Calculus of Variations (in dire need)

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SUMMARY

The discussion centers on the calculus of variations, specifically solving a problem involving maximizing the volume of a solid obtained by rotating a curve about the x-axis. The problem requires finding the function y(x) that represents the curve between the points (0,0) and (a,0) with a length of 2a. Key insights include the use of Lagrange multipliers to handle constraints and the realization that the maximum volume does not correspond to a sphere due to the length restriction. The integral to maximize is identified as ∫0a π(y(x))2 dx.

PREREQUISITES
  • Understanding of calculus, particularly integration and differentiation.
  • Familiarity with the concept of Lagrange multipliers for constrained optimization.
  • Basic knowledge of the calculus of variations and its applications.
  • Experience with solid geometry, particularly volume calculations of solids of revolution.
NEXT STEPS
  • Study the calculus of variations, focusing on Euler-Lagrange equations.
  • Learn about Lagrange multipliers and their application in optimization problems.
  • Explore solid geometry, particularly the relationship between surface area and volume.
  • Review advanced integration techniques, especially in the context of physical applications.
USEFUL FOR

This discussion is beneficial for undergraduate physics students, mathematics students, and anyone interested in advanced calculus and optimization techniques in physics and engineering contexts.

avocadogirl
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Please allow me to preface:

I'm an undergraduate physics student at a small school where upper-level courses are on a two year rotation. So, I'm currently in an advanced Mathematical Methods course for which I lack prerequisites. I'm only concurrently enrolled in differential equations, I've a poor background in calculus and, I've yet to take modern physics--I've only completed the introductory level mechanics and electromagnetism courses.

After a fellow student explained calculus of variations to me, briefly, I realized how startlingly wrong and almost juvenile was my attempted approach to one of the problems. What only adds to my aggravation is that, this course is a distance learning course broadcast from another small school several hundred miles away. So, given all of my previous positive experiences, I'm referring to Physics Forum with hopes of a clear, concise tutorial.

For all those who may contribute, you've my deep and sincere gratitude. Thank you.


Homework Statement



A curve y(x) of length 2a is drawn between the points (0,0) and (a,0) in such a way that the solid obtained by rotating the curve about the x-axis has the largest possible volume. Find y(1/2a)

Homework Equations



\partialF/\partialy - [(d/dx) *(\partialF/\partialy' =0 ?

The Attempt at a Solution



∫ 2piF(x) dx - I’m integrating some function, which would just be the function of the curve, so, I’d have to multiply it times two, and then pi to represent the rotation?

Making it a sphere would maximize the volume so, the angle at which the curve would intersect the x-axis would be 90 degrees, correct?

So, if I attempt to find a function to represent the changing segments of length of the curve, and I use the formula to take the square root of (1+ y’^2), then make the substitution of y’ = ctn(theta), would my theta be 90?

Or would I try to solve it parametrically?

Or, would I try to solve it like the Courant and Robbins problem where you try to find the maximum path around a sphere?

I would think the diameter of a would be relevant, and the curve length of 2a would be the value of the integral for determining a component of the y(x), and maybe I would use a Lagrane multiplier, but, I don't really understand how the variables relate to one another to know how to use a Lagrange multiplier.

Thank you, again.
 
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Your integral is close. The function y represents the radius of some circular cross-section, so the integral to maximize is \int^{a}_{0}\pi \left(y(x)\right)^{2}~\textrm{d}x

With this integral, you should be able to use your relevant equation, though keep in mind that the total length of y is 2a, so you'll have an upper bound. Upper bounds can be dealt with with a Lagrange multiplier like you suggested. If you get stuck you might want to take a look at http://www.mpri.lsu.edu/textbook/Chapter8-b.htm for suggestions on how to approach constrained problems like this.

From this you can see that the maximum volume won't be from a sphere, because in order to have a sphere between 0 and a, your radius has to be (1/2)*a, which makes the circumference of the circular cross-section through the middle \pi{}a, and that makes the length of the curve \frac{1}{2\pi}a (since it's half of the circumference).

The sphere is a maximum volume given a particular surface area, but it's not a maximal solution for any given situation. If you didn't have a restriction on curve length in this problem, what would be the maximum area? Would there be one?EDIT: I just realized that this problem is made significantly easier if you consider the area under y(x), rather than the volume integral I mentioned. What is the relationship between the area under y(x) and the volume of the solid?
 
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