Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the volume of the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+\) \(z^{2} / c^{2}=1\) by making the change of variables \(x=u a, y=v b\), and \(z=c w\). Also, find the moment of inertia of this solid about the \(z\)-axis assuming that it has constant density \(k\).

Step-by-Step Solution

Answer

Volume: \( \frac{4}{3} \pi abc \). Moment of Inertia: \( kabc \left(\frac{4}{9} \pi (a^2 + b^2)\right) \).

Step 1: Substitute the Change of Variables

We are given the change of variables: \( x = ua \), \( y = vb \), and \( z = wc \). Substitute these into the ellipsoid equation: \[ \left( \frac{ua}{a} \right)^2 + \left( \frac{vb}{b} \right)^2 + \left( \frac{wc}{c} \right)^2 = 1. \] Simplifying this yields \( u^2 + v^2 + w^2 = 1 \), which is the equation of a unit sphere.

Step 2: Calculate the Jacobian of the Transformation

The transformation from ellipsoid coordinates \( (x, y, z) \) to spherical coordinates \( (u, v, w) \) involves scaling by \( a \), \( b \), and \( c \). The volume element \( dV = dx \, dy \, dz \) transforms according to the absolute value of the Jacobian determinant which is \( a \times b \times c \). Thus, \( dV = a b c \ du \, dv \, dw \).

Step 3: Calculate the Volume of the Sphere

We now compute the volume integral over the sphere \( u^2 + v^2 + w^2 \leq 1 \) using spherical coordinates. The volume of the unit sphere is \( \frac{4}{3} \pi \). Therefore, the volume of the ellipsoid is \( \frac{4}{3} \pi \times a b c \).

Step 4: Formulate the Equation for Moment of Inertia

The moment of inertia \( I \) about the \( z \)-axis is given by \[ I = \int_V (x^2 + y^2) \, \rho \, dV. \] Substituting \( x = ua \) and \( y = vb \), the equation becomes \[ I = \int_V \rho \left((ua)^2 + (vb)^2\right) abcdudvdw. \] Simplifying, \[ I = \int_V \rho (a^2u^2 + b^2v^2) abcdudvdw. \]

Step 5: Evaluate the Moment of Inertia Integral

Transform the integral into spherical coordinates over the unit sphere: \[ I = kabc \int_V (a^2u^2 + b^2v^2) dudvdw. \] Due to symmetry, \( \int u^2 dudvdw = \int v^2 dudvdw = \frac{1}{3} \int dudvdw \). Hence, \[ I = k a^2 b c \frac{1}{3} \int dudvdw + kab^2 c \frac{1}{3} \int dudvdw = kabc \left(\frac{1}{3} a^2 + \frac{1}{3} b^2 \right) \int dudvdw. \] The integral \( \int dudvdw = \frac{4}{3} \pi \), leading to \[ I = kabc \left(\frac{4}{9} \pi \right) (a^2 + b^2). \]

Key Concepts

Jacobian Determinant

When performing a change of variables in multivariable calculus, the Jacobian determinant plays a crucial role. It helps to convert volume elements from one set of variables to another. In this exercise, we transform the ellipsoid to a unit sphere by introducing new variables: \( x = ua \), \( y = vb \), \( z = wc \). This transformation involves scaling, and the scaling factor, which impacts the volume element, is represented by the Jacobian determinant.

• The transformation's effects are captured by the determinant: \( a \times b \times c \).
• The volume element \( dV = dx \, dy \, dz \) transforms as \( |Jacobian| \ du \, dv \, dw \).
• In our case, \( dV = abc \, du \, dv \, dw \).

The Jacobian determinant provides the necessary factor to adjust the integration process according to the new variable set. By capturing how volume stretches or shrinks during transformation, it ensures accurate computation of integrals in the new coordinate system.

Moment of Inertia

The moment of inertia quantifies an object's resistance to rotational motion around an axis. For the given ellipsoid, we're interested in its moment of inertia about the \( z \)-axis. Using the change of variables, we express the inertia integral over the ellipsoid.

Understanding the Inertia Equation

• Moment of inertia, \( I \), integrates \( (x^2 + y^2) \) over the volume \( V \) with a constant density \( \rho \).
• After substituting \( x = ua \) and \( y = vb \), the formula adapts to: \[I = \int_V \rho (a^2u^2 + b^2v^2) abcdudvdw.\]
• This reformulation highlights the integral over the sphere using scaled variables.

Symmetry simplifies the problem and ensures the integration over \( u^2 \) and \( v^2 \) results in symmetric values. The integral value \( \int dudvdw = \frac{4}{3} \pi \) helps finalize the calculation, yielding the inertia as a product of fundamental constants and geometrical properties of the ellipsoid.

Spherical Coordinates

Spherical coordinates simplify the integration process when dealing with problems involving spheres or ellipsoids, as they account for the symmetry inherently present.

• In spherical coordinates, points are described by the radius, polar angle, and azimuthal angle.
• Converting the ellipsoid to a unit sphere uses spherical coordinates: \( u^2 + v^2 + w^2 = 1 \).

This system is particularly advantageous because elementary shapes like spheres become straightforward to analyze, reducing a complex 3D shape to a simple boundary \( = 1 \) condition. As such, the volume and inertia computations become manageable, adhering closely to known results, like the volume being \( \frac{4}{3} \pi r^3 \).

Change of Variables

The change of variables is a fundamental technique for simplifying integral computations over complex regions. Here, it transforms an ellipsoid into a manageable unit sphere, making integration straightforward.

Applying the Technique

• Transpose the coordinates from \( (x, y, z) \) to scaled spherical variables: \( x = ua \), \( y = vb \), \( z = wc \).
• Transforming the region narrows the original complex shape into a comprehensible unit sphere: \( u^2 + v^2 + w^2 = 1 \).
This method simplifies the integration by utilizing intrinsic symmetry and uniformity. It effectively converts the problem into a familiar geometric scenario where traditional spherical integration techniques apply. Any measure of volume or distribution, such as the moment of inertia, profits from this simplification, as it links a difficult 3D problem to a simple mathematical form.