Problem 42 - Calculus Early Transcendentals

Question

Suppose that a surface \(\sigma\) in 3 -space and a function \(f(x, y, z)\) are described geometrically. For example, \(\sigma\) might be the sphere of radius 1 centered at the origin and \(f(x, y, z)\) might be the distance from the point \((x, y, z)\) to the \(z\) -axis. How would you explain to a classmate a procedure for evaluating the surface integral of \(f\) over \(\sigma ?\)

Step-by-Step Solution

Answer

The integral evaluates to \( \pi^2 \).

Step 1: Understand the Problem

We need to evaluate the surface integral of the function \( f(x, y, z) \) over a surface \( \sigma \). Here, \( \sigma \) is the sphere of radius 1 centered at the origin. The function \( f \) is defined as the distance from a point \((x, y, z)\) to the \( z \)-axis. We will use this setup to compute the integral.

Step 2: Define the Surface Parameterization

The sphere of radius 1 centered at the origin can be parameterized using spherical coordinates. The parameterization \( \vec{r}(u, v) \) is given by: \( x = \sin u \cos v, \ y = \sin u \sin v, \ z = \cos u \), where \( 0 \leq u \leq \pi \) and \( 0 \leq v < 2\pi \).

Step 3: Calculate the Function on the Surface

The function \( f(x, y, z) \) is the distance from the point \( (x, y, z) \) to the \( z \)-axis. The distance to the \( z \)-axis is given by \( f(x, y, z) = \sqrt{x^2 + y^2} \). Substituting the parameterization, \( f(\sin u \cos v, \sin u \sin v, \cos u) = \sqrt{(\sin u \cos v)^2 + (\sin u \sin v)^2} = \sin u \).

Step 4: Compute the Surface Element

The differential surface element \( dS \) on the parameterized surface \( \vec{r}(u, v) \) is given by the magnitude of the cross product \( \left| \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} \right| dudv \). Calculating these partial derivatives and their cross product gives us \( \sin u \ dudv \) for the magnitude.

Step 5: Set Up the Surface Integral

The surface integral is \( \int_{\sigma} f(x, y, z) \, dS = \int_0^{2\pi} \int_0^{\pi} (\sin u) (\sin u) \, dudv \). This simplifies to \( \int_0^{2\pi} \int_0^{\pi} \sin^2 u \, dudv \).

Step 6: Evaluate the Integral

Evaluate the integral first with respect to \( u \): \( \int_0^{\pi} \sin^2 u \, du = \frac{\pi}{2} \). Then with respect to \( v \): \( \int_0^{2\pi} 1 \, dv = 2\pi \). Multiplying both results gives the final answer \( \pi^2 \).

Key Concepts

Spherical Coordinates

Spherical coordinates are a system for describing points in three-dimensional space. They are especially useful for surfaces that have symmetry around the origin, such as spheres.In spherical coordinates, instead of using the traditional Cartesian coordinates defined by \((x, y, z)\), we use three parameters: \( (r, \theta, \phi) \).

• \( r \) is the radius, the distance from the origin to the point. For a sphere of radius 1, \( r = 1 \).
• \( \theta \) (the azimuthal angle) is the angle in the xy-plane from the positive x-axis. It varies from \(0\) to \(2\pi\).
• \( \phi \) (the polar angle) is the angle from the positive z-axis. It varies from \(0\) to \(\pi\).

This system allows us to represent the sphere with simple parameterizations, facilitating the process in surface integrals.

Parameterization

Parameterization is a way to express surfaces in terms of two parameters. This helps in calculations by transforming a surface into a shape where mathematical operations are easier to perform.For our sphere of radius 1, parameterizing using spherical coordinates gives\[\begin{align*} x &= \sin u \cos v, \ y &= \sin u \sin v, \ z &= \cos u.\end{align*}\]Here, \(u\) plays the role of the polar angle \(\phi\), and \(v\) is the azimuthal angle \(\theta\).

• \(u \) ranges from \(0 \) to \(\pi\), dealing with vertical movement along the sphere.
• \(v \) ranges from \(0 \) to \(2\pi\), wrapping around the sphere horizontally.

Using this method is essential to simplify the evaluation of surface integrals, especially for complex surfaces.

Surface Element

The surface element is the small patch of surface area over which integration occurs in a surface integral. To find it, we calculate the cross product of partial derivatives of the parameterization.For our sphere, the parameterization \(\vec{r}(u, v)\) is: \[\begin{align*} \frac{\partial \vec{r}}{\partial u} &= (\cos u \cos v, \cos u \sin v, -\sin u), \ \frac{\partial \vec{r}}{\partial v} &= (-\sin u \sin v, \sin u \cos v, 0).\end{align*}\]The surface element \(dS\) is found from \[\left| \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} \right| dudv\].The cross product calculation yields \( \sin u \ dudv \),indicating each small piece of the sphere's surface. This is a crucial stepbecause it sets up the integration process correctly, allowing us to integrate functions over the surface.

Cross Product

The cross product is an operation on two vectors in three-dimensional space. It results in a third vector perpendicular to both original vectors. The magnitude of this cross product provides the area of the parallelogram that the two vectors span.For a parameterized surface, this property is used to find the surface element by calculating the cross product of the partial derivatives of the parametrization.Given two vectors \( \vec{a} = (a_1, a_2, a_3) \) and \( \vec{b} = (b_1, b_2, b_3) \), the cross product \( \vec{a} \times \vec{b} \) is:\[\begin{align*} (a_2b_3 - a_3b_2, \ a_3b_1 - a_1b_3, \ a_1b_2 - a_2b_1)\end{align*}\]In our sphere problem, this mathematical tool allows us to compute the differential area of tiny segments of the sphere surface. By multiplying this area by the function value at corresponding points, we efficiently conduct surface integration.