Problem 36 - Thomas Calculus

Question

\(r=f(\boldsymbol{\theta})\) vs. \(\boldsymbol{r}=2 \boldsymbol{f}(\boldsymbol{\theta}) \quad\) The curves \(r=f(\theta), \alpha \leq \theta \leq \beta,\) and \(r=2 f(\theta), \alpha \leq \theta \leq \beta,\) are revolved about the \(x\) -axis to generate surfaces. Can anything be said about the relative areas of these surfaces? Give reasons for your answer.

Step-by-Step Solution

Answer

The surface area of the revolution of the second curve is twice that of the first curve.

Step 1: Understand the setup

We have two curves, \(r = f(\theta)\) and \(r = 2f(\theta)\), where \(\alpha \leq \theta \leq \beta\). These curves are revolved about the \(x\)-axis to form surfaces of revolution.

Step 2: Formula for surface area of revolution

The surface area \(A\) of a surface of revolution created by rotating a curve \(y = f(x)\) from \(x = a\) to \(x = b\) around the x-axis is given by:\[ A = \int_a^b 2\pi f(x) \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] Convert for polar coordinates to get:\[ A = \int_\alpha^\beta 2\pi r(\theta) \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \]

Step 3: Substitute the first curve

For \(r = f(\theta)\), the surface area is:\[ A_1 = \int_\alpha^\beta 2\pi f(\theta) \sqrt{ \left( \frac{df}{d\theta} \right)^2 + (f(\theta))^2 } \, d\theta \]

Step 4: Substitute the second curve

For \(r = 2f(\theta)\), the surface area is:\[ A_2 = \int_\alpha^\beta 2\pi (2f(\theta)) \sqrt{ \left( \frac{d(2f)}{d\theta} \right)^2 + (2f(\theta))^2 } \, d\theta \]This simplifies to:\[ A_2 = \int_\alpha^\beta 4\pi f(\theta) \sqrt{ 4\left( \frac{df}{d\theta} \right)^2 + 4(f(\theta))^2 } \, d\theta \]\[ A_2 = 2 \int_\alpha^\beta 2\pi f(\theta) \sqrt{ \left( \frac{df}{d\theta} \right)^2 + (f(\theta))^2 } \, d\theta \]

Step 5: Compare the areas

From the expressions for \(A_1\) and \(A_2\), we find:\[ A_2 = 2A_1 \]This shows that the surface area of the revolution of the second curve is twice that of the first curve.

Key Concepts

Polar Coordinates Integration

Polar coordinates provide a unique way of defining the location of a point in a plane using a radius and an angle, typically denoted as \(r\) and \(\theta\) respectively. This becomes particularly useful when dealing with problems involving circular or spiral patterns, such as surfaces of revolution.
When performing integration in polar coordinates, it's important to remember that the polar area element differs from the rectangular one, and is expressed as \(r\,d\theta\). This subtle difference is crucial when setting up integrals, especially for calculating surface areas.
In our problem, we want to integrate over \(\theta\) to calculate the surface area of curves being revolved around the x-axis. This requires the use of polar coordinates since the curves are expressed as functions of \(\theta\). By converting our integral parameter from \(x\) to \(\theta\), we are able to tailor the integration process to fit the unique setup of the problem. Thus, polar integration becomes an indispensable tool, allowing a seamless translation from abstract mathematical formulas into tangible geometric analyses for these rotational surfaces.

Surface Area Calculation

Calculating the surface area of an object formed by rotating a curve around an axis is often tricky. However, it's made easier with a specific formula. It's vital to adapt the general formula for surface area to the context of polar coordinates.
The formula for the surface area of a surface of revolution, when a curve is rotated around the x-axis, can be adapted for polar coordinates as follows:

• Start with the surface area formula for a function \(y = f(x)\): \[ A = \int_a^b 2\pi f(x) \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]
• Convert to polar coordinates: \[ A = \int_\alpha^\beta 2\pi r(\theta) \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \]

With these formulas, for our two curves \(r = f(\theta)\) and \(r = 2f(\theta)\), it's possible to determine each corresponding surface area. By performing this calculation correctly, the numerical comparison of \(A_1 \) and \(2A_1 \) naturally follows, confirming that the surface formed by \(r = 2f(\theta)\) is double that of the original curve. Understanding this formula is crucial for solving a wide range of geometric integration problems.

Curve Transformation

Transformation of curves involves changing a curve's representation—such as scaling or translating the curve—to meet specific analytical needs. In the context of this exercise, curve transformation is essential to understand how scaling affects the resulting surface area.
The curves given are \(r = f(\theta)\) and \(r = 2f(\theta)\). The transformation here is simple yet critical: the second curve is a scaled version of the first by a factor of 2.

• This doubling of the radius means every point on the curve \(r = f(\theta)\) is expanded outwards.
• This scaling factor affects all dimensions, thus, the transformation noticeably impacts characteristics like surface area significantly.

The geometric effect of this transformation is demonstrated in the calculation steps, showing that the resulting surface area is adjusted by a factor of two, emphasizing how transformations heavily influence the surface properties of revolved objects. Recognizing these transformations is pivotal for dissecting and solving more complex problems involving curve transformations and their geometric consequences.