{\pi}{2}$, etc?

Finally, when they set up the integral, they split the region in two parts and doubled the integral, which is similar to what you would do to an even function. Because this is a parametric equation, how would you know that?

 

[Fantini]

Consider a simpler example: you want to obtain the unit sphere as the surface of revolution of a circle. You take $x=\cos (\theta), y = \sin (\theta)$ for $- \pi/2 \leq \theta \leq \pi/2$ and rotate it once. If you take the whole interval $- \pi/2 \leq \theta \leq 3 \pi/2$ you will obtain the volume twice.

As for graphing, I wouldn't because I don't have a clue. :( I wish we had more tools that helped the visual intuition about how these curves are. However, the advantage of integration is that you don't have to be a geometric trickster and instead solve things mechanically (advantage and disadvantage, really).

The sphere once again proves a suitable case for discussion: due to symmetry regarding $- \pi/2 \leq \theta \leq 0$ and $0 \leq \theta \leq \pi/2$ you can compute only half of the area and double it. The same happens with the hypocycloid. :)

 

[Dethrone]

Thanks for the response!

1. In this question, we benefited from knowing that it is an hypocycloid. But in a scenario in which we were given a random parametric equation, how will we know that it is symmetric, or that it is an even function? I'm used to splitting this way:
$$\int_{-2}^{2} f(x)\,dx=2\int_{0}^{2} f(x)\,dx$$
But in this parametric equation, we're not splitting the function by the origin, we're splitting $0$ to $\pi$ -> $0$ to $\frac{\pi}{2}$ and $\frac{\pi}{2}$ to $\pi$. I guess the main question is how do we know that the equation is symmetry about $\frac{\pi}{2}$.

2. Also, this question is also a case in which the graph is symmetric about the x-axis, as you noted. Thus, we only need to rotate the upper portion, lest we double the area. In a situation in which it was not symmetric about the x-axis, will we need to rotate the whole curve?

I've noticed that in solids of revolution, when the graph is symmetric about the x-axis, such as a circle, we only need the upper portion to rotate in order to obtain a sphere. If it were not symmetric that way, we'd need to rotate the whole portion.

The thing is, I'm not familiar with parametric equations, so if i did not graph this, I would not have known any of these symmetric properties. Surely there must be a mathematical way...?

Sorry for the massive wall of text :(.

 

[Fantini]

1. We won't, unless we graph it and/or know its properties beforehand. There may be principles which apply in order to figure out geometric properties from algebraic equations alone, but I do not know them. For the particular case at hand, I find it hard to describe other than geometric intuition.

2. Possibly. The difficulties arising from considering very general curves (but not that general so we may apply usual methods) are that they do not possesses nice geometric properties, such as symmetry. Without these, it's hard to tell whether we'd need to rotate all of it, just half, two thirds or what else. There is no mathematical way to tell beforehand, which is why my experience tells me physicists tend to fare better than us at it: unshackled by mathematical generalities, they blaze through interesting cases using their intuition.

I apologize, too, for the massive wall, and for not being so helpful in these matters. :(

Best wishes! :)

 

[Dethrone]

Not, it's completely alright. The example that I'm reading from my workbook seems to imply that it's like common-sense, so I thought I was missing something. (I'm glad that I'm not) It seems that they're probably using the known properties of the hypocycloid to their advantage, which is what you did. I mean, I guess it's possible to graph it, but you'd have to try out different values of $\theta$ and connect the dots, which is kind of tedious.

Thanks for the help! I'll do more of these questions, and I'll let you know if I have any further inquiries :D

 

[I like Serena]

When you run $\theta$ in the negative direction, you can see that:
$$x(-\theta) = x(\theta),\ y(-\theta)=-y(\theta)$$
So you get the same x-values and opposite y-values.
Therefore the graph is symmetric in the x-axis.

Similarly:
$$x(\pi-\theta) = -x(\theta),\ y(\pi-\theta)=y(\theta)$$
Therefore the graph is symmetric in the y-axis.

