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Chapter 6: Problem 57

Find the volumes of the solids obtained by rotating the region bounded by the given curves about the \(x\) -axis. In each case, sketch the region together with a typical disk element. \(y=x^{3}, y=x^{2}, 0 \leq x \leq 1\)

Short Answer

Expert verified
The volume is \(\frac{2\pi}{35}\).

Step by step solution

01

Understand the Problem

We have two curves, \(y = x^3\) and \(y = x^2\), and we are asked to find the volume of the solid formed by rotating the region bounded by these curves around the \(x\)-axis. We need to find the region between these curves from \(x = 0\) to \(x = 1\).
02

Sketch the Region and Disk Element

Draw the graphs of \(y = x^3\) and \(y = x^2\) in the \(xy\)-plane with the \(x\)-axis from 0 to 1. The curve \(y = x^2\) is above \(y = x^3\) from \(x = 0\) to \(x = 1\). A typical disk element from the solid when rotating around the \(x\)-axis has outer radius \(R = x^2\) and inner radius \(r = x^3\).
03

Express the Volume Using the Washer Method

The volume of the solid can be found using the washer method, where each disk (washer) has volume \(dV = \pi (R^2 - r^2) dx\). Here, \(R = x^2\) and \(r = x^3\), so the volume is:\[ V = \int_{0}^{1} \pi ((x^2)^2 - (x^3)^2) \, dx = \pi \int_{0}^{1} (x^4 - x^6) \, dx.\]
04

Integrate to Find the Volume

Calculate the integral to find the volume:\[ V = \pi \left[ \frac{x^5}{5} - \frac{x^7}{7} \right]_{0}^{1} = \pi \left( \frac{1}{5} - \frac{1}{7} \right).\] Simplifying gives: \[ V = \pi \left( \frac{7 - 5}{35} \right) = \frac{2\pi}{35}.\]
05

Conclusion

The volume of the solid obtained by rotating the region bounded by \(y = x^3\), \(y = x^2\), and \(x = 0\) to \(x = 1\) around the \(x\)-axis is \(\frac{2\pi}{35}\).

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Key Concepts

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

Washer Method
The washer method is a technique used in calculus to find the volume of a solid of revolution. When you have a region bounded by two curves and you rotate it around an axis, you often get a shape that resembles a washer.
A washer is like a disk with a hole in the center, and to find its volume, you calculate the area of the outer circle and subtract the area of the inner circle.
  • Outer radius: This is the radius of the bigger circle, represented by \(R\).
  • Inner radius: This is the radius of the smaller circle, represented by \(r\).
  • The volume of a thin washer can be given by the formula: \(dV = \pi (R^2 - r^2) \, dx\).
By integrating these washer volumes over the interval of interest, from where the region starts and ends on the axis, you can find the total volume of the solid.
Bounded Region
A bounded region refers to the area enclosed by curves on a graph. It is the space "caught" between the given curves.
In our problem, this region is defined by the curves \(y = x^3\) and \(y = x^2\), from \(x = 0\) to \(x = 1\). This region is like a slice of the graph that we will rotate.
  • To identify it, determine where the curves meet or intersect. In this example, both curves intersect at \(x = 0\) and \(x = 1\).
  • Visualizing this section on a graph helps in setting our limits for integration, which runs from the starting to the ending intersection points on the \(x\)-axis.
When the bounded region rotates around the \(x\)-axis, it forms a solid of revolution whose volume can then be calculated.
Integral Calculus
Integral calculus is the branch of calculus concerned with finding the total size or value, such as areas under a curve or volumes. In this exercise, we use integration to find the volume.
  • We set up an integral that represents the volume of the solid formed by the bounded region rotated about the axis.
  • The integral is \( V = \int_{a}^{b} \pi (R^2 - r^2) \, dx\), where \(R\) and \(r\) are the outer and inner radius functions, respectively.
  • In this case, \(R = x^2\) and \(r = x^3\).
By calculating the integral, we obtain the volume. This process systematically adds together an infinite number of those tiny washer volumes.
Curve Rotation
Curve rotation is a process where a 2D region is revolved around a line (axis) to create a 3D object.
This concept is fundamental in finding the volume of solids of revolution.
  • Decide the axis of rotation: For this problem, it's the \(x\)-axis.
  • Determine the curves being rotated: Here, \(y = x^3\) and \(y = x^2\) are rotated.
  • The generated 3D solid will have different cross-sections along the rotation axis, forming what looks like a stack of washers.
Understanding how each curve contributes to the shape of the solid helps set up the integration correctly. Curve rotation transforms the area between curves into a measurable volume.

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Most popular questions from this chapter

Evaluate the definite integrals. $$ \int_{0}^{\pi / 4} \sin \left(x-\frac{\pi}{4}\right) d x $$

Find the areas of the regions bounded by the lines and curves. \(y=x^{2}, y=\frac{1}{x}, x=1, x=2\)

Find the volumes of the solids obtained by rotating the region bounded by the given curves about the \(x\) -axis. In each case, sketch the region together with a typical disk element. \(y=1-x^{2}, y=1,-1 \leq x \leq 1\)

Find the length of the quarter-circle $$ y=\sqrt{1-x^{2}} $$ for \(0 \leq x \leq 1\), by each of the following methods: (a) a formula from geometry (b) the integral formula from Subsection \(6.3 .6\)

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