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Chapter 7: Problem 12

In Exercises 11-20, find the volume of the solid generated by revolving the region bounded by the lines and curves about the x-axis. $$y=x^{3}, \quad y=0, \quad x=2$$

Short Answer

Expert verified
The volume of the solid generated is \(\frac{128\pi}{7}\) cubic units.

Step by step solution

01

Identify the function and the interval

First, identify the function to be integrated and the interval. The function is \(y = x^{3}\) and the interval for \(x\) is from 0 to 2.
02

Set up the volume integral

Using the method of disks/washers, the volume \(V\) of a solid of revolution about the x-axis is given by \(V = \pi \int_{a}^{b} [f(x)]^{2} dx\), where \(f(x)\) is the function and \([a, b]\) is the interval. So we have \(V = \pi \int_{0}^{2} (x^{3})^{2} dx\).
03

Compute the integral

Simplify and compute the integral. \(V = \pi \int_{0}^{2} x^{6} dx = \pi [ \frac{1}{7}x^{7} ]_{0}^{2} = \pi (\frac{1}{7}(2)^{7} - \frac{1}{7}(0)^{7})\). The integral becomes \(V = \frac{128\pi}{7}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume Calculation
Calculating the volume of a solid of revolution can seem tricky at first, but it's just a few logical steps. Imagine the region you are interested in is spun around an axis, like a sushi roll being made.

The volume is determined by integrating the area of circular slices that stack together to form the 3D shape.
  • First, you need to find the function representing the shape. For this problem, it's the curve specified by the equation \( y = x^3 \).
  • Next, check out the range for your variable, which is from where to where that shape stretches. In this case, it's between 0 and 2 on the x-axis.
  • Finally, plug these details into the integral formula that helps us calculate volume, ensuring the shape accurately transforms into the solid with depth when revolved.
This set of steps turns the process of volume determination into a repeatable formula that you can apply to various shapes.
Disk Method
The Disk Method is an essential tool in calculus for finding volumes, particularly for solids of revolution. This method helps visualize the process by considering the solid as a collection of disks or washers stacked along the axis of revolution.

Here's how it works:
  • Consider each small slice of the solid as a disk with some thickness.
  • The radius of each disk is defined by the function \( f(x) \), identifying how far the shape extends per slice.
  • The volume of a small disk is \( \pi \times [f(x)]^2 \times dx \), where \( dx \) is a small change in \( x \).
This approach makes it much easier to visualize how 3D volumes are calculated, and by integrating these small volumes, you get the entire solid's volume.

When the example curve \( y = x^3 \) is revolved around the x-axis, the radius of the disks shall grow with \( x \), expanding the volume as \( x \) increases from 0 to 2.
Integral Calculus
Integral calculus helps us solve problems involving accumulation, such as calculating areas and volumes. When we talk about solids of revolution, integral calculus is our primary tool to integrate these volumes.

The general idea involves:
  • Setting up the integral based on the function and limits. For example, if you have \( V = \pi \int_{0}^{2} (x^3)^2 \, dx \), you compute the accumulation of disk areas from \( x = 0 \) to \( x = 2 \).
  • Using the power rule, integrate the function yielding \( \pi [ \frac{1}{7}x^{7} ]_{0}^{2} \).
  • Applying the Fundamental Theorem of Calculus, evaluate the definite integral to get the solid's volume, which computes here as \( \frac{128\pi}{7} \).
This process reveals the power of integral calculus in transitioning from abstract mathematics to real-world applications, allowing us to efficiently calculate and visualize volumes.

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