Question

A surface of revolution, whose equation in cylindrical polar coordinates is $\rho =$ $\rho (z)$, is bounded by the circles $\rho =a,z=\pm c(a>c).$ Show that the function that makes the surface integral $I=\int {\rho }^{-1/2}\overline{d}S$ stationary with respect to small variations is given by $\rho (z)=k+z^2/(4k)$, where $k=[a\pm {(a^2-c^2)}^{1/2}]/2$

Short answer

ρ(z) = k + z²/(4k) where k = [a ± √(a² - c²)]/2.

Step-by-step solution

Identify the variables and boundaries

Given the surface of revolution $\rho =\rho (z)$, bounded by $\rho =a$ and $z=\pm c$ where $a>c$ . Understand the integration limits and the function to be maximized, $I=\int {\rho }^{-1/2}\overline{d}S$ .

Express the surface element

In cylindrical coordinates, the surface element $\overline{d}S$ can be expressed in terms of $\rho (z)$ and its first derivative ${\rho }^{\prime }(z)$. Given by: $$\overline{d}S=\sqrt{1+({\rho }^{\prime }{)}^2}\overline{d}z$$ .

Apply Required Conditions

Considering the boundaries, use the calculus of variations: seek a function $\rho (z)$ such that $I=\int {\rho }^{-1/2}\sqrt{1+({\rho }^{\prime }{)}^2}\overline{d}z$ remains stationary by applying the Euler-Lagrange equation to the integrand F = $({\rho }^{-1/2}\sqrt{1+({\rho }^{\prime }{)}^2})$.

Simplify Euler-Lagrange Equation

Solve the Euler-Lagrange equation: obtaining $\rho (z)=k+\frac{z^2}{4k}$, after differentiating and simplifying under the stationary condition.

Define the constant k

Identify the constant $k$ using the boundary conditions: After solving, \k = \frac {a \pm \sqrt{a^2 - c^2}}{2} .

Key concepts

Surface of Revolution

A surface of revolution is created by rotating a curve around an axis. In this exercise, we are dealing with a surface described in cylindrical polar coordinates by the function $\rho =\rho (z)$. This surface is bounded by circles at $\rho =a$ and $z=\pm c$, where $a>c$.
To visualize, imagine rotating a curve that depends on $z$ around the vertical axis, creating a 3D surface.
This concept is foundational for understanding how the given function $\rho (z)$ and its boundaries come into play in the problem.

Euler-Lagrange Equation

The Euler-Lagrange equation is a key tool in the calculus of variations. It helps us find the function that makes an integral stationary (meaning it doesn't change for small variations in the function).
In this case, our goal is to find the function $\rho (z)$ that makes the surface integral $I=\int {\rho }^{-1/2}\overline{d}S$ stationary.
The general form of the Euler-Lagrange equation for a function $F$ depending on $\rho$, ${\rho }^{\prime }$ (the first derivative of $\rho$), and $z$ is:
$$\frac{\overline{d}}{\overline{d}z}\frac{\overline{d}F}{\overline{d}{\rho }^{\prime }}-\frac{\overline{d}F}{\overline{d}\rho }=0$$ Applying this equation to our integrand $F={\rho }^{-1/2}\sqrt{1+({\rho }^{\prime }{)}^2}$, we find the function $\rho (z)$.

Cylindrical Coordinates

Cylindrical coordinates are useful for problems with circular symmetry. They are defined by three variables: $\rho$ (the radial distance from the axis), $\theta$ (the angle around the axis), and $z$ (the height along the axis).
In this exercise, we use cylindrical coordinates to describe a surface of revolution.
The surface element $\overline{d}S$ in cylindrical coordinates is given by:
$$\overline{d}S=\sqrt{1+({\rho }^{\prime }{)}^2}\overline{d}z$$ This element takes into account the slope of the surface, affecting the integral we need to solve.
Understanding cylindrical coordinates helps us express the surface element and makes the problem more manageable.

Stationary Integral

A stationary integral is an integral whose value does not change for small variations in the function being integrated.
In our case, we want the surface integral $I=\int {\rho }^{-1/2}\sqrt{1+({\rho }^{\prime }{)}^2}\overline{d}z$ to be stationary.
This requires us to find the function $\rho (z)$ that satisfies this condition.
By applying the Euler-Lagrange equation to the integrand, we find that the stationary function is $\rho (z)=k+\frac{z^2}{4k}$.
Here, $k$ is a constant determined by the boundary conditions, specifically:
$$k=\frac{a\pm \sqrt{a^2-c^2}}{2}$$ Finding this function ensures that the integral remains stationary, solving the problem effectively.