A curve \(y=y(x),\) joining two points \(x_{1}\) and \(x_{2}\) 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 \(y f\left(y, x^{\prime}, \lambda\right)=C .\) since \(y=0\) at the endpoints, \(C=0 .\) Then either \(y=0\) for all \(x,\) or \(f=0 .\) But \(y \equiv 0\) gives zero volume of the solid of revolution, so for maximum volume you want to solve \(f=0\) at the endpoints, \(C=0 .\) Then either \(y=0\) for all \(x,\) or \(f=0 .\) But \(y \equiv 0\) gives zero volume of the solid of revolution, so for maximum volume you want to solve \(f=0\).

The Euler-Lagrange equation is a fundamental tool in the calculus of variations. This equation helps find the function that optimizes a certain quantity, such as length, area, or, in our case, volume. For a given functional, say \(J[y] = \int_{x_1}^{x_2} F(y, y', x) \, dx\), the Euler-Lagrange equation is derived by setting the first variation to zero:
\[ \frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) - \frac{\partial F}{\partial y} = 0. \]
This equation forms the basis for solving optimization problems involving integrals. In our exercise, it's used to derive the necessary conditions a curve \(y=y(x)\) must satisfy to either maximize or minimize a given functional. Essentially, it translates the optimization problem into a differential equation that we need to solve for the function \(y(x)\).

The Lagrange multiplier method is a strategy for finding the local maxima and minima of a function subject to equality constraints. In our context, we use this method to maximize the volume of revolution given a fixed surface area. Here’s how it works in principle:

• Define the main functional you want to optimize (e.g., volume)
• Introduce a constraint functional (e.g., surface area)
• Combine them into a single functional using a Lagrange multiplier, \(\lambda\). This gives us a new 'Lagrangian':
  
  \[L = \pi \int_{x_1}^{x_2} y^2 \, dx + \lambda \left( 2 \pi \int_{x_1}^{x_2} y \sqrt{1 + (y')^2} \, dx - S \right)\]
  
  
• Apply the Euler-Lagrange equation to the Lagrangian. This step will yield a differential equation that considers both the volume and surface area constraints.

By solving this new equation, we obtain a solution that satisfies both the original functional's optimization and the constraint.

The calculus of variations is a field of mathematical analysis that deals with optimizing functionals, which are mappings from a set of functions to the real numbers. These functionals often take the form of integrals. The core question it addresses is: 'What function makes a particular integral reach an extreme value (minimum or maximum)?'
To solve problems in this field, we rely on several techniques, including:

• Deriving the Euler-Lagrange equation, which aids in identifying potential optimizing functions
• Using boundary conditions to narrow down valid solutions
• Applying constraints via Lagrange multipliers if there are additional conditions to be satisfied

In the given exercise, we utilize the calculus of variations to determine the shape of a curve that maximizes the volume of a solid of revolution subject to a fixed surface area. By setting up the appropriate functional and then applying the Euler-Lagrange equation and Lagrange multipliers, we can systematically find the optimal solution.

A solid of revolution is created by rotating a curve around an axis. In our exercise, we revolve a curve \(y=y(x)\) around the \(x\)-axis to form such a solid. The volume and surface area of these solids are critical metrics.
The volume \(V\) of a solid of revolution generated by rotating the curve \(y=y(x)\) about the \(x\)-axis from \(x_1\) to \(x_2 \) is given by:
\[ V = \pi \int_{x_1}^{x_2} y^2 \, dx. \]
Meanwhile, the surface area \(S\) can be calculated as:
\[ S = 2 \pi \int_{x_1}^{x_2} y \sqrt{1 + (y')^2} \, dx. \]
These integrals represent the core quantities we aim to optimize. In this exercise, we maximize the volume while keeping the surface area constant. By formulating our problem through these integrals and applying methods from the calculus of variations, we find the curve \(y=y(x)\) that meets our optimization criteria.