Chapter 14: Problem 44
Use the double integral in polar coordinates to derive the formula $$A=\int_{\alpha}^{\beta} \frac{1}{2} r^{2} d \theta$$ for the area of the fan-shaped region between the origin and polar curve \(r=f(\theta), \alpha \leq \theta \leq \beta\).
Short Answer
Expert verified
The area is given by \(A = \int_{\alpha}^{\beta} \frac{1}{2} r^{2} \, d\theta\).
Step by step solution
01
Understand the Problem
We are tasked with converting a polar coordinate problem into a double integral to find the area of a fan-shaped region defined by a given polar curve \(r = f(\theta)\) for angles between \(\alpha\) and \(\beta\).
02
Set Up the Area in Polar Coordinates
The area \(A\) of a region in polar coordinates can be found using the double integral formula \(A = \int_{\alpha}^{\beta} \int_{0}^{f(\theta)} r \, dr \, d\theta\). This integral allows us to compute the area by summing up the infinitesimally small areas of differential sectors (which look like small slices of a circle).
03
Write the Area as a Double Integral
The integral limits reflect the radial distance from the origin (from 0 to \(f(\theta)\)) and the angular sweep from \(\alpha\) to \(\beta\). Thus, the expression for the area is:\[A = \int_{\alpha}^{\beta} \int_{0}^{f(\theta)} r \, dr \, d\theta.\]
04
Evaluate the Inner Integral
The inner integral \(\int_{0}^{f(\theta)} r \, dr\) is computed as follows:1. Antiderivative of \(r\) is \(\frac{1}{2}r^{2}\).2. Evaluate from \(0\) to \(f(\theta)\):\[\left[ \frac{1}{2}r^{2} \right]_{0}^{f(\theta)} = \frac{1}{2}(f(\theta))^{2} - \frac{1}{2}(0)^{2}= \frac{1}{2}(f(\theta))^{2}.\]
05
Substitute Back to the Outer Integral
Substitute the result from the inner integral back into the outer integral:\[A = \int_{\alpha}^{\beta} \frac{1}{2} (f(\theta))^{2} \, d\theta.\]
06
Simplify the Integral Expression
Notice that \((f(\theta))^{2}\) corresponds to \(r^{2}\) since \(r = f(\theta)\). Therefore, the formula becomes:\[A = \int_{\alpha}^{\beta} \frac{1}{2} r^{2} \, d\theta,\] which matches the required expression for the area of the fan-shaped region in polar coordinates.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Double Integral
A double integral is a powerful mathematical tool used to compute the volume under a surface or the area of a region in a plane. It is essentially an extension of a regular integral, with one more dimension. For the scenario of calculating areas in polar coordinates, double integrals are quite handy.
- A double integral over a polar region simplifies the process of finding areas where the geometry is naturally suited to polar coordinates, like circular or fan-shaped areas.
- In essence, we calculate the sum of infinitely small areas by integrating radial and angular components separately.
- By integrating over both the radial distance and the angle, a double integral accounts for both dimensions effectively.
Area Calculation
Calculating the area of a fan-shaped region, as described in the exercise, involves understanding how polar coordinates simplify this task. Essentially, we use integration to accumulate the tiny sector areas—the 'slices' of the fan.
- Imagine cutting the fan into many narrow slices along the radial lines. Each slice resembles a triangle with a very small base (an arc) and height (the radius).
- By summing these infinitesimal slices, using integration, we build up the total area.
- The integration bounds for the radius extend from 0 to \( f(\theta) \), while the angle, or theta, varies from \( \alpha \) to \( \beta \).
Polar Curves
Polar curves represent a type of graph defined in terms of radii and angles rather than traditional x and y coordinates. This way of expressing curves is particularly useful for shapes that have circular symmetry, such as the fan we are analyzing.
- Phrases like "polar curves" describe functions where the radius \( r \) varies as a function of the angle \( \theta \).
- They can depict a variety of intricate and complex shapes by defining \( r \) as a function of \( \theta \), noted as \( r = f(\theta) \).
- Working with polar curves in exercises involves often translating between the radial function and its equivalent Cartesian coordinates if needed.