Problem 30 - University Calculus: Early Transcendentals

Question

Find the area of the region enclosed by the positive \(x\) -axis and spiral \(r=4 \theta / 3,0 \leq \theta \leq 2 \pi .\) The region looks like a snail shell.

Step-by-Step Solution

Answer

The area is \( \frac{64\pi^3}{27} \).

Step 1: Identify the Problem

We need to find the area of the region enclosed by a polar curve, specifically the spiral given by the equation \( r = \frac{4\theta}{3} \) over the interval \(0 \leq \theta \leq 2\pi\).

Step 2: Recall Polar Area Formula

The area \( A \) enclosed by a polar curve \( r = f(\theta) \) from \( \theta=a \) to \( \theta=b \) is given by the integral formula: \[ A = \frac{1}{2} \int_{a}^{b} r^2 \, d\theta. \]

Step 3: Substitute Into the Formula

Substitute \( r = \frac{4\theta}{3} \) into the formula to get: \[ A = \frac{1}{2} \int_{0}^{2\pi} \left(\frac{4\theta}{3}\right)^2 \, d\theta. \] This simplifies to: \[ A = \frac{1}{2} \int_{0}^{2\pi} \frac{16\theta^2}{9} \, d\theta. \]

Step 4: Simplify the Integral

Factor constants out of the integral: \[ A = \frac{8}{9} \int_{0}^{2\pi} \theta^2 \, d\theta. \]

Step 5: Evaluate the Integral

Compute the integral \( \int \theta^2 \, d\theta \) to find \( \frac{\theta^3}{3} \), then evaluate it from 0 to \( 2\pi \): \[ \frac{8}{9} \left[ \frac{(2\pi)^3}{3} - \frac{0^3}{3} \right]. \]

Step 6: Calculate the Result

Calculate the expression: \( \frac{8}{9} \cdot \frac{8\pi^3}{3} = \frac{64\pi^3}{27} \).

Key Concepts

Polar Area Formula

When we talk about finding areas in polar coordinates, we use a special method called the polar area formula. Polar coordinates describe a point in the plane using the distance from a reference point and an angle from a reference direction. Finding the area enclosed by a polar curve is quite different from the traditional method used in Cartesian coordinates.

The formula for calculating the area, \( A \), bounded by a polar curve \( r = f(\theta) \) from angle \( \theta = a \) to \( \theta = b \) is given by:

• \[ A = \frac{1}{2} \int_{a}^{b} r^2 \, d\theta \]

This formula is derived from the concept of breaking the area into small, thin sectors. Each sector is like a tiny slice of pie, whose area is easier to calculate.
Understanding this formula is crucial as it simplifies an otherwise complex problem. It allows us to integrate polar functions and find the area under the curve, even for complicated shapes like spirals.

Spiral Curve

A spiral curve is an intriguing type of polar graph that winds around a point, getting further away as it winds along. The specific spiral in this problem is described by the equation \( r = \frac{4\theta}{3} \), which represents a spiral starting at the pole and increasing outward as \( \theta \) increases.

Spirals like this are fascinating because they show how polar coordinates can elegantly describe curves that are cumbersome in Cartesian coordinates.

• One noteworthy feature is that the radius, \( r \), is directly proportional to the angle, \( \theta \). This means as \( \theta \) increases, so does the distance from the origin.
• This type of spiral is often referred to as an Archimedean spiral, characterized by the property that the distance between turns of the spiral is constant.

The problem asked us to find the area enclosed by this spiral from \( \theta = 0 \) to \( \theta = 2\pi \). Knowing the geometry of a spiral assists in visualizing and solving the problem.

Integral Calculus

Integral calculus is a branch of mathematics focusing on finding areas, volumes, and accumulations. In this problem, we're using integral calculus to evaluate the area under a spiral curve.

By setting up an integral using the polar area formula, you turn a geometric problem into a calculus one. In our specific example, we calculated the area enclosed by the spiral from \( \theta = 0 \) to \( \theta = 2\pi \).To do this:

• We substitute the given function \( r = \frac{4\theta}{3} \) into the polar area formula.
• This gives us \[ A = \frac{1}{2} \int_{0}^{2\pi} \left(\frac{4\theta}{3}\right)^2 \, d\theta \].
• After simplification, the integral becomes \[ A = \frac{8}{9} \int_{0}^{2\pi} \theta^2 \, d\theta \].

Solving this integral involves using the power rule to find the antiderivative, \( \int \theta^2 \, d\theta = \frac{\theta^3}{3} \), which is then evaluated over the bounds of the interval.

This integral ultimately gives us the area, illustrating how integral calculus transforms the problem of calculating areas into a manageable sequence of mathematical steps.