Problem 11 - University Calculus: Early Transcendentals

Question

Polar moment of inertia of an elliptical plate \(\quad\) A thin plate of constant density covers the region bounded by the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1, a>0, b>0,\) in the \(x y\) -plane. Find the first moment of the plate about the origin. (Hint: Use the transformation \(x=a r \cos \theta, y=b r \sin \theta .\)

Step-by-Step Solution

Answer

The first moment of the plate about the origin is \( \frac{2\pi \rho \cdot ab}{3} \).

Step 1: Polar Coordinate Transformation

First, realize that the problem involves an elliptical region. Use the given transformation equations to convert Cartesian coordinates to polar coordinates: \( x = a r \cos \theta \) and \( y = b r \sin \theta \). The differential area element in polar coordinates for an ellipse becomes \( dx \ dy = ab \ r \, dr \, d\theta \).

Step 2: Set Up the Integral for the First Moment

The first moment about the origin can be computed as the double integral of \( r \cdot dA \), where \( dA \) is the differential area element. Hence, \[ M_O = \int_{0}^{2\pi} \int_{0}^{1} \rho \cdot r \cdot ab \, r \, dr \, d\theta.\]Here, \( \rho \) is the constant density. Be sure to set the bounds of \( r \) from 0 to 1 and \( \theta \) from 0 to \( 2\pi \).

Step 3: Solve the Inner Integral

Calculate the inner integral with respect to \( r \):\[ \int_{0}^{1} \rho \cdot ab \, r^2 \, dr = \rho \cdot ab \cdot \left[ \frac{r^3}{3} \right]_0^1 = \frac{\rho \cdot ab}{3}.\]

Step 4: Solve the Outer Integral

Compute the outer integral with respect to \( \theta \):\[ \int_{0}^{2\pi} \frac{\rho \cdot ab}{3} \, d\theta = \frac{\rho \cdot ab}{3} \cdot [\theta]_0^{2\pi} = \frac{\rho \cdot ab \cdot 2\pi}{3}.\]

Step 5: Simplify the Answer

The first moment of the plate about the origin is simplified to:\[ M_O = \frac{2\pi \rho \cdot ab}{3}.\]This result represents the first moment of the elliptical plate about the origin.

Key Concepts

Polar Coordinates

Polar coordinates, a system used in mathematics, can be particularly handy when dealing with problems involving curves or shapes like circles and ellipses. Instead of describing a location in terms of fixed distances along two perpendicular axes, polar coordinates express points based on a distance from a reference point (the origin) and an angle from a reference direction.

Key components of polar coordinates include:

• **Radius (\( r \) )**: The radial distance from the origin to the point.
• **Angle (\( \theta \) )**: The angle measured from a fixed direction, usually the positive x-axis, to the line connecting the origin to the point.

The transformation between Cartesian and polar coordinates can often simplify calculus problems, just like in this exercise. Here, we use:

• \( x = a r \cos \theta \)
• \( y = b r \sin \theta \)

These equations allow us to navigate complex shapes like ellipses more efficiently, and transform differential elements in Cartesian planes into simpler forms in polar coordinates, as shown by \( dx \, dy = ab \, r \, dr \, d\theta \). This simplicity in integration is a pivotal feature of polar coordinates.

Elliptical Region

Understanding elliptical regions requires the basic comprehension of an ellipse, a common shape in mathematics. An ellipse looks like a squashed or elongated circle, characterized by two axes, namely the major axis (longest) and minor axis (shortest). The general equation of an ellipse in Cartesian coordinates is:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

• **\( a \)**: Represents the semi-major axis length.
• **\( b \)**: Represents the semi-minor axis length.

In physics and engineering, elliptical regions frequently appear when analyzing rotations, stress distributions, or other phenomena. Switching from Cartesian to polar coordinates for ellipses simplifies calculations significantly. This conversion involves substituting parameters into polar forms, which turns the elliptical boundary conditions into manageable integrals, as demonstrated by:

• \( x = a r \cos \theta \)
• \( y = b r \sin \theta \)

Thus, for our exercise, this approach simplifies the integral calculations over the elliptical region, allowing us to derive moments of inertia with ease.

Integral Calculus

Integral calculus focuses on accumulation and areas under curves, playing a vital role in calculating moments of inertia, areas, and volumes. This exercise involves applying integral calculus to compute the first moment of an elliptical plate about its center of mass.

Integral calculus comes with two main types:

• Ifinite Integrals: Used to find areas and accumulations on a particular range.
• Definite Integrals: Provide precise values over closed intervals.

For the problem at hand:

• The aim is to compute the first moment (\( M_O \) ) about the origin.
• We establish a double integral over the elliptical region: \[ M_O = \int_{0}^{2\pi} \int_{0}^{1} \rho \, ab \, r^2 \, dr \, d\theta \]

This integral takes into account the density (\( \rho \)) and the area element (\( ab \, r \, dr \, d\theta \)), calculated in steps as follows:

• Inner integral with respect to \( r \): \[ \int_{0}^{1} \rho \, ab \, r^2 \, dr = \frac{\rho \, ab}{3} \]
• Outer integral with respect to \( \theta \): \[ \int_{0}^{2\pi} \frac{\rho \, ab}{3} \, d\theta = \frac{\rho \, ab \, 2\pi}{3} \]

Thus, through integral calculus, we conclude with \( M_O = \frac{2\pi \rho \, ab}{3} \), elegantly summarizing the accumulation of density across the elliptical plate.