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Mobile App Chapter 4: Problem 4
Berechnen Sie den Flächeninhalt, den die Archimedische Spirale mit der Funktionsgleichung \(r(\varphi)=a \varphi\) im Intervall \(0 \leqslant \varphi<2 \pi\) einschlieBt \((a:\) Konstante, \(a>0\) ).
Short Answer
Expert verified
The area is \(\frac{4 a^2 \pi^3}{3}\).
Step by step solution
01
- Understand the Problem
The problem requires you to find the area enclosed by the Archimedean spiral described by the polar equation \(r(\varphi) = a \varphi\) where \(\varphi\) ranges from 0 to \(2\pi\).
02
- Set Up the Area Integral in Polar Coordinates
The area \(A\) in polar coordinates is given by \(A = \frac{1}{2} \int_{0}^{2\pi} r(\varphi)^2 d\varphi\). Substitute \(r(\varphi) = a \varphi\) into the formula.
03
- Substitute and Simplify the Integrand
Substituting \(r(\varphi) = a \varphi\) into the area formula gives \( A = \frac{1}{2} \int_{0}^{2\pi} (a \varphi)^2 d\varphi = \frac{a^2}{2} \int_{0}^{2\pi} \varphi^2 d\varphi\).
04
- Compute the Integral
Now compute the integral \(\int_{0}^{2\pi} \varphi^2 d\varphi\). The antiderivative of \(\varphi^2\) is \(\frac{\varphi^3}{3}\). Evaluate it from 0 to \(2\pi\): \[\int_{0}^{2\pi} \varphi^2 d\varphi = \frac{(2\pi)^3}{3} - \frac{0^3}{3} = \frac{8\pi^3}{3}\]
05
- Finalize the Area Calculation
Multiply the result by \(\frac{a^2}{2}\): \[ A = \frac{a^2}{2} \cdot \frac{8\pi^3}{3} = \frac{4 a^2 \pi^3}{3} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polar coordinates
In polar coordinates, we describe points on a plane using a radius and an angle, unlike the Cartesian system which uses x and y coordinates. The radius (\r) indicates the distance from the origin, and the angle (\theta) is the angle between the positive x-axis and the line connecting the point to the origin.
To convert polar coordinates to Cartesian coordinates, use these formulas: \[x = r \theta \quad \text{and} \quad y = r \sin \theta \]
The Archimedean spiral, given by the equation \(r(\varphi) = a \varphi\), naturally lends itself to polar coordinates because it describes a rotational progress, spiraling outward or inward relative to the origin.
To convert polar coordinates to Cartesian coordinates, use these formulas: \[x = r \theta \quad \text{and} \quad y = r \sin \theta \]
The Archimedean spiral, given by the equation \(r(\varphi) = a \varphi\), naturally lends itself to polar coordinates because it describes a rotational progress, spiraling outward or inward relative to the origin.
Integral calculus
Integral calculus is a branch of mathematics focusing on the accumulation of quantities and the areas under and between curves. In this exercise, we use integral calculus to find the area enclosed by the Archimedean spiral.
We achieve this by setting up an integral in polar coordinates, where the area \(A\) of a region defined by a curve \(r(\varphi)\) from \(\varphi = \alpha\) to \(\varphi = \beta\) is given by: \[A = \frac{1}{2} \int_{\alpha}^{\beta} r(\varphi)^2 d\varphi\]
The integral takes into account the incremental areas formed by the radial slices and sums them over the specified angle interval.
We achieve this by setting up an integral in polar coordinates, where the area \(A\) of a region defined by a curve \(r(\varphi)\) from \(\varphi = \alpha\) to \(\varphi = \beta\) is given by: \[A = \frac{1}{2} \int_{\alpha}^{\beta} r(\varphi)^2 d\varphi\]
The integral takes into account the incremental areas formed by the radial slices and sums them over the specified angle interval.
Archimedean spiral
The Archimedean spiral is a spiral named after the ancient Greek mathematician Archimedes. It's defined by the polar equation \(r(\varphi) = a \varphi\), where \(a\) is a positive constant. This spiral has the unique property that the distance between turns is constant.
This makes the Archimedean spiral different from other spirals like the logarithmic or hyperbolic spirals. It can be visualized as a point moving away from the origin at a uniform speed while also rotating at a constant angular velocity.
The starting point of the spiral in this problem is at the origin (\r = 0) when \(\varphi = 0\), and it progresses outward as \(\varphi\) increases.
This makes the Archimedean spiral different from other spirals like the logarithmic or hyperbolic spirals. It can be visualized as a point moving away from the origin at a uniform speed while also rotating at a constant angular velocity.
The starting point of the spiral in this problem is at the origin (\r = 0) when \(\varphi = 0\), and it progresses outward as \(\varphi\) increases.
Area calculation
The primary goal here is to find the enclosed area of the Archimedean spiral between \(\varphi = 0\) and \(\varphi = 2\pi\). Follow these steps:
Ensure each step is carried out with precision to achieve accurate results.
- Set up the area integral with the limits of integration from \(0\) to \(2\pi\)
- Substitute the polar equation \(r(\varphi) = a \varphi\)
- Simplify the integrand: \[A = \frac{1}{2} \int_{0}^{2\pi} (a \varphi)^2 d\varphi = \frac{a^2}{2} \int_{0}^{2\pi} \varphi^2 d\varphi\]
- Compute the integral of \(\varphi^2\)
- Evaluate the antiderivative from 0 to \(2\pi\), resulting in \(\frac{8\pi^3}{3}\)
- Multiply the result by \(\frac{a^2}{2}\) to get the final enclosed area: \[A = \frac{4 a^2 \pi^3}{3}\]
Ensure each step is carried out with precision to achieve accurate results.
Mathematical problem solving
Mathematical problem solving involves identifying and understanding the problem, setting up the relevant mathematical representation, and systematically solving it. In this exercise:
This structured approach helps in breaking down complex problems into manageable steps, ensuring a clear path to the solution.
- Read and comprehend the requirements: Find the area enclosed by a spiral.
- Translate into mathematical terms: Use the formula for area in polar coordinates and the given spiral equation.
- Set up the integral with proper limits of integration and substitute the equation.
- Simplify the integrand and solve the integral systematically.
- Check the steps and ensure each calculation is accurate.
This structured approach helps in breaking down complex problems into manageable steps, ensuring a clear path to the solution.
