Question

Find the area of the surface generated by revolving the given curve about the \(x\) -axis \(x=1-t^{2}, y=2 t, 0 \leq t \leq 1\)

Short answer

The surface area is \(\frac{8\pi}{3}(2\sqrt{2} - 1)\).

Step-by-step solution

Identify the Parametric Equations

We are given the parametric equations for the curve: \(x = 1 - t^2\) and \(y = 2t \). The parameter \(t\) ranges from 0 to 1.

Set Up the Surface Area Integral

The formula for the surface area of a curve revolved about the \(x\)-axis is \( A = \int_{a}^{b} 2\pi y(t) \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \). Here, \(y(t) = 2t\), \(\frac{dx}{dt} = -2t\), and \(\frac{dy}{dt} = 2\). Thus, the integral becomes \( 2\pi \int_{0}^{1} 2t \sqrt{(-2t)^2 + 2^2} \, dt \).

Simplify the Expression Inside the Integral

Calculate \((-2t)^2 + 2^2 = 4t^2 + 4\). This simplifies to \(4(t^2 + 1)\), thus the expression under the square root becomes \(\sqrt{4(t^2 + 1)} = 2\sqrt{t^2 + 1}\).

Compute the Surface Area Integral

Plug the simplified expression back into the integral: \[ A = 2\pi \int_{0}^{1} 2t \cdot 2\sqrt{t^2 + 1} \, dt = 8\pi \int_{0}^{1} t\sqrt{t^2 + 1} \, dt \]Now, use a substitution method where \(u = t^2 + 1\), hence \(du = 2t \, dt\), transforming the integral into:\[ 8\pi \int_{1}^{2} \frac{1}{2} \sqrt{u} \, du = 4\pi \int_{1}^{2} \sqrt{u} \, du \]

Integrate and Evaluate

The integral \(\int \sqrt{u} \, du = \frac{2}{3}u^{3/2}\). Evaluating from 1 to 2:\[ 4\pi \left[ \frac{2}{3}u^{3/2} \right]_{1}^{2} = \frac{8\pi}{3}\left[ (2^{3/2}) - (1^{3/2}) \right] = \frac{8\pi}{3}\left( 2\sqrt{2} - 1 \right) \]

Finalize the Result

The computed surface area integral results in: \( A = \frac{8\pi}{3}(2\sqrt{2} - 1) \). This represents the total area of the surface generated by revolving the given curve around the \(x\)-axis.

Key concepts

Parametric Equations

Parametric equations are a powerful way to represent curves by expressing the coordinates of the points on the curve as functions of a variable, often denoted as a parameter, like \(t\). This method allows for a more flexible way of tracing curves, especially useful when the curve has complex behaviors or when dealing with sections that can cross back over themselves. In this particular exercise, we are given the parametric equations \(x = 1 - t^2\) and \(y = 2t\), with the parameter \(t\) ranging from 0 to 1. By using parametric equations, we can describe the evolution of the curve as the parameter \(t\) changes, offering a dynamic representation. It is important to substitute these equations properly into any formulas for calculating surface area or arc length to ensure accuracy. This method is particularly useful for visualizing how the curve interacts with the coordinate axes.

Calculus Integration

Calculus integration is the mathematical process that allows us to calculate various properties of curves and shapes, such as area, volume, and, in this exercise, the surface area generated by revolving a curve. The integral used in this problem involves integrating the function describing the surface area around the \(x\)-axis. The setup is based on a specific integral formula:

• \( A = \int_{a}^{b} 2\pi y(t) \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \)

With integration, we essentially sum up an infinite number of infinitely small elements to find the total area. Calculus integration supports complex computations through systematic processes of breaking down and evaluating areas under curves or spaces within surfaces. It's a cornerstone in finding the solution to such problems in mathematics, requiring substantial practice to master.

Curve Revolution

A curve revolution involves rotating a curve around a fixed line (axis), forming a three-dimensional surface. In this problem, we are rotating the curve defined by the parametric equations around the \(x\)-axis. This method helps transform a two-dimensional problem into a three-dimensional problem by using integration to calculate the generated surface area. The revolution of curves is a concept applied widely to solve problems in engineering and physics, where estimating the surface area or volume of complex shapes is essential.
When setting up the problem, understand the direction and axis of revolvement, as it affects how the integral for the surface area's formula is derived and applied. We use parametric equations to describe such curves because they can sometimes provide a simpler path to setting up and solving integrals that involve rotation.

Substitution Method

The substitution method is a powerful tool in calculus used to simplify the integration process by changing variables. In this exercise, after setting up the surface area integral using parametric equations, we apply the substitution method to simplify the complex square root expression. This involves replacing a complicated part of the integral with a new variable that makes the integration process easier.

• Initially, we had \(8\pi \int_{0}^{1} t\sqrt{t^2 + 1} \, dt\).
• By substituting \(u = t^2 + 1\), the derivative \(du = 2t \, dt\) simplifies the expression.
• The range changes to \(u = 1\) to \(u = 2\).

Replacing \(t\sqrt{t^2 + 1}\) with \(\frac{1}{2} \sqrt{u}\) allows us to evaluate the integral more straightforwardly. This method is often used when dealing with integrals that initially appear intractable or overly complex.