Problem 11 - The Calculus with Analytic Geometry

Question

Find the volume of the solid generated by revolving about the \(x\) axis the region bounded by the curve \(y=x^{3}\) and the lines \(y=0\) and \(x=2\)

Step-by-Step Solution

Answer

\( \frac{128\pi}{7} \)

Step 1 - Understand the Problem

We need to find the volume of the solid formed by revolving the region bounded by the curve \( y = x^3 \), \( y = 0 \), and \( x = 2 \) about the \( x \)-axis.

Step 2 - Set Up the Integral

To find the volume of a solid of revolution around the \( x \)-axis, we use the disk method. The formula is \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]Here, \( f(x) = x^3 \), \( a = 0 \), and \( b = 2 \).

Step 3 - Write the Integral

Substitute \( f(x) = x^3 \), \( a = 0 \), and \( b = 2 \) into the volume formula: \[ V = \pi \int_{0}^{2} (x^3)^2 \, dx \]Which simplifies to \[ V = \pi \int_{0}^{2} x^6 \, dx \]

Step 4 - Evaluate the Integral

To evaluate the integral, find the antiderivative of \( x^6 \): \( \int x^6 \, dx = \frac{x^7}{7} \)Now evaluate from 0 to 2: \[ V = \pi \left[ \frac{x^7}{7} \right]_{0}^{2} = \pi \left( \frac{2^7}{7} - \frac{0^7}{7} \right) \]

Step 5 - Simplify the Result

Calculate the powers and simplifications: \[ V = \pi \left( \frac{128}{7} \right) = \frac{128\pi}{7} \]

Key Concepts

disk method

The disk method is used to find the volume of a solid generated by revolving a region around an axis.
Imagine slicing the solid perpendicular to the axis of rotation into thin disks. Each disk has a small thickness, and the sum of their volumes approximates the total volume of the solid.
To apply the disk method:

• Identify the function that describes the boundary of the region.
• Square the function to get the area of a cross-sectional disk.
• Integrate this area over the given interval to find the total volume.

In the exercise, we rotated the region bounded by the curve \(y = x^3\) around the \(x\)-axis, creating a solid of revolution.

definite integral

A definite integral calculates the accumulated value of a function over a specific interval.
It's the area under the curve of the function between two points. The volume of a solid of revolution is found by integrating the area of disks within the specified bounds.
In the exercise, the definite integral used was:
$$\begin{equation*} V = \frac{128\text{\textgreek{p}}}{7} \text{ cubic units} \text{ using the integral } \underline{\phantom{xxx}} V = \text{\textgreek{p}} \text{} \begin{equation*} V = \text{\textgreek{p}}\text{} \text{}\text{ -- interpreting -- } V = \text{\textgreek{p}}\text{}\text{ -- interpreted -- }\text{} V = \text{\textgreek{p}}\text{} \begin{equation*} V = \text{\textgreek{p}} \text{}
Each disk has a radius equal to the value of the function at that point. Integrating these areas from \(x = 0\) to \(x = 2\) gives the final volume.

antiderivative

An antiderivative is a function that reverses the process of differentiation. It helps in finding the original function given its derivative.
In integral calculus, the antiderivative is used to compute the value of definite integrals.
When finding the volume of a solid of revolution, we integrate the square of the function describing the shape. To solve this integral, we need the antiderivative of the integrand.

• In the exercise, we found the antiderivative of \(x^6\).
• The antiderivative of \(x^6\) is \(\frac{x^7}{7}\).
• Evaluating this antiderivative from 0 to 2 gives the result \(\frac{128}{7}\).

Using this antiderivative, we determined the final volume:
\[ V = \frac{128\pi}{7} \]It's a crucial step in solving integral problems, especially when dealing with volumes of solids of revolution.