Problem 20 - Advanced Engineering Mathematics

Question

Find the indicated moment of inertia of the lamina that has the given shape and density. Outside \(r=1\) and inside \(r=2 \sin 2 \theta\), first quadrant; \(\rho(r, \theta)=\sec ^{2} \theta ; I_{y}\)

Step-by-Step Solution

Answer

To find \( I_y \), integrate \( r^3 \sin^2\theta \sec^2\theta \) from \( r = 2\sin(2\theta) \) to \( r = 1 \) and \( \theta = 0 \) to \( \frac{\pi}{2} \).

Step 1: Understand the Problem

We are asked to find the moment of inertia about the y-axis, denoted as \( I_y \), for a region defined in the first quadrant by the lemniscate \( r = 2\sin(2\theta) \) and the circle \( r = 1 \). The density function is given as \( \rho(r, \theta) = \sec^2(\theta) \).

Step 2: Establish Integral for Moment of Inertia

The moment of inertia \( I_y \) for a polar region is calculated using the integral \( I_y = \int \int_R r^2 \sin^2 \theta \cdot \rho(r, \theta) \cdot r \, dr \, d\theta \). Here, the extra \( r \) comes from converting to polar coordinates.

Step 3: Determine Limits of Integration

Since we are in the first quadrant, \( \theta \) ranges from 0 to \( \frac{\pi}{2} \). For \( r \), it ranges from the lemniscate \( 2\sin(2\theta) \) to the circle \( 1 \). Thus, the integration bounds for \( r \) are from \( 2\sin(2\theta) \) to \( 1 \).

Step 4: Setup the Double Integral

The double integral setup is as follows:\[ I_y = \int_{0}^{\frac{\pi}{2}} \int_{2\sin(2\theta)}^{1} r^3 \sin^2 \theta \sec^2 \theta \, dr \, d\theta.\] Here, \( \rho(r, \theta) = \sec^2(\theta) \) and the \( r^3 \) accounts for \( r^2 \sin^2 \theta \cdot r \).

Step 5: Integrate with Respect to r

First, integrate with respect to \( r \):\[ \int_{2\sin(2\theta)}^{1} r^3 \, dr = \left[ \frac{r^4}{4} \right]_{2\sin(2\theta)}^{1} = \frac{1}{4} - \frac{(2\sin(2\theta))^4}{4}.\]

Step 6: Integrate with Respect to \( \theta \)

Substitute the results from the \( r \)-integral back into the \( \theta \)-integral:\[ I_y = \int_{0}^{\frac{\pi}{2}} \left( \frac{1}{4} - \frac{16 \sin^4(2\theta)}{4} \right) \sin^2(\theta) \sec^2(\theta) \, d\theta.\] Simplify and integrate this expression, calculating each portion separately.

Key Concepts

Polar Coordinates

Polar coordinates are a common system used to describe points in a plane using a radius and an angle. Instead of the familiar Cartesian coordinates, which describe a location using x and y values, polar coordinates use:

• \( r \) representing the radius or distance from the origin.
• \( \theta \) representing the angle measured from the positive x-axis.

This system is particularly useful when dealing with circles or curves that naturally fit the description of radius and angle, such as lemniscates or spirals.
In our example, the region of interest is defined by two equations in polar coordinates: a circle \( r = 1 \) and a lemniscate \( r = 2\sin(2\theta) \). To solve the problem, the switch from Cartesian to polar coordinates simplifies the process as the shapes align more naturally with the coordinate system.

Double Integral

When calculating the moment of inertia for regions in a plane, double integrals are used. These integrals allow us to consider not just two dimensions separately, but how they work together within a specific area. A double integral over a region \( R \) is given as \[\int \int_R \, f(x, y) \, dA,\]where \( f(x, y) \) is a function defined over \( R \), and \( dA \) is an element of area.
In polar coordinates, this becomes \[\int \int_R \, f(r, \theta) \, r \, dr \, d\theta,\]where \( r \) appears as a factor due to the conversion from Cartesian to polar coordinates.
In this exercise, the double integral is set up to find the moment of inertia \( I_y \), with components \( r^2 \sin^2(\theta) \cdot \rho(r, \theta) \cdot r \). The comparable length of the formula requires an extra \( r \) factor.

Density Function

A density function \( \rho(r, \theta) \) reflects how mass is distributed over a given area. It's crucial in calculating properties such as the moment of inertia, as it affects how mass contributes to the property in question.
In this problem, the density function is \( \rho(r, \theta) = \sec^2(\theta) \). It means that the density of the lamina changes with the angle \( \theta \), following the secant-squared relationship.

• Secant, \( \sec\theta \), is the reciprocal of the cosine function.
• Therefore, \( \sec^2\theta \) is more significantly impactful when \( \cos\theta \) is small, typically when \( \theta \) is approaching \( \frac{\pi}{2} \).

This variability influences the computation of the moment of inertia, contributing uniquely based on the angle.

Lemniscate

The lemniscate is a particular kind of curve resembling the infinity symbol (∞), often described in polar coordinates. In this exercise, the lemniscate is defined by the equation \( r = 2\sin(2\theta) \).

• This equation describes a figure-eight shape that is symmetric about both axes.
• The specific formulation \( 2\sin(2\theta) \) also relates to its name, stemming from the Latin word "lemniscus," meaning a type of ribbon.

In the problem, this lemniscate is used to define one of the boundaries of the region within which the moment of inertia is calculated.
Understanding its properties is crucial when dealing with polar coordinates, as it defines integration limits within the first quadrant where \( \theta \) ranges from 0 to \( \frac{\pi}{2} \).

Integration Limits

Integration limits are essential for computing integrals, as they specify the boundaries within which the integral is calculated. They define the beginnings and ends of the area of interest.
For this exercise, we determine limits in both \( r \) and \( \theta \):

• \( \theta \) ranges from 0 to \( \frac{\pi}{2} \), confining the problem to the first quadrant.
• \( r \) ranges between the curves described by the lemniscate \( r = 2\sin(2\theta) \) and the circle \( r = 1 \).

These specific limits ensure the integration correctly captures the desired region, accommodating the necessary components for calculating the moment of inertia \( I_y \). Proper definition of integration limits is fundamental to solving problems involving areas or volumes in calculus, particularly those expressed in polar coordinates.