Problem 33 - Thomas Calculus

Question

a. Parametrization of an ellipsoid The parametrization \(x=a \cos \theta, \quad y=b \sin \theta, \quad 0 \leq \theta \leq 2 \pi\) gives the ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1 .\) Using the angles \(\theta\) and \(\phi\) in spherical coordinates, show that $$\mathbf{r}(\theta, \phi)=(a \cos \theta \cos \phi) \mathbf{i}+(b \sin \theta \cos \phi) \mathbf{j}+(c \sin \phi) \mathbf{k}$$ is a parametrization of the ellipsoid \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)+\) \(\left(z^{2} / c^{2}\right)=1\) b. Write an integral for the surface area of the ellipsoid, but do not evaluate the integral.

Step-by-Step Solution

Answer

This parametric form correctly describes the ellipsoid. Set up the integral for surface area using cross products of partial derivatives.

Step 1: Understanding the Given Parametrization of an Ellipsoid

The given parametrization is for an ellipsoid expressed in three dimensions: \( x = a \cos \theta \cos \phi \), \( y = b \sin \theta \cos \phi \), and \( z = c \sin \phi \), where \( 0 \leq \theta \leq 2\pi \) and \( -\frac{\pi}{2} \leq \phi \leq \frac{\pi}{2} \). It needs to represent the ellipsoid equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \).

Step 2: Verification of Parametrization

Substitute the parametrization expressions into the ellipsoid equation:- Substitute \( x = a \cos \theta \cos \phi \) into \( \frac{x^2}{a^2} \), which simplifies to \( \cos^2 \theta \cos^2 \phi \).- Substitute \( y = b \sin \theta \cos \phi \) into \( \frac{y^2}{b^2} \), which simplifies to \( \sin^2 \theta \cos^2 \phi \).- Substitute \( z = c \sin \phi \) into \( \frac{z^2}{c^2} \), which simplifies to \( \sin^2 \phi \).Thus, the equation becomes \( \cos^2 \theta \cos^2 \phi + \sin^2 \theta \cos^2 \phi + \sin^2 \phi = 1 \).

Step 3: Simplifying Using Trigonometric Identity

Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \), we have:\( \cos^2 \theta \cos^2 \phi + \sin^2 \theta \cos^2 \phi + \sin^2 \phi = \cos^2 \phi + \sin^2 \phi \).Thus, the equation simplifies to \( \cos^2 \phi + \sin^2 \phi = 1 \), proving the expression is a valid parametrization of the ellipsoid.

Step 4: Setting Up the Surface Area Integral

To find an expression for the surface area of the ellipsoid, set up the integral over the surface parametrized by \( \mathbf{r}(\theta, \phi) \).The surface area \( S \) for a parametrized surface \( \mathbf{r}(u, v) \) is given by:\[ S = \int \int_{D} \left| \frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial \phi} \right| \, d\theta \, d\phi \]Compute the partial derivatives \( \frac{\partial \mathbf{r}}{\partial \theta} \) and \( \frac{\partial \mathbf{r}}{\partial \phi} \), then find their cross product's magnitude, integrating over the domain \( 0 \leq \theta \leq 2\pi \) and \( -\frac{\pi}{2} \leq \phi \leq \frac{\pi}{2} \).

Key Concepts

Spherical Coordinates

Spherical coordinates are a system used to define the position of a point in three-dimensional space with the help of three parameters: radius, polar angle, and azimuthal angle. This system extends polar coordinates to three dimensions by adding an angle to measure the elevation from the reference plane.
Spherical coordinates are especially handy for parametrizing surfaces such as ellipsoids because they naturally accommodate varying radial distances and angles, which align well with the symmetrical nature of these shapes. In our case,

• the azimuthal angle \( \theta \) varies from 0 to \( 2\pi \) and represents the rotation around the vertical (or \( z \)-axis),
• while the polar angle \( \phi \) varies from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \), indicating the inclination from the horizontal plane.

By using these angles, we can express a point on an ellipsoid with the parametrization formulas: \( x = a \cos \theta \cos \phi \), \( y = b \sin \theta \cos \phi \), and \( z = c \sin \phi \). These account for the ellipsoid's radii \( a \), \( b \), and \( c \), along their respective axes.

Surface Area Integral

Calculating the surface area of a parametrized surface like an ellipsoid involves setting up an integral. This integral covers the entire surface defined by the parametrization, using spherical coordinates. The fundamental formula for surface area over a parametrized surface \( \mathbf{r}(u, v) \) is given by:
\[ S = \int \int_{D} \left| \frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial \phi} \right| \, d\theta \, d\phi \] Here, the term \( \left| \frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial \phi} \right| \) represents the magnitude of the cross product of the partial derivatives of the parametrization, which gives us the infinitesimal area element at each point on the surface.

This integral thus sums all these infinitesimal areas over the region \( D \) defined by the parameter ranges \( 0 \leq \theta \leq 2\pi \) and \( -\frac{\pi}{2} \leq \phi \leq \frac{\pi}{2} \), covering the entire surface of the ellipsoid. Although evaluating this integral can be complex and often requires numerical computation, setting it up provides a structured pathway to understanding the geometry of the surface.

Trigonometric Identity

At the heart of verifying the correct parametrization of an ellipsoid is the use of a key trigonometric identity: \( \cos^2 \theta + \sin^2 \theta = 1 \). This well-known identity applies to any angle and forms the basis for simplifying expressions in many mathematical problems.
When examining the components of the parametrization, we find that\( \cos^2 \theta \cos^2 \phi + \sin^2 \theta \cos^2 \phi + \sin^2 \phi \) transforms to \( \cos^2 \phi + \sin^2 \phi \) using this identity. This step is crucial as it ensures the entire expression simplifies to 1, confirming it's valid for an ellipsoidal surface \( \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \).

In this way, the trigonometric identity helps verify that the parametrized coordinates indeed describe the geometry of the ellipsoid correctly. Mastery of such identities not only simplifies complex equations but also fosters deeper understanding of underlying geometrical relationships.