Easiest to visualize it, is W|A:
x=cos^3 theta, y=sin^3 theta - Wolfram|Alpha Results

You can also find this by analyzing the properties of the graph.
Where is $x'(\theta)=0$?
And $y'(\theta)=0$?
These are horizontal and vertical tangents.
In this case there are no asymptotes.
Filling in a couple of points, then allows you to draw the graph.

 

[Dethrone]

Hi ILY!

I tried to come up with a set of similar reasoning as you have above, but instead of testing the effects of a negative and positive parameter like you did, I tried out positive and negative x-values, which didn't get me anywhere. I'm glad you mentioned it!

I'm trying to digest this:
$$(-\theta) = x(\theta),\ y(-\theta)=-y(\theta)$$

A negative parameter indicates that you're going in the opposite direction on the curve, am I correct? So if you're going in the opposite direction, and the x-values stay the same while the y-values are negative, then you get a bottom curve that is purely symmetric to the curve above, meaning that it is symmetric in the x-axis.

$$x(\pi-\theta) = -x(\theta),\ y(\pi-\theta)=y(\theta)$$

This is very clever!

Though, looking back at the problem, if we know that it is symmetric about/in the y-axis, then we have to find the value of the parameter $\theta$ that corresponds to when $x=0$ so that when can split the integral in two. We have $x=a \cos^3\left({\pi}\right) = 0$, $\theta = \frac{\pi}{2}$ or $\frac{3\pi}{2}$. Combining the knowledge above, we know that it must be $\frac{\pi}{2}$. Which finally tells us this:

$$S=2\pi \int_{0}^{\pi} \sqrt{(\d{x}{\theta}+\d{y}{\theta}})\,d\theta= (2)2\pi \int_{0}^{\pi /2} \sqrt{(\d{x}{\theta}+\d{y}{\theta}})\,d\theta$$

If you're not familiar with the properties of a hypocycloid, this does seem like a lot of work to prove the above. It'd be much easier if I were to keep the integral from $0$ to $\pi$. (Wondering)Thanks ILY!

 

[I like Serena]

$$S=2\pi \int_{0}^{\pi} \sqrt{(\d{x}{\theta}+\d{y}{\theta}})\,d\theta= (2)2\pi \int_{0}^{\pi /2} \sqrt{(\d{x}{\theta}+\d{y}{\theta}})\,d\theta$$

If you're not familiar with the properties of a hypocycloid, this does seem like a lot of work to prove the above. It'd be much easier if I were to keep the integral from $0$ to $\pi$. (Wondering)

There is a bit of a problem with the integral you mentioned.
For starters, it's a weird integral.
I don't understand how it is supposed to give the surface area of the body of revolution.

More importantly, the argument of the square root is negative on a significant part of the domain.

The proper formula for the surface area should be:
$$S = \int 2\pi y \sqrt{dx^2+dy^2}
= 2\pi \int_0^\pi y \sqrt{\left(\!\d x \theta\!\right)^{\!2} + \left(\!\d y \theta\!\right)^{\!2}} \,d\theta$$

 

[Dethrone]

You are right, I mistyped it because my brain was malfunctioning yesterday. But applying the above formula, you obtain:

$$S=4 \pi \int_0^{\pi/2} a\sin^3\left({\theta}\right)(3a\cos\left({\theta}\right)\sin\left({\theta}\right))d\theta$$

Which should simplify even further to the integral in my first post. Thanks for noticing, I would have missed it! :D

 

[I like Serena]

You are right, I mistyped it because my brain was malfunctioning yesterday. But applying the above formula, you obtain:

$$S=4 \pi \int_0^{\pi/2} a\sin^3\left({\theta}\right)(3a\cos\left({\theta}\right)\sin\left({\theta}\right))d\theta$$

Which should simplify even further to the integral in my first post. Thanks for noticing, I would have missed it! :D

And there is the reason why you would want to integrate up to $\frac \pi 2$.

You have used that $\sqrt{\cos^2\theta \sin^2\theta} = \cos\theta \sin\theta$.

That is only true if $0\le \theta \le\frac \pi 2$.

 

[Dethrone]

Let me think that through:

Strictly speaking, $$\sqrt{\cos^2\theta \sin^2\theta} = \mid\cos\theta \sin\theta\mid$$

$$\mid\cos\theta \sin\theta\mid=\begin{cases}\cos\left({\theta}\right)\sin\left({\theta}\right) \text{if } 0\le \theta \le\frac \pi 2, & \\[3pt] -\cos\left({\theta}\right)\sin\left({\theta}\right) \text{if } \frac{\pi}{2}\le \theta \le \pi. \\ \end{cases}$$

We must split the integral this way because the function is different depending on which domain it lies on, right? In questions like these, I tend to overlook the absolute value because it's annoying. It's like in trig substitution, most people just skip the absolute value because reasoning it out is annoying each time.

 

[I like Serena]

Let me think that through:

Strictly speaking, $$\sqrt{\cos^2\theta \sin^2\theta} = |\cos\theta \sin\theta|$$

$$|\cos\theta \sin\theta|=\begin{cases}\phantom-\cos\left({\theta}\right)\sin\left({\theta}\right) &\text{if } 0\le \theta \le\frac \pi 2, \\
-\cos\left({\theta}\right)\sin\left({\theta}\right) &\text{if } \frac{\pi}{2}\le \theta \le \pi. \\ \end{cases}$$

We must split the integral this way because the function is different depending on which domain it lies on, right?

Yep! (Nod)

 

[Dethrone]

Also, I've always questioned what the best way to solve integrals such as these,

Say we want to integrate
$$\int_{0}^{\pi} |\cos\theta \sin\theta|\,d\theta$$

Using what we said above,

$$=\int_{0}^{\pi/2} \cos\theta \sin\theta\,d\theta + \int_{\pi/2}^{\pi} -\cos\theta \sin\theta\,d\theta$$

Clearly these areas cancel out, one is above the x-axis and one is below. Usually, I use my common sense and say "let's add absolute values to both integrals so that they become positive". Is there not be a better way?

An problem I find is integrating $\sin\left({\theta}\right)$ from $0$ to $2\pi $. I know that the integral must be split in between, or else the areas will cancel. What is the best way to tackle it? I've seen some textbooks take the absolute value of sinx and integrate that instead which makes sense; is that the best way to approach it?

 

[I like Serena]

Also, I've always questioned what the best way to solve integrals such as these,

Say we want to integrate
$$\int_{0}^{\pi} |\cos\theta \sin\theta|\,d\theta$$

Using what we said above,

$$=\int_{0}^{\pi/2} \cos\theta \sin\theta\,d\theta + \int_{\pi/2}^{\pi} -\cos\theta \sin\theta\,d\theta$$

Clearly these areas cancel out, one is above the x-axis and one is below.

Those areas do not cancel out...

Usually, I use my common sense and say "let's add absolute values to both integrals so that they become positive". Is there not be a better way?

An problem I find is integrating $\sin\left({\theta}\right)$ from $0$ to $2\pi $. I know that the integral must be split in between, or else the areas will cancel. What is the best way to tackle it? I've seen some textbooks take the absolute value of sinx and integrate that instead which makes sense; is that the best way to approach it?

That depends on the problem statement.
If the total area is requested, absolute value symbols are needed.
If they ask for the integral, areas must cancel as is appropriate.

 

[Dethrone]

Oh okay! I guess I jumped conclusions too fast, haha. They don't cancel, which makes sense now.

If the question were to just evaluate, then

$$\int_{0}^{2\pi} \sin\left({x}\right)\,dx=0$$If they wanted the total area, then we take the absolute value of the entire function so that we're only integrating positive areas, right?

$$\int_{0}^{2\pi} |\sin\left({x}\right)|\,dx$$

Which we would then split it accordingly.

 

[I like Serena]

Oh okay! I guess I jumped conclusions too fast, haha. They don't cancel, which makes sense now.

If the question were to just evaluate, then

$$\int_{0}^{2\pi} \sin\left({x}\right)\,dx=0$$If they wanted the total area, then we take the absolute value of the entire function so that we're only integrating positive areas, right?

$$\int_{0}^{2\pi} |\sin\left({x}\right)|\,dx$$

Which we would then split it accordingly.

Exactly! (Wink)