Question

Find the volume of the following solids using the method of your choice. The solid formed when the region bounded by \(y=x^{3}\), the \(x\) -axis, and \(x=2\) is revolved about the \(x\) -axis

Short answer

Answer: The approximate volume of the solid is \(57.24 \pi\).

Step-by-step solution

Set up the integral using the disk method formula

The disk method formula for finding the volume of a solid revolves around the x-axis is: \(V = \pi \int_a^b [f(x)]^2\,dx\) In our case, the function f(x) is given by the equation: \(y = x^3\). We will integrate the square of this function with respect to x, from 0 to 2 (since x=2 is given as the boundary).

Plug the function into the formula

Now, we will substitute y = x^3 into the formula and integrate from 0 to 2: \(V = \pi \int_0^2 (x^3)^2\, dx\)

Simplify and integrate

Now, we will simplify the expression inside the integral and then integrate it with respect to x: \(V = \pi \int_0^2 x^6\, dx\) To integrate this, we will use the power rule: \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\) Therefore, \(V = \pi [\frac{x^7}{7} \Big|_0^2]\)

Compute the limits of the integral

Now, we will substitute the limits of integration into the expression and subtract the lower limit value from the upper limit value: \(V = \pi[(\frac{2^7}{7}) - (\frac{0^7}{7})]\)

Simplify and find the volume

Now, we will simplify the expression to find the volume of the solid: \(V = \pi(\frac{128}{7})\) \(V \approx 57.24 \pi\) Therefore, the volume of the solid formed when the region bounded by the curve \(y = x^3\), the x-axis, and x=2 is revolved around the x-axis is approximately \(57.24 \pi\).

Key concepts

Disk Method

When calculating the volume of solids of revolution, the disk method is a powerful technique. This method comes into play when you revolve a region around the x-axis or y-axis, creating a symmetrical solid or shape. You imagine slicing the solid into thin, disk-like slices, which add up to form the entire solid.
For solids revolving around the x-axis, each slice is a disk with a small thickness, noted as \(dx\). The radius of each disk is determined by the function value, \(f(x)\), at that precise point.

• The formula for finding the volume using the disk method is: \(V = \pi \int_a^b [f(x)]^2\,dx\).
• Here, \([f(x)]^2\) represents the area of each disk as \([f(x)]^2\) is the radius squared.
• The integration summates these infinitesimal disk volumes to obtain the total volume.

In our original example, the function \(f(x) = x^3\) quantifies the radius, and we integrate from 0 to 2 to get all the disks' cumulative volume.

Integral Calculus

Integral calculus is fundamental in finding areas under curves, among other applications. It enables us to transition from a geometric perspective into analytical solutions. When applied to finding a solid's volume, integral calculus allows us to consider microscopic changes continuously, offering precise results.
In essence, integration is the reverse process of differentiation, helping us calculate not only areas but volumes as in our case of solids of revolution.

• The integral \( \int f(x) \, dx \) allows us to calculate areas, or in the case of volume, to accumulate infinite thin slices (volumes).
• Definite integrals, bounded between two limits \( a \) and \( b \), give us the total area or volume for a specific interval.
• In the disk method, integration allows us to smoothly sum the infinite disks formed by revolving the area around an axis.

Integral calculus hence provides us with the tools to measure continuous totals, whether regarding areas or volumes.

Power Rule

The power rule is a basic yet vital tool in integral calculus. It's a straightforward rule to integrate monomials, which are polynomial terms like \(x^n\). This rule becomes crucial when integrating functions during the computation of areas or volumes. It provides a quick mechanism to find antiderivatives of powers of \(x\).

For integration, the power rule states: if \(neq -1\), the integral of \(x^n\) is \(\frac{x^{n+1}}{n+1} + C\), where \(C\) is the integration constant.

• In our example, employing the power rule, we calculate: \int x^6 \, dx = \frac{x^7}{7}\.
• The choice of the upper and lower limits removes the need for the constant, focusing on the definite integral's numeric value.
• This method simplifies complex polynomial expressions into manageable forms within integrals, making it ideal for calculating volumes like our exercise.

The power rule thus significantly eases the process of finding volumes by simplifying polynomials during integration.