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Magazine Chapter 15: Problem 30
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.
Short Answer
Expert verified
The area is \( \frac{64\pi^3}{27} \).
Step by step solution
01
Understand the Problem
We are given a spiral described by the polar equation \( r = \frac{4\theta}{3} \) for \( 0 \leq \theta \leq 2\pi \). We need to find the area enclosed by this spiral and the positive x-axis.
02
Set Up the Integral for Area
The area \(A\) in polar coordinates between \(\theta = \alpha\) and \(\theta = \beta\) is given by the formula:\[ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \]For our problem, \( r = \frac{4\theta}{3} \), \( \alpha = 0 \), and \( \beta = 2\pi \). Substitute these into the formula to set up the integral for the area.
03
Substitute and Simplify
Substitute \( r = \frac{4\theta}{3} \) into the formula:\[ A = \frac{1}{2} \int_{0}^{2\pi} \left( \frac{4\theta}{3} \right)^2 \, d\theta \]Simplify the expression:\[ A = \frac{1}{2} \int_{0}^{2\pi} \frac{16\theta^2}{9} \, d\theta = \frac{8}{9} \int_{0}^{2\pi} \theta^2 \, d\theta \]
04
Evaluate the Integral
Evaluate the integral \( \int_{0}^{2\pi} \theta^2 \, d\theta \):Using the power rule for integration, \( \int \theta^n \, d\theta = \frac{\theta^{n+1}}{n+1} + C \), we have:\[ \int_{0}^{2\pi} \theta^2 \, d\theta = \left[ \frac{\theta^3}{3} \right]_{0}^{2\pi} = \frac{(2\pi)^3}{3} - \frac{0^3}{3} = \frac{8\pi^3}{3} \]
05
Final Calculation
Substitute back the result of the evaluated integral into our expression for \( A \):\[ A = \frac{8}{9} \times \frac{8\pi^3}{3} \]Simplify the expression:\[ A = \frac{64\pi^3}{27} \]
06
Conclusion: Area Enclosed by the Spiral
The area enclosed by the spiral \( r = \frac{4\theta}{3} \) and the positive x-axis over the interval \( 0 \leq \theta \leq 2\pi \) is \( \frac{64\pi^3}{27} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Area Calculation
Area calculation in polar coordinates involves a special formula. When we want to compute the area enclosed by a spiral curve and another boundary, we apply this formula to integrate over the curve's path.
In polar coordinates, the area enclosed by a curve, from an angle \( \theta = \alpha \) to \( \theta = \beta \), is calculated using:
\[ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \]
It starts with setting up the integral, substituting the given polar equation, and solving the resulting integral.
In polar coordinates, the area enclosed by a curve, from an angle \( \theta = \alpha \) to \( \theta = \beta \), is calculated using:
\[ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \]
- This formula considers each infinitesimal sector of angle \( d\theta \) and area \( \frac{1}{2} r^2 \, d\theta \).
- The factor of \( \frac{1}{2} \) accounts for the circular nature of polar sectors.
It starts with setting up the integral, substituting the given polar equation, and solving the resulting integral.
Spiral Curve
A spiral curve in the context of polar coordinates is a fascinating path that contains loops or coils around a central point. Spiral equations often depend directly on the angle variable \( \theta \), making the radius \( r \) grow or shrink as \( \theta \) changes.
For example, the spiral equation given in the problem, \( r = \frac{4\theta}{3} \), means the radius increases linearly with \( \theta \).
For example, the spiral equation given in the problem, \( r = \frac{4\theta}{3} \), means the radius increases linearly with \( \theta \).
- This results in a spiral rather than a circular or elliptical path.
- As \( \theta \) changes from 0 to \( 2\pi \), the spiral wraps around the origin forming a snail-like shape.
Integral Evaluation
Evaluating integrals is a key skill in calculating areas, especially for more complex functions like in polar equations. Integrals calculate areas by summing up infinitesimally small contributions across an interval.
In our problem, the integral evaluated was:
\[ \int_{0}^{2\pi} \theta^2 \, d\theta \]To solve it:
This evaluates to:
\[ \frac{\theta^3}{3} \bigg|_{0}^{2\pi} = \frac{(2\pi)^3}{3} \]
This simplifying step is vital in figuring out the enclosed area correctly.
In our problem, the integral evaluated was:
\[ \int_{0}^{2\pi} \theta^2 \, d\theta \]To solve it:
- We used the power rule, \( \int \theta^n \, d\theta = \frac{\theta^{n+1}}{n+1} + C \).
- Substitute \( n = 2 \), and calculate from 0 to \( 2\pi \).
This evaluates to:
\[ \frac{\theta^3}{3} \bigg|_{0}^{2\pi} = \frac{(2\pi)^3}{3} \]
This simplifying step is vital in figuring out the enclosed area correctly.
Polar Equations
Polar equations express relationships in terms of \( r \) and \( \theta \), where \( r \) (radius) can be any function of \( \theta \). They offer a great way to represent curves that loop around a central point, such as spirals.
Understanding the properties of polar equations is essential in solving such mathematical problems, as it dictates the entire integration setup and approach.
- In polar coordinates, patterns often not easily represented in Cartesian coordinates become straightforward.
- The spiral in the problem, \( r = \frac{4\theta}{3} \), showcases a direct coupling between radius and angle.
Understanding the properties of polar equations is essential in solving such mathematical problems, as it dictates the entire integration setup and approach.
