Problem 51 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the area of the surface generated by revolving the curve \(x=2+\cos t, y=1+\sin t\), for \(0 \leq t \leq 2 \pi\) about the \(x\) -axis.

Step-by-Step Solution

Answer

The surface area is \(4\pi^2\).

Step 1: Understand the problem

We need to find the area of the surface generated when the given parametric curve is revolved around the x-axis.

Step 2: Recall the formula for surface area of revolution

The formula to calculate the surface area of revolution about the x-axis for a parametric curve is \( A = \int_{a}^{b} 2\pi y \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt \), where \( y = f(t) \) and \( x = g(t) \).

Step 3: Compute derivatives

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

Step 4: Plug derivatives into surface area formula

Substitute the derivatives and the parametric equations into the formula:\[A = \int_{0}^{2\pi} 2 \pi (1 + \sin t) \sqrt{ (-\sin t)^2 + (\cos t)^2 } \, dt \]

Step 5: Simplify the integral expression

Since \((-\sin t)^2 + (\cos t)^2 = \sin^2 t + \cos^2 t = 1\), the integral simplifies to:\[A = \int_{0}^{2\pi} 2 \pi (1 + \sin t) \, dt \]

Step 6: Perform Integration

Calculate the integral:\[A = 2\pi \left[ \int_{0}^{2\pi} (1 + \sin t) \, dt \right] = 2\pi \left[ \int_{0}^{2\pi} \, dt + \int_{0}^{2\pi} \sin t \, dt \right] \].

Step 7: Calculate each part of the integral

The first part: \(\int_{0}^{2\pi} \, dt = [t]_{0}^{2\pi} = 2\pi\). The second part: \(\int_{0}^{2\pi} \sin t \, dt = [-\cos t]_{0}^{2\pi} = 0\).

Step 8: Calculate the total surface area

Combining the results, the area is:\[A = 2\pi (2\pi + 0) = 4\pi^2\].

Key Concepts

Parametric Equations

Parametric equations are a powerful mathematical tool used to express coordinates in a plane in terms of a third parameter, often denoted as \( t \). These equations are particularly useful for describing curves with intricate shapes. In the problem given, the curve is defined by the parametric equations \( x = 2 + \cos t \) and \( y = 1 + \sin t \). Here, \( t \) is the parameter which varies from \( 0 \) to \( 2\pi \).

• The parametric form allows us to express curves that cannot be represented as functions \( y = f(x) \).
• Each value of \( t \) gives a unique point \((x, y)\) on the curve.
• They can represent a variety of shapes like ellipses, circles, and spirals depending on the equations.

This flexibility is essential when working with rotations and revolutions, as it simplifies the process of calculating geometrical properties like length, area, and in this case, surface area.

Integration Techniques

Integration is a fundamental technique used to calculate the exact value of areas, volumes, and in this case, surfaces. To find the surface area of a curve revolved around an axis, we make use of an integration formula tailored for parametric functions. The formula for surface area of revolution when a curve \( y = f(t) \) is revolved around the \( x \)-axis is:
\[A = \int_{a}^{b} 2\pi f(t) \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt\]
To solve the problem:

• First, compute the derivatives: \( \frac{dx}{dt} = -\sin t \) and \( \frac{dy}{dt} = \cos t \).
• Since \( \sin^2 t + \cos^2 t = 1 \), the integrand simplifies, making the integral much easier to compute.
• The integration is separated into two parts, one for the constant term and another for the \( \sin t \) term.

By integrating from \( t = 0 \) to \( t = 2\pi \), we can accurately find the area of the surface generated by the curve revolving around the \( x \)-axis. Utilizing these techniques ensures the exact shape and extent of the surface is captured.

Calculus Applications

Calculus is a dynamic field of mathematics essential for solving real-world problems where change is involved. One application is finding the surface area of a revolution, which is crucial in engineering and physics. When a curve is revolved about an axis, it creates a 3D surface, and calculus helps us find the surface area accurately.

• This involves setting up an integral that accounts for every tiny piece of the curve as it moves through a full rotation.
• The setup involves derivatives which provide the rate of change, essential to understanding how the shape evolves as \( t \) changes.
• Using calculus in this manner allows for deeper insights into shapes and volumes, paving the way for advancements in technology and design.

The problem illustrated how calculus takes abstract parametric equations and, through a series of systematic steps, reveals the surface area generated by a rotated curve. It highlights the power of integral calculus in unraveling complex geometrical shapes and making them manageable and understandable.