# Calculus of Variations, Isoperimetric, given surface area max volume

#### mishima

Problem Statement
A curve y = y(x), joining two points x1 and x2 on the x axis, is revolved around
the x axis to produce a surface and a volume of revolution. Given the surface area,
find the shape of the curve y = y(x) to maximize the volume. Hint: You should
find a first integral of the Euler equation of the form yf(y, x', λ) = C. Since y = 0
at the endpoints, C = 0. Then either y = 0 for all x, or f = 0. But y ≡ 0 gives zero
volume of the solid of revolution, so for maximum volume you want to solve f = 0.
Relevant Equations
F+λG, Euler's Equations
My volume integral is...

$$\pi\int y^2 dx$$

My surface area integral is...

$$2\pi\int y \sqrt {1+x'^2} dy$$

I'm fairly sure the variable of integration on my volume and surface area integrals has to be the same, is that right? But when I change the variable in the surface area integral to $2\pi\int y \sqrt {1+y'^2} dx$ I'm not seeing how to get the f(y, x', λ) mentioned in the hint...

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#### mishima

Instead, what I do get for my F+λG after changing variable of integration to dx in the surface area integral:

$$y^2+\lambda y \sqrt {1+y'^2}$$

Then putting the Euler's equation together...

$$\frac {\partial (F+λG)} {\partial y'}=\frac {\lambda yy'} {\sqrt {1+y'^2}}$$

and

$$\frac {\partial (F+λG)} {\partial y}=2y+\lambda \sqrt {1+y'^2}$$

#### Orodruin

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The hint is to look at a first integral to the problem rather than the EL equations. Your integrand does not depend explicitly on x. What first integral does this imply?

• mishima

#### mishima

I must then just be completely misunderstanding the term "first integral"...I thought you needed the EL equations for that. I'm used to doing problems where the partial with respect to y is zero, and the partial with respect to y' is implied to be constant. This latter type of term is what I'm thinking a first integral is?

#### Orodruin

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I must then just be completely misunderstanding the term "first integral"...I thought you needed the EL equations for that. I'm used to doing problems where the partial with respect to y is zero, and the partial with respect to y' is implied to be constant. This latter type of term is what I'm thinking a first integral is?
A first integral is any integral of the EL equations giving you a function of y, y’, and x equal to an integration constant. There are some standard cases, the integrand being independent of x is one of them and the integrand being independent of y is another. You should be able to find both in any textbook covering variational calculus.

• mishima

#### mishima

Alright, I think I see what you are saying. I don't need the EL equations because instead I am looking at making

$$\int (F+λG) dx$$

stationary, and F+λG doesn't depend on x explicitly. So, as far as the integral is concerned, F+λG is constant. Then I can just factor the y and use the fact that y'=1/x' to get the form yf(y, x', λ) = C .

#### Orodruin

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So, as far as the integral is concerned, F+λG is constant.
No, this is not correct. As far as the integral is concerned, $y$ and $y'$ are functions of $x$. However, as the integrand is not an explicit function of $x$, there exists is a particular first integral. This first integral should be covered in your textbook.

Edit: If, against all odds, your textbook does not cover this, then you should throw it away and have a look here or in a real textbook.

#### mishima

Actually I think I was just making this too hard. I just needed to change the integration variable of my volume integral instead of my surface area integral. So

$$V = \pi \int y^2 x' dy$$

That means my F+λG is a comfortable function without any x

$$y^2x'+\lambda y \sqrt {1+x'^2}$$

Then it makes total sense that $\frac {\partial }{\partial x}$ would be zero and my partial derivative to x' is a constant.

$$y^2+\frac {\lambda y x'}{\sqrt{1+x'^2}}=K$$

Then applying the hint,

$$y+\frac {\lambda x'}{\sqrt {1+x'^2}}=0$$

Which can be reduced to

$$x'=\frac {y}{\sqrt{\lambda^2 + y^2}}$$

Giving the equation of a circle for the unknown function.

This was from Boas' chapter on calculus of variations. The first integral is covered and it seems my original interpretation was correct (the expression that results when the integrand does not contain the dependent variable). I just chose dx when I should have chosen dy for my volume and surface area integrals, from what I can tell.

#### Orodruin

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This change of variables is a bit dubious because you will no longer be looking for a 1-to-1 function $x(y)$. The better course of action is to use the Beltrami identity.

• mishima

#### mishima

Thanks, I have stumbled across the Beltrami identity online but it is not covered in Boas. Good to know an example of why it is necessary.

#### Orodruin

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Well, she is essentially covering it by rewriting the problem in the same fashion as you have presented here. However, this will not be possible in the same way when you have functionals depending on several functions and so it does not necessarily generalise well. In the region where $y$ and $x$ are one-to-one, it holds that
$$\int f(y,y') dx = \int f(y,1/x') x' dy \equiv \int F(x',y) dy,$$
with appropriate boundary conditions and $F(x',y) = x' f(1/x',y)$. You then have the first integral
$$\partial_{x'} F(x',y) = C$$
since $F$ does not depend explicitly on $x(y)$. This leads to
\begin{align*}
C &= \frac{\partial F}{\partial x'} = \frac{\partial}{\partial x'}[x' f(y,1/x')] \\
&= f(y,y') + x' \partial_{x'} f(y,1/x') = f(y,y') - (1/x') \left.\frac{\partial f}{\partial y'}\right|_{y' = 1/x'}\\
&= f(y,y') - y' \frac{\partial f}{\partial y'},
\end{align*}
which is the Beltrami identity.

Shame on Boas for not writing this important identity out! It does appear in most texts covering variational calculus - including this one ... • mishima

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