Problem 28 - Thomas Calculus

Question

Polar moment of a cardioid region Find the polar moment of inertia about the origin of a thin plate enclosed by the cardioid \(r=1+\cos \theta\) if the plate's density function is \(\delta(x, y)=1\)

Step-by-Step Solution

Answer

The polar moment of inertia is \( \frac{15\pi}{8} \).

Step 1: Convert the Polar Equation to Cartesian Coordinates

The cardioid is given in polar coordinates as \( r = 1 + \cos \theta \). To find the polar moment of inertia, we first convert this equation into Cartesian coordinates. The equations for conversion are \( x = r \cos \theta \) and \( y = r \sin \theta \). Substitute \( r \) to express them in terms of \( \theta \).

Step 2: Write the Polar Moment of Inertia Integral

The polar moment of inertia about the origin in polar coordinates is given by the integral: \[ I = \int_{0}^{2\pi} \int_{0}^{1 + \cos \theta} r^3 \delta(r \cos \theta, r \sin \theta) \, dr \, d\theta \]Here, \( \delta(x, y) = 1 \), so the integral simplifies to \( I = \int_{0}^{2\pi} \int_{0}^{1 + \cos \theta} r^3 \, dr \, d\theta \).

Step 3: Evaluate the Inner Integral

Calculate the inner integral with respect to \( r \): \[ \int_{0}^{1+\cos\theta} r^3 \, dr \]. Using the power rule for integration, we have: \[ \left[ \frac{r^4}{4} \right]_{0}^{1+\cos\theta} = \frac{(1+\cos\theta)^4}{4} \].

Step 4: Evaluate the Outer Integral

Now, substitute back into the outer integral over \( \theta \): \[ I = \int_{0}^{2\pi} \frac{(1+\cos\theta)^4}{4} \, d\theta \]. Use trigonometric identities and symmetry to solve this integral, as it is periodic with the period of the cosine function. Calculate it to find the moment of inertia.

Step 5: Simplify the Integral using Trigonometric Identities

Expand \((1+\cos\theta)^4\) to simplify the integration: \[ (1 + 4\cos\theta + 6\cos^2\theta + 4\cos^3\theta + \cos^4\theta) \cdot \frac{1}{4} \]. This is split into separate integrals that can be evaluated using known integrals of trigonometric functions.

Step 6: Calculate Each Part of the Integral

For each term in the expanded form, calculate:- \( \int_{0}^{2\pi} 1 \, d\theta = 2\pi \)- \( \int_{0}^{2\pi} \cos\theta \, d\theta = 0 \)- \( \int_{0}^{2\pi} \cos^3\theta \, d\theta = 0 \)- \( \int_{0}^{2\pi} \cos^2\theta \, d\theta = \pi \)- \( \int_{0}^{2\pi} \cos^4\theta \, d\theta = \frac{3\pi}{4} \)Combine these results to obtain the final value.

Step 7: Combine Results to Find the Polar Moment of Inertia

Now substitute all the calculated integrals back into the expanded equation for \( I \): \[ I = \frac{1}{4}(2\pi + 0 + 6\pi + 0 + \frac{3\pi}{4}) \].This simplifies to give: \[ I = \frac{15\pi}{8} \]. Thus, the polar moment of inertia is \( \frac{15\pi}{8} \).

Key Concepts

Polar Coordinates

In mathematics, polar coordinates provide a fascinating way to describe the position of a point in a plane. Unlike the traditional Cartesian coordinates that use a grid of squares, polar coordinates utilize circles centered around the origin (0,0). A point in polar coordinates is determined by:

• A distance from the origin called the radial coordinate (r)
• An angle from the positive x-axis, measured in radians (θ)

Together, these two values give us a unique way to specify any point in a plane. For example, the cardioid curve given by the equation \( r = 1 + \cos \theta \) is a result of how the distance r changes as the angle θ varies. The beauty of polar coordinates is how they simplify problems involving circular or rotational symmetry. This transformation simplifies integration over regions bounded by curves like cardioids.

Density Function

Density functions play a crucial role in determining physical properties such as mass or inertia in physics. In this context, density represents the mass per unit area of an object. For example, if we have a thin plate, the density function \( \delta(x, y) \) describes how weight is distributed across its surface.

In the given exercise, the density function is constant at \( \delta(x, y) = 1 \), which simplifies calculations. This uniform density means every part of the plate has the same mass per unit area, making it easier to compute the polar moment of inertia. Understanding density functions allows us to evaluate integral expressions that accumulate total properties over complex shapes and designs, such as the area enclosed by a cardioid.

Trigonometric Identities

Trigonometric identities are mathematical tools that help simplify expressions involving trigonometric functions like sine and cosine. In the realm of calculus and integration, these identities are vital for transforming complex expressions into manageable forms.

For example, to integrate the function \((1+\cos \theta)^4\), one can expand it using identities such as:

• \( \cos^2 \theta = \frac{1+\cos 2\theta}{2} \)
• \( \cos^3 \theta \) is often simplified using power-reduction formulas

These identities make it possible to break down a challenging integral into a series of simpler ones. Here, after expanding \((1+\cos \theta)^4\), each new integral can be solved independently, ultimately leading to an easier calculation of the overall integral.

Integral Evaluation

Integral evaluation is a fundamental aspect of calculus that helps us find solutions to problems involving accumulation. An integral calculates the total of a function over a specified interval, which can represent various physical concepts like areas, volumes, or moments of inertia.

In this exercise, the polar moment of inertia is expressed as an integral in polar coordinates:\[I = \int_{0}^{2\pi} \int_{0}^{1 + \cos \theta} r^3 \ dr \ d\theta\]This double integral accumulates contributions from small elements within the region defined by the cardioid curve. By evaluating these inner and outer integrals separately, each in terms of \( r \) and \( \theta \), we combine the results to find the total moment of inertia. Mastery of integral evaluation techniques allows us to solve complex physical problems that would otherwise be unsolvable.