Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

Recall the formula \(A=\frac{1}{2} r^{2} \theta\) for the area of the sector of a circle of radius \(r\) and central angle \(\theta\) radians (Section 10.7). Use this to obtain the formula $$ A=\frac{r_{1}+r_{2}}{2}\left(r_{2}-r_{1}\right)\left(\theta_{2}-\theta_{1}\right) $$ for the area of the polar rectangle \(\left\\{(r, \theta): r_{1} \leq r \leq r_{2}\right.\), \(\left.\theta_{1} \leq \theta \leq \theta_{2}\right\\} .\)

Step-by-Step Solution

Answer

The area of the polar rectangle is given by \(A=\frac{r_1 + r_2}{2}(r_2 - r_1)(\theta_2 - \theta_1)\).

Step 1: Understanding the Polar Rectangle

A polar rectangle is defined by the range of radius values from \( r_1 \) to \( r_2 \) and the range of angle values from \( \theta_1 \) to \( \theta_2 \). Our goal is to find the area of this region.

Step 2: Visualizing the Sector

Imagine the polar rectangle as a combination of two sectors with radii \( r_1 \) and \( r_2 \), and the angle range \( \theta_2 - \theta_1 \). The area of each sector can be calculated using the formula \( A = \frac{1}{2} r^2 \theta \).

Step 3: Calculating Sector Areas

Calculate the area of the sector that makes up the larger circle (radius \( r_2 \)): \[ A_2 = \frac{1}{2} r_2^2 (\theta_2 - \theta_1) \]Similarly, calculate for the smaller circle (radius \( r_1 \)):\[ A_1 = \frac{1}{2} r_1^2 (\theta_2 - \theta_1) \]

Step 4: Determining the Area of the Polar Rectangle

To find the area of the polar rectangle, subtract the area \( A_1 \) of the inner sector from the area \( A_2 \) of the outer sector. This gives:\[ A = A_2 - A_1 = \frac{1}{2} r_2^2 (\theta_2 - \theta_1) - \frac{1}{2} r_1^2 (\theta_2 - \theta_1) \]

Step 5: Simplifying the Expression

Factor out the common terms:\[ A = \frac{1}{2} (r_2^2 - r_1^2) (\theta_2 - \theta_1) \]Recognize the difference of squares: \[ r_2^2 - r_1^2 = (r_2 + r_1)(r_2 - r_1) \]Substitute back into the equation:\[ A = \frac{1}{2} (r_2 + r_1)(r_2 - r_1) (\theta_2 - \theta_1) \]

Step 6: Final Simplification to Obtain Desired Formula

Distribute the \( \frac{1}{2} \) into the first term:\[ A = \frac{r_2 + r_1}{2} (r_2 - r_1)(\theta_2 - \theta_1) \]This is the required area formula for the polar rectangle.

Key Concepts

Area of a Sector

The area of a sector is a segment of a circle that resembles a pizza slice. If we know the radius of the circle and the angle of the sector, we can find its area using the formula: \[ A = \frac{1}{2} r^{2} \theta \]Here, \( r \) stands for the radius of the circle, and \( \theta \) represents the central angle in radians. This formula is derived from the fact that if the entire circle were one whole sector, its area would be \( \pi r^2 \). Since we're only interested in a fraction of that circle, corresponding to the angle \( \theta \), we scale the total area by \( \frac{1}{2} \theta / \pi \). This allows us to compute the specific area of our sector easily.

Polar Rectangle

A polar rectangle is a concept in polar coordinates, which can be tricky at first. In polar coordinates, we describe locations with a radius and an angle, unlike the usual x-y coordinate system.A polar rectangle is defined by:

• Radial boundaries, from \( r_1 \) to \( r_2 \)
• Angular boundaries, from \( \theta_1 \) to \( \theta_2 \)

Imagine it as a doughnut ring sector. This is because it covers a sweep of angles, much like a sector, but stretches between two different radii. When tackling problems involving polar rectangles, we find the area between these boundaries, essentially carving out a slice of the circle.

Radius and Angle in Polar Coordinates

In polar coordinates, every point is defined by two values: a radius \( r \) and an angle \( \theta \). This system is particularly useful for problems involving circles and other round shapes, as it represents positions in terms of circles rather than boxes.Here’s how it works:

• The radius \( r \) tells you how far from the origin the point is.
• The angle \( \theta \) tells you the direction from the origin, typically measured from the positive x-axis.

This way of defining points makes circular problems, like finding sectors or areas defined by radii and angles much simpler, by avoiding the complexity of rectangular coordinates.

Difference of Squares

The difference of squares is a valuable algebraic tool, seen as \( a^2 - b^2 = (a + b)(a - b) \). This identity is like a secret key that simplifies certain expressions.In the context of polar geometry, it helps us simplify the area expression for a polar rectangle. When we take the area of two sectors and subtract them to find the region between, we get a term like \( r_2^2 - r_1^2 \). Recognizing this as a difference of squares allows re-arrangement to:

• \( r_2^2 - r_1^2 = (r_2 + r_1)(r_2 - r_1) \)
• This factorization transforms a complex arithmetic task into a simple multiplication problem.

Using the difference of squares tactically breaks the problem into more manageable parts, great for simplifying equations involved in polar coordinates. This lets us arrive at solutions efficiently and elegantly.