Problem 47 - Essential Calculus

Question

In Exercises \(47-50\), the integral represents the volume of a solid of revolution. Identify (a) the plane region that is revolved and (b) the axis of revolution. $$ 2 \pi \int_{0}^{2} x^{3} d x $$

Step-by-Step Solution

Answer

The plane region is the area enclosed by \(f(x) = x^{3}\), the x-axis, and the lines \(x=0\) and \(x=2\). The axis of revolution is the x-axis.

Step 1: Identify the function and limits

We are given the integral \(2 \pi \int_{0}^{2} x^{3} d x\). Here, the function \(f(x)\) is \(x^{3}\) and the limits of integration are \(a = 0\) and \(b = 2\).

Step 2: Identify the plane region

The plane region is the region bound by the curve of the function \(f(x) = x^{3}\), the x-axis (the line \(y=0\)), and the lines \(x=a=0\) and \(x=b=2\). This is a region in the first quadrant, under the curve \(f(x) = x^{3}\), and between \(x=0\) and \(x=2\).

Step 3: Identify the axis of revolution

From the form of the integral, we can infer that the region is revolved around the x-axis to form the solid of revolution. This is because the integral is in the form of the Disk Method, which is typically used to find volume when the solid is revolved around the x-axis.

Key Concepts

Volume of Solid of Revolution

Calculating the volume of a solid of revolution involves integrating a region around an axis. This concept arises when you have a two-dimensional shape that you want to "spin" around a line, creating a three-dimensional object.

To find the volume of such a solid, you'll need to look at the shape formed and understand how it sweeps out a space in 3D. In many problems, this revolves around either the x-axis or y-axis.

Once you identify the region and axis, you apply an appropriate integration method. The result of this integral gives you the volume of the entire solid formed by this revolution.

Disk Method

The Disk Method is a straightforward way to find the volume when the solid is created by rotating around an axis.

This technique involves slicing the solid into thin disks or washers and summing their volumes from start to end of the region's bounds.

• Each disk's volume is calculated using the formula \(V = \pi r^2 h\), where \(r\) is the radius of the disk, and \(h\) is its thickness.
• The radius here is usually the value of the function (or distance to the axis of rotation), and the thickness is an infinitesimal change in the variable of integration.

The integral \(2\pi \