Problem 44 - University Calculus: Early Transcendentals

Question

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 \(r=f(\theta), \alpha \leq \theta \leq \beta\).

Step-by-Step Solution

Answer

The area is given by \(A = \int_{\alpha}^{\beta} \frac{1}{2} r^{2} \, d\theta\).

Step 1: 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 \(\beta\).

Step 2: 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).

Step 3: 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:\[A = \int_{\alpha}^{\beta} \int_{0}^{f(\theta)} r \, dr \, d\theta.\]

Step 4: 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 \(f(\theta)\):\[\left[ \frac{1}{2}r^{2} \right]_{0}^{f(\theta)} = \frac{1}{2}(f(\theta))^{2} - \frac{1}{2}(0)^{2}= \frac{1}{2}(f(\theta))^{2}.\]

Step 5: Substitute Back to the Outer Integral

Substitute the result from the inner integral back into the outer integral:\[A = \int_{\alpha}^{\beta} \frac{1}{2} (f(\theta))^{2} \, d\theta.\]

Step 6: 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.

Key Concepts

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.

Transforming to polar coordinates changes the representation of the function, making it often easier to solve certain integrals. When dealing with such polar curves, we use the formula for a region in polar coordinates, which is set up in the step-by-step solution: \[A = \int_{\alpha}^{\beta} \int_{0}^{f(\theta)} r \ dr \ d\theta\]This nested integral first accounts for the radius from the origin out to the curve, then sums these slices over the given angle range.

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 \( \beta \).

The inner integral calculates the area of these tiny segments:\[\int_{0}^{f(\theta)} r \, dr\]When solving, the result gives the area of each slice in terms of \( \theta \), represented as:\[\frac{1}{2} (f(\theta))^2\]The outer integral then collects all these segments as \( \theta \) sweeps from \( \alpha \) to \( \beta \), giving us the entire area in a coherent manner.

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.

Consider the fan-shaped region determined by the curve. Its area can be articulated simply by examining how far out from the origin \( r \) extends at each angle and over what range \( \theta \) spans. This exercise beautifully illustrates how the use of polar curves and integrals can simplify what might otherwise be a complex calculation in Cartesian coordinates.