Problem 69 - Calculus Early Transcendentals

Question

Suppose the density of a thin plate represented by the region \(R\) is \(\rho(r, \theta)\) (in units of mass per area). The mass of the plate is \(\iint_{R} \rho(r, \theta) d A .\) Find the mass of the thin half annulus \(R=\\{(r, \theta): 1 \leq r \leq 4,0 \leq \theta \leq \pi\\}\) with a density \(\rho(r, \theta)=4+r \sin \theta\)

Step-by-Step Solution

Answer

Based on the given conditions, the mass of the thin half annulus is found by using double integral in polar coordinates of the density function, ρ(r, θ) = 4 + r*sin(θ). The mass is calculated as M = 24 + 28π units.

Step 1: Set up the double integral

To set up the double integral, we will integrate \(\rho(r, \theta)\) over the given region \(R\), using polar coordinates: $$ M = \iint_R \rho(r, \theta) dA = \int_{0}^{\pi} \int_{1}^{4} (4 + r\sin\theta) r\, dr\, d\theta $$

Step 2: Integrate with respect to \(r\)

Integrate the inner integral with respect to \(r\): $$ \int_{1}^{4} (4 + r\sin\theta)r\, dr = \int_{1}^{4} (4r + r^2\sin\theta) dr $$ Now, integrate both terms: $$ \int_{1}^{4} (4r + r^2\sin\theta) dr = \left[2r^2 + \frac{1}{3}r^3\sin\theta\right]_{1}^{4} $$ Simplify: $$ \left[2(4^2) + \frac{1}{3}(4^3)\sin\theta - \left(2(1^2) + \frac{1}{3}(1^3)\sin\theta\right)\right] = 12\sin\theta + 28 $$

Step 3: Integrate with respect to \(\theta\)

Now integrate the outer integral with respect to \(\theta\): $$ \int_{0}^{\pi} (12\sin\theta + 28) d\theta $$ Integrate both terms: $$ \int_{0}^{\pi} (12\sin\theta + 28) d\theta = \left[-12\cos\theta + 28\theta\right]_{0}^{\pi} $$ Simplify: $$ \left[-12\cos(\pi) + 28(\pi) - (-12\cos(0) + 28(0))\right] = 12(2) + 28\pi = 24 + 28\pi $$ The mass of the thin half annulus is \(24 + 28\pi\) units.

Key Concepts

Double Integrals

Double integrals are a powerful mathematical tool used to calculate areas, volumes, and in this case, mass across regions in two dimensions. They function by multiplying together small quantities, summed across a given region, to compute a particular property of interest. In the context of our problem, they help us determine the total mass of a plate with varying density over a specific region.

In this exercise, we have the double integral set up as:

• First, an integral over radius \( r \) from 1 to 4,
• and second, an integral over angle \( \theta \) from 0 to \( \pi \).

Each integral handles a different dimension in our region, allowing us to consider all of the points \( (r, \theta) \) in the area described by the problem.

The function \( \rho(r, \theta) \) is the density function, and when integrated over the complete region \( R \), it gives us the total mass therein. Each little piece \( dA \) of the annulus contributes to the mass according to how thickly \( \rho \) lays across it.

Polar Coordinates

Polar coordinates are a very handy way of analyzing regions and functions that have radial symmetry or circular shapes. Unlike the Cartesian coordinate system that uses \( x \) and \( y \) axes, polar coordinates use \( r \) (the radius) and \( \theta \) (the angle) to define positions in the plane.

In our setup, the region is a half annulus, which is perfectly suited to polar coordinates. The boundaries can be easily described as:

• anging from 1 to 4 units from the origin (the radii), and
• an angle \( \theta \) that ranges from 0 to \( \pi \).

This makes the integration process more streamlined and in line with the region's shape. Polar coordinates simplify the problem by aligning the area and coordinate descriptions, often leading to elegant and more manageable integrals.

Density Function

The density function \( \rho(r, \theta) = 4 + r \sin \theta \) describes how mass is distributed over the plate. It is given as a function of both \( r \) and \( \theta \), meaning the density varies with position.

This density function tells us that at any given point, the mass per unit area is influenced by:

• a constant amount of 4 units,
• plus a variable amount depending on both the radius \( r \) and angle \( \theta \).

The \( r \sin \theta \) part means that areas further from the origin and with certain angles will have more mass per unit area. When performing integration, this function helps weigh different parts of the region accordingly. The regions where \( \rho \) is higher contribute more heavily to the total mass.

Integration Techniques

The integration techniques used here are essentially split into two steps: integrating with respect to \( r \) and with respect to \( \theta \).

• **Integration over \( r \):** Considered within the radial bounds \( r = 1 \) to \( r = 4 \), this is calculated first. The operation must be done carefully by treating every term separately, including terms like \( 4r \) and \( r^2 \sin \theta \).
• **Integration over \( \theta \):** Following \( r \), integrate over the angle \( \theta \) from \( 0 \) to \( \pi \). This step incorporates angular density variations into the mass calculation.

Handling each integral calls for both fundamental and specialized integration skills. You might encounter the product rule and substitution occasionally, but here the focus lies on straightforward evaluations both analytically and in terms of interpreting results. Therefore, the techniques used in this problem are accessible, yet deliver insightful understanding on how integration in polar coordinates aids in calculating exact values for practical scenarios like mass distributions.