Question

Find the area of the surface generated by revolving \(x=t^{2}, y=2 t, 0 \leq t \leq 4\) about the \(x\) -axis.

Short answer

The surface area is approximately 346.36 square units.

Step-by-step solution

Understand the Surface Revolution Problem

We need to find the area of a surface generated when a curve is revolved around the x-axis. The curve is defined parametrically by the equations \( x = t^2 \) and \( y = 2t \), for \( 0 \leq t \leq 4 \).

Recall the Surface Area Formula for Parametric Equations

The surface area \( S \) of a curve \( y = f(x) \) revolved about the x-axis is given by \[ S = \int_{a}^{b} 2\pi y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]. In our case, \( y = 2t \), so \( y(t) = 2t \).

Compute Derivatives

First, find the derivatives needed: \( \frac{dx}{dt} = \frac{d}{dt}(t^2) = 2t \) and \( \frac{dy}{dt} = \frac{d}{dt}(2t) = 2 \).

Plug into Surface Area Formula

Substitute into the surface area formula: \[ S = \int_0^4 2\pi(2t) \sqrt{(2t)^2 + (2)^2} \, dt \]. Simplify the expression to get \[ S = 4\pi \int_0^4 t \sqrt{4t^2 + 4} \, dt \].

Simplify the Expression Inside the Integral

We simplify \( \sqrt{4t^2 + 4} \) to \( \sqrt{4(t^2 + 1)} = 2\sqrt{t^2 + 1} \). Hence, the integral becomes \[ S = 8\pi \int_0^4 t \sqrt{t^2 + 1} \, dt \].

Evaluate the Integral

Use substitution to evaluate the integral: let \( u = t^2 + 1 \), then \( du = 2t \, dt \), or \( t \, dt = \frac{1}{2} du \). The limits change from \( t = 0 \) to \( u = 1 \) and from \( t = 4 \) to \( u = 17 \). The integral becomes \[ S = 8\pi \int_1^{17} \frac{1}{2} u^{1/2} \, du = 4\pi \int_1^{17} u^{1/2} \, du \].

Integrate and Substitute Limits

Integrate \( \int u^{1/2} \, du = \frac{2}{3} u^{3/2} \). Evaluate from \( u=1 \) to \( u=17 \). Calculate: \[ \frac{2}{3}(17^{3/2} - 1^{3/2}) \].

Calculate the Final Result

Substitute into the expression for \( S \): \( S = 4\pi \times \frac{2}{3} ((17^{3/2}) - 1) = \frac{8\pi}{3} ((17^{3/2}) - 1) \). Computing values gives \( S \approx 346.36 \).

Key concepts

Parametric Equations

Parametric equations are a way to represent a curve by using one or more parameters. In the problem, the curve is represented by two equations: \( x = t^2 \) and \( y = 2t \), where \( t \) is the parameter. This means that as \( t \) varies from 0 to 4, we trace the curve in a plane, plotting the coordinates \((x, y)\).
The beauty of parametric equations is that they allow us to describe more complex curves than we could with a single function \( y = f(x) \). For instance, using parametric forms, we can easily model paths that loop back on themselves or that change direction.
Understanding the concept of parametric equations is essential if you wish to explore problems involving curves in fields like physics and engineering. They provide a versatile method for describing motion along a path or the shape of an object.

Calculus Integration

Calculus integration is the process of finding the integral of a function. It is used to determine areas under curves, among other things. In our problem, we integrate to find the surface area generated by revolving a parametric curve around the x-axis.
In this context, integration allows us to sum infinitely small pieces of surface area to get the total area. We used the formula:
\[ S = \int_{a}^{b} 2\pi y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]
This integral sums the ring-shaped slices created as the parametric curve revolves. The function within the integral represents the surface area of these slices as we move along the curve.

• The limits \( a = 0 \) and \( b = 4 \) represent the parameter \( t \) values to evaluate over.
• The expression \( \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \) is key. It accounts for how much \( y \) and \( x \) change with \( t \), which affects how curved the surface is.

This integration process requires both understanding the underlying geometry and applying calculus techniques to compute the total area.

Differential Calculus

Differential calculus is all about finding the rate at which things change. For this exercise, it is crucial to compute derivatives to later substitute into the surface area formula. By taking derivatives of the parametric equations \( x = t^2 \) and \( y = 2t \), we understand how the parameters change.
For \( x = t^2 \), the derivative \( \frac{dx}{dt} = 2t \) tells us how the x-coordinate changes as \( t \) changes. Similarly, for \( y = 2t \), \( \frac{dy}{dt} = 2 \) indicates how the y-coordinate changes.
Finding these derivatives gives insight into the curve's behavior, such as its speed and direction at any point \( t \). These derivatives are then crucial in the integral, determining the surface's curvature and slope.
Using differential calculus in problems like this also sharpens numerical problem-solving skills. It goes beyond finding solutions into understanding the dynamic aspects of curves. This foundational knowledge can then be applied to countless scenarios in real-world situations.