Problem 35 - Essential Calculus

Question

In Exercises \(35-38\), set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the \(x\) -axis. $$ y=\frac{1}{3} x^{3}, \quad 0 \leq x \leq 3 $$

Step-by-Step Solution

Answer

The surface area of the surface generated by revolving the curve \(y=\frac{1}{3}x^{3}\), when \(0 \leq x \leq 3\), about the x-axis is \(81\pi\).

Step 1: Set Up the Integral

To solve for the surface area of revolution, the formula is: \[ A = 2\pi \int_{a}^{b} y \sqrt{1+ \left(\frac{dy}{dx}\right)^{2}} dx \]Let's first find the derivative of \(y = \frac{1}{3} x^{3}\) which is:\(\frac{dy}{dx} = x^{2}\)Then substitute \(y\) and \(\frac{dy}{dx}\) into the formula, the lower limit \(a\) is 0 and the upper limit \(b\) is 3, which gives us the integral to find the surface area.

Step 2: Compute the Integral

The set up integral becomes:\[ A = 2\pi \int_{0}^{3} \frac{1}{3} x^{3} \sqrt{1+ (x^{2})^{2}} dx \]By simplifying the integral, we get:\[ A = 2\pi \int_{0}^{3} \frac{1}{3} x^{5} dx \]Then, use the power rule for integration (which says that the integral of x^n dx is \(\frac{1}{n+1}x^{n+1}\)) to evaluate the integral from 0 to 3.

Step 3: Evaluate the Integral

Evaluate the integral to find the surface area:\[ A = 2\pi [\frac{1}{18} x^{6}]_{0}^{3} \]This simplifies to:\[ A = 2\pi (\frac{1}{18}*3^{6} - 0) \]This simplifies to:\[ A = 2\pi*\frac{729}{18} \]Finally, this simplifies to:\[ A = 81\pi \]

Key Concepts

Definite Integral

A definite integral is a crucial concept in calculus used to find the area under a curve over a specific interval. In this context, it's utilized to calculate the surface area of a solid obtained by rotating a function about an axis.
When you see a problem asking for the area using a definite integral, it refers to integrating between two specific boundary points, or limits. Here, the limits are provided as 0 and 3, which means we calculate the integral from the function defined from 0 to 3.
The specific formula used for the surface area of revolution is:

• \[ A = 2\pi \int_{a}^{b} y \sqrt{1+ \left(\frac{dy}{dx}\right)^{2}} dx \]

This formula helps find the total surface area by accounting for the curve's length and revolving it around the axis.
This method breaks down continuous curves into infinitesimally small segments, calculates their contribution, and then sums them to get the total surface area.

Derivative

In calculus, a derivative represents how a function changes as its input changes. It is the foundational concept that prepares us for integrals.
To solve our exercise, we first need to differentiate the curve function, which is \(y=\frac{1}{3}x^3\). This process is straightforward through finding its derivative \(\frac{dy}{dx}\). For our function, it results in \(x^2\).
Taking derivatives is a key step towards forming the integral for surface area calculation, as it gives us an understanding of the curve's slope at any point. This data is crucial for determining the impact of rotation around the axis, essentially defining the radial extent of each infinitesimal slice of the curve in 3D space. Understanding derivatives will transform how you view functions and their varying rates over domains.

Power Rule for Integration

The power rule for integration is a handy technique used to find integrals, especially when dealing with polynomial functions.
The rule states that to integrate \(x^n\), you can use the formula:

• \(\int x^n dx = \frac{1}{n+1}x^{n+1} + C\)

C is the constant of integration, important in indefinite integrals. However, in definite integrals, we focus on calculating values at the upper and lower bounds.
In solving our problem, once the integral is simplified to \(\int \frac{1}{3} x^5 dx\), we apply the power rule. The process changes the expression to \(\frac{1}{18} x^6\), evaluated from 0 to 3.
Mastering the power rule simplifies the process of finding areas, and helps tackle complex integrals by reducing them into manageable forms.