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Magazine Chapter 8: Problem 61
(a) Write an integral which represents the circumference of a circle of radius \(r\). (b) Evaluate the integral, and show that you get the answer you expect.
Short Answer
Expert verified
The integral \( \int_0^{2\pi} r \, d\theta = 2\pi r \) represents the circle's circumference.
Step by step solution
01
Define the Problem
We need to find an integral that represents the circumference of a circle with radius \( r \). Recall that the circumference \( C \) of a circle is given by the formula \( C = 2 \pi r \). Our task involves deriving this using integration.
02
Set Up the Integral
To represent the circumference as an integral, consider the parametric equations for a circle: \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). For the circle, the range for \( \theta \) is from \( 0 \) to \( 2\pi \). Use the formula for arc length in parametric form: \( \int_a^b \sqrt{(dx/d\theta)^2 + (dy/d\theta)^2} \, d\theta \).
03
Differentiate the Parametric Equations
Calculate the derivatives: \( \frac{dx}{d\theta} = -r \sin(\theta) \) and \( \frac{dy}{d\theta} = r \cos(\theta) \). Substitute these into the arc length formula: \[ \int_0^{2\pi} \sqrt{(-r \sin(\theta))^2 + (r \cos(\theta))^2} \, d\theta \].
04
Simplify the Integral
Inside the integral, simplify the expression: \((-r \sin(\theta))^2 + (r \cos(\theta))^2 = r^2 \sin^2(\theta) + r^2 \cos^2(\theta) = r^2 (\sin^2(\theta) + \cos^2(\theta)) = r^2\). Thus, the integral becomes \[ \int_0^{2\pi} \sqrt{r^2} \, d\theta = \int_0^{2\pi} r \, d\theta \].
05
Evaluate the Integral
Evaluate the integral: \( \int_0^{2\pi} r \, d\theta \). Since \( r \) is a constant, this evaluates to \( r [\theta]_0^{2\pi} = r \times 2\pi - r \times 0 = 2\pi r \).
06
Confirm the Result
The result \( 2\pi r \) matches the known formula for the circumference of a circle. This confirms that the integral correctly represents the circumference.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equations
Parametric equations are a powerful tool in mathematics used to represent curves in a plane. Instead of describing a curve through a direct relationship between x and y, parametric equations use an independent parameter, usually denoted as \( \theta \). For a circle of radius \( r \), the parametric equations are:
By manipulating these equations, we can explore various properties of curves, such as their lengths and areas, through integration.
- \( x = r \cos(\theta) \)
- \( y = r \sin(\theta) \)
By manipulating these equations, we can explore various properties of curves, such as their lengths and areas, through integration.
Arc Length
The concept of arc length involves finding the length of a curve defined by parametric equations. For curves expressed in parametric form, the formula to calculate arc length is:
\[\int_a^b \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \, d\theta\]This integral evaluates the length of the curve between two points determined by the parameter \( \theta \), specifically from \( a \) to \( b \).
\[\int_a^b \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \, d\theta\]This integral evaluates the length of the curve between two points determined by the parameter \( \theta \), specifically from \( a \) to \( b \).
- \( \frac{dx}{d\theta} = -r \sin(\theta) \)
- \( \frac{dy}{d\theta} = r \cos(\theta) \)
Circumference of a Circle
The circumference of a circle is a key geometric concept and can also be derived through integration. It is usually expressed by the formula \( C = 2\pi r \), where \( r \) is the radius of the circle.
Using the parametric equations \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \), the circumference can be represented as an integral using:
\[\int_0^{2\pi} \sqrt{(-r \sin(\theta))^2 + (r \cos(\theta))^2} \, d\theta\]Simplifying the integrand leads to \( \int_0^{2\pi} r \, d\theta \), representing the actual circumference which evaluates to \( 2\pi r \), confirming the familiar formula.
Using the parametric equations \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \), the circumference can be represented as an integral using:
\[\int_0^{2\pi} \sqrt{(-r \sin(\theta))^2 + (r \cos(\theta))^2} \, d\theta\]Simplifying the integrand leads to \( \int_0^{2\pi} r \, d\theta \), representing the actual circumference which evaluates to \( 2\pi r \), confirming the familiar formula.
Integral Evaluation
Integral evaluation involves solving the integral expression to find a numerical result. In the context of finding a circle's circumference, the integral simplifies to \( \int_0^{2\pi} r \, d\theta \).
Recognizing \( r \) as a constant, the integral evaluation becomes straightforward:
\[\int_0^{2\pi} r \, d\theta = r \times [\theta]_0^{2\pi} = r \times (2\pi - 0) = 2\pi r\]This process verifies that the mathematical interpretation aligns with the known formula for a circle's circumference. It's important when using integrals to ensure the limits and expressions are carefully evaluated to reflect the accurate geometric properties.
Recognizing \( r \) as a constant, the integral evaluation becomes straightforward:
\[\int_0^{2\pi} r \, d\theta = r \times [\theta]_0^{2\pi} = r \times (2\pi - 0) = 2\pi r\]This process verifies that the mathematical interpretation aligns with the known formula for a circle's circumference. It's important when using integrals to ensure the limits and expressions are carefully evaluated to reflect the accurate geometric properties.
