Calculus of Variations, Isoperimetric, given surface area max volume

In summary: K$$In summary, the hint was to look at a first integral to the problem rather than the EL equations. When changing the variable of integration to dx in the surface area integral, I get:$$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)}
  • #1
mishima
556
34
Homework 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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  • #2
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

$$\frac {\partial (F+λG)} {\partial y}=2y+\lambda \sqrt {1+y'^2}$$
 
  • #3
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?
 
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  • #4
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?
 
  • #5
mishima said:
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.
 
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  • #6
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 .
 
  • #7
mishima said:
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.
 
  • #8
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.
 
  • #9
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.
 
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  • #10
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.
 
  • #11
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 ... :oldeyes:
 
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  • #12
Boas covers the Beltrami identity but leaves it as an exercise for the student - in section 8 Miscellaneous Problems Q1.
 

1. What is the Calculus of Variations?

The Calculus of Variations is a mathematical theory that deals with finding the optimal solution to a mathematical problem, typically involving a functional (a function of a function) rather than a simple function. It involves finding the function that minimizes or maximizes a given functional.

2. What is the Isoperimetric Problem?

The Isoperimetric Problem is a classic problem in the Calculus of Variations that involves finding the shape with the largest area for a given perimeter. In other words, it is the problem of finding the shape that encloses the most area with a fixed boundary length.

3. How is the Isoperimetric Problem solved using the Calculus of Variations?

The Isoperimetric Problem is solved by setting up a functional that represents the area of the shape in terms of its perimeter. Then, using the Euler-Lagrange equation, the optimal shape can be found by finding the function that satisfies the necessary condition for a minimum or maximum.

4. Can the Isoperimetric Problem be extended to finding the maximum volume for a given surface area?

Yes, the Isoperimetric Problem can be extended to finding the maximum volume for a given surface area. This is known as the Isoperimetric Problem of Maximal Volume and is solved using a similar approach as the traditional Isoperimetric Problem.

5. What are some real-world applications of the Isoperimetric Problem and the Calculus of Variations?

The Isoperimetric Problem and the Calculus of Variations have various applications in physics, engineering, and economics. Some examples include finding the optimal shape for a soap bubble, the shape of a hanging chain, and the path of a light beam between two points.

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