Problem 20 - Thomas Calculus

Question

Snail shell 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 enclosed by the spiral is \( \frac{64\pi^3}{27} \).

Step 1: Understand the Spiral Equation

The given equation is in polar form: \( r = \frac{4 \theta}{3} \). As \( \theta \) varies from \( 0 \) to \( 2\pi \), the spiral unwinds from the origin outward, tracing a path described by \( r \).

Step 2: Determine the Integration Limits

The limits for \( \theta \) are given as [0, 2\pi], meaning the spiral rotates completely around the origin as \( \theta \) moves from 0 to \( 2\pi \).

Step 3: Set Up the Integral for Area in Polar Coordinates

The formula for the area \( A \) in polar coordinates is given by \( A = \int_{a}^{b} \frac{1}{2} r^2 \, d\theta \). Substituting the known \( r \) and limits: \( A = \int_{0}^{2\pi} \frac{1}{2} \left( \frac{4 \theta}{3} \right)^2 \, d\theta \).

Step 4: Simplify the Integral Expression

Simplify the integrand: \( \left( \frac{4 \theta}{3} \right)^2 = \frac{16 \theta^2}{9} \). The integral becomes \( A = \int_{0}^{2\pi} \frac{1}{2} \times \frac{16 \theta^2}{9} \, d\theta = \frac{8}{9} \int_{0}^{2\pi} \theta^2 \, d\theta \).

Step 5: Integrate the Function

Find the integral of \( \theta^2 \): \( \int \theta^2 \, d\theta = \frac{\theta^3}{3} \). Apply the integration limits: \( \left[ \frac{\theta^3}{3} \right]_{0}^{2\pi} = \frac{(2\pi)^3}{3} - \frac{0^3}{3} = \frac{8\pi^3}{3} \).

Step 6: Calculate the Final Area

Multiply by the constant factor from the integral: \( A = \frac{8}{9} \times \frac{8\pi^3}{3} = \frac{64\pi^3}{27} \). Thus, the area of the region enclosed by the spiral and the positive x-axis is \( \frac{64\pi^3}{27} \).

Key Concepts

Spiral Equations

Understanding spiral equations is key to mastering problems like the one involving the snail shell spiral. In this context, a spiral equation in polar coordinates specifies how the radius, \(r\), changes with the angle, \(\theta\). The formula \( r = \frac{4 \theta}{3} \) indicates that as \( \theta \) increases, so does \( r \), causing the spiral to wind outward from the origin. This behavior is typical of an Archimedean spiral, where the distance between successive turns remains constant. These spirals are commonly described by equations of the form \( r = a + b\theta \), where \(a\) and \(b\) are constants. Here, the lack of an intercept term (a constant part) means the spiral starts right at the origin. Recognizing the type of spiral helps in visualizing the problem and setting up subsequent calculations.

• An Archimedean spiral has equal spacing between turns.
• The starting point of such spirals is usually the origin.
• Visualizing the spiral can aid in understanding the physical layout of the problem.

Area Integration

Finding areas using integration, especially in polar coordinates, involves breaking down the region into small sections and summing them up. In the context of polar coordinates, these sections are typically circular wedges, and the formula for the area from \( \theta = a \) to \( \theta = b \) is \(A = \int_{a}^{b} \frac{1}{2} r^2 \, d\theta.\)This integral sums up the infinitesimal areas represented by each small wedge as \( \theta \) varies. To integrate, recognize that each element of area is akin to a sector of a circle with radius \( r \) and angle \( d\theta \). The process involves:

• Identifying the function \( r(\theta) \), which governs the region's boundary.
• Setting appropriate limits for \( \theta \), as they define the extent of the region.
• Squaring the radius to align with how circular sectors contribute to total area.
• Using the integral to capture the area by evaluating it over the defined limits.

Breaking the area problem into these steps ensures clarity when transitioning from geometric interpretation to direct integration.

Polar Integrals

Polar integrals are a powerful tool for evaluating areas and other physical properties directly in polar coordinates. A polar integral differs from its Cartesian counterpart in that it inherently respects the radial nature of the problem. When integrating, the integrand often involves expressions dependent on \( r \) and \( \theta \). For example, the integral \( A = \int_{a}^{b} \frac{1}{2} r^2 \, d\theta \) is set up to calculate an area by weighing each infinitesimal sector according to its contribution. Understanding polar integrals involves understanding how polar coordinates work:

• Polar coordinates describe points based on distance from the origin and angle from a reference direction.
• When setting up a polar integral, integrate over \( \theta \), as \( r \) typically varies with \( \theta \).
• Sectors calculated in polar integrals inherently account for the varying radius, unlike Cartesian rectangles.

By using polar integrals, you can efficiently analyze shapes and regions that naturally correspond to curves and circular arcs, simplifying calculations that might be cumbersome in Cartesian coordinates.