Problem 31 - Thomas Calculus

Question

Find the areas of the surfaces generated by revolving the curves about the indicated axes. $$ x=\cos t, \quad y=2+\sin t, \quad 0 \leq t \leq 2 \pi ; \quad x-\text {axis} $$

Step-by-Step Solution

Answer

The surface area is \(8\pi^2\).

Step 1: Understand the Problem

We have a parametric curve \((x, y) = (\cos t, 2 + \sin t)\) and we need to find the area of the surface generated when this curve is revolved around the x-axis for \(t\) in the interval \([0, 2\pi]\).

Step 2: Use the Parametric Surface Area Formula

The surface area \(S\) of a curve \(y = f(x)\) rotated about the x-axis can be calculated using the formula: \[ S = \int_{a}^{b} 2\pi y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \] For the given parametric equations, we have \((x, y) = (\cos t, 2 + \sin t)\).

Step 3: Find Derivatives

Calculate the derivatives of the parametric equations: 1. \( \frac{dx}{dt} = -\sin t \)2. \( \frac{dy}{dt} = \cos t \)

Step 4: Compute the Integrand

Substitute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) into the integrand: \[ \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \sqrt{(-\sin t)^2 + (\cos t)^2} = \sqrt{\sin^2 t + \cos^2 t} = \sqrt{1} = 1 \]

Step 5: Set Up the Integral

Now, replace \(y\) in the surface area formula and set up the integral for the surface area: \[ S = \int_{0}^{2\pi} 2\pi(2 + \sin t) \cdot 1 \, dt \] which simplifies to \[ S = 2\pi \int_{0}^{2\pi} (2 + \sin t) \, dt \]

Step 6: Integrate

Evaluate the integral: \[ S = 2\pi \left[ \int_{0}^{2\pi} 2 \, dt + \int_{0}^{2\pi} \sin t \, dt \right] \] 1. \( \int_{0}^{2\pi} 2 \, dt = 2t \Big|_0^{2\pi} = 4\pi \)2. \( \int_{0}^{2\pi} \sin t \, dt = -\cos t \Big|_0^{2\pi} = 0 \)Thus, \[ S = 2\pi (4\pi + 0) = 8\pi^2 \]

Step 7: Conclusion

The total surface area generated by revolving the curve about the x-axis is \(8\pi^2\).

Key Concepts

Understanding Parametric Equations

Parametric equations represent a way of defining a curve using a parameter, often denoted as \( t \). In the given problem, the curve is represented as \( (x(t), y(t)) = (\cos t, 2 + \sin t) \). These equations link the position on the x-axis and y-axis to the parameter \( t \).
This approach is often useful in scenarios where the behavior of the curve can be more easily described through a parametric form rather than as an explicit function of x or y.
To work with parametric equations:

• Determine how x and y vary with respect to the parameter \( t \).
• Understand that the interval \( [0, 2\pi] \) in this problem indicates that both x and y are traced as \( t \) moves from 0 to \( 2\pi \).

Rotating a Curve around the X-Axis

To find out what happens when a curve is rotated around an axis, think of the original plot making a 3D surface or shell. In this exercise, we deal with rotation about the x-axis.
Visualize the curve defined by the parametric equations spinning around the x-axis like a jacket sleeve rolling over a cylindrical arm.
The surface made by this rotation can have its area determined using integral calculus.

• Imagine each tiny segment of the curve becoming a circular strip when rotated.
• These strips together form the complete surface around the x-axis.
• The formula used involves calculating the circumference of these circles and summing them over the path of the curve.

Using Integral Calculus for Surface Area

Integral Calculus allows us to sum up infinitesimal quantities to find entire areas. In this case, the problem involves calculating the surface area generated by a curve's rotation.
The integral formula used is: \[ S = \int_{a}^{b} 2\pi y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]This complex equation finds:

• All tiny surfaces created by the curve's rotation around the x-axis.
• \( y \) represents the radius of rotation, coming directly from the parametric form \( y = 2 + \sin t \).
• \( \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = 1 \) confirms these lengths sum up to match classic circle properties over each small arc.

Here's how it's usually done:- First, calculate derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).- Substitute into the integrand to get the complete expression.- Integrate over the given interval \([0, 2\pi]\) to find the total surface area.Ultimately, the outcome \( 8\pi^2 \) represents the entire area wrapped around the x-axis for this curve, showcasing how math elegantly calculates such spatial rotations.