Problem 39 - Calculus Early Transcendentals

Question

Sketch each region and use a double integral to find its area. $$\text { The annular region }\\{(r, \theta): 1 \leq r \leq 2,0 \leq \theta \leq \pi\\}$$

Step-by-Step Solution

Answer

In summary, the annular region is a semiannular shape enclosed between two concentric circles with radii 1 and 2 in the first half of the plane. The area of this region can be calculated using a double integral in polar coordinates with the proper limits, resulting in an area of $\frac{3\pi}{2}$.

Step 1: Sketch the annular region

To sketch the annular region, consider the bounds for \(r\) and \(\theta\). The bound for \(r\) is \(1 \leq r \leq 2\). This bound indicates that the annular region will fall between two concentric circles with radii 1 and 2. The bound for \(\theta\) is \(0 \leq \theta \leq \pi\). This means that the annular region will be limited within the first half of the plane, indicating the region will be a semiannular shape. Sketch this annular region on an \(xy\)-plane with the two concentric circles and the semiannular shape enclosed between them.

Step 2: Set up the double integral

To find the area of the annular region using a double integral, we will need to set up the integral with suitable limits and then evaluate it. As stated before, the double integral in polar coordinates is expressed as \(\int \int _R r drd\theta\). For our problem, the limits for \(r\) are given by \(1 \leq r \leq 2\), and the limits for \(\theta\) are given by \(0 \leq \theta \leq \pi\). Therefore, the double integral can be written as: $$\int_{0}^{\pi} \int_{1}^{2} r dr d\theta$$

Step 3: Evaluate the inner integral

We'll now evaluate the inner integral with respect to \(r\). This requires integrating \(r\) with the given limits for \(r\): $$\int_{1}^{2} r dr = \frac{1}{2}r^2 \bigg|_{1}^{2} = \frac{1}{2}(2^2 - 1^2) = \frac{1}{2}(4-1) = \frac{3}{2}$$ We can now substitute this value back into the double integral: $$\int_{0}^{\pi} \frac{3}{2} d\theta$$

Step 4: Evaluate the outer integral

Finally, evaluate the outer integral with respect to \(\theta\) with the given limits for \(\theta\): $$\int_{0}^{\pi} \frac{3}{2} d\theta = \frac{3}{2}\theta\bigg|_{0}^{\pi} = \frac{3}{2}(\pi - 0) = \frac{3\pi}{2}$$ The area of the annular region is, therefore, \(\frac{3\pi}{2}\).

Key Concepts

Understanding Polar Coordinates

Polar coordinates offer an alternative way to describe locations on a plane. Unlike the traditional Cartesian system, which uses \(x\) and \(y\) values, polar coordinates define a point by its distance from a reference point (the origin) and the angle it makes with a reference direction (usually the positive x-axis). In polar coordinates, each point is represented as \((r, \theta)\):

• \(r\) is the radius or the distance from the origin.
• \(\theta\) is the angle in radians, measured from the positive x-axis.

For our specific problem, we are given \(1 \leq r \leq 2\) and \(0 \leq \theta \leq \pi\). This means the points we are interested in fall between two circles with radii 1 and 2, but only on the upper half of the plane. This makes our region semiannular. Utilizing polar coordinates makes it easier to handle problems involving circles or rotation because calculations with angles and radii often become simpler than using Cartesian coordinates. You can think of polar coordinates as a more intuitive way to describe circular regions.

Finding the Area of Annular Regions

An annular region is the area trapped between two concentric circles. It looks like a ring, or annulus, and is formed by specifying an inner and outer radius. Here's a simple way to visualize this:

• Imagine two circles centered at the same point.
• Their radii are different, with one being smaller and completely inside the other.

The space between these two circles forms the annular region. In this exercise, the annular region is defined by \( 1 \leq r \leq 2 \) and \( 0 \leq \theta \leq \pi \). This tells us:

• The annulus is only half of the complete ring because \(\theta\) only ranges from 0 to \(\pi\), indicating a semiannular shape.
• The limits on \(r\) signal that we should consider the area between two radii: 1 and 2.

Realizing the region as semiannular is crucial. The shape will influence the setup and evaluation of our integral when calculating the area.

Determining Integration Boundaries

Integration boundaries play a key role in defining the limits where integration occurs. These boundaries tell us the extent of the region over which we want to integrate. They are crucial for setting up the equations especially when calculating areas like that of an annular region. For a double integral in polar coordinates:

• The inner boundary, here \(1 \leq r \leq 2\), dictates the radial distance over which we integrate.
• The outer boundary, \(0 \leq \theta \leq \pi\), defines the angular section of the circle.

Together, these boundaries describe the specific semiannular sector we need to find the area of. In this problem:

• We calculated the integral of \(r\) first, because the innermost integration is always with respect to the radial coordinate in polar expressions.
• Then, we integrate over \(\theta\), which dictates how far round the circle we go, completing the evaluation of the double integral.

By effectively setting and using these boundaries, the process of evaluating the area through integration becomes systematic and solves the problem accurately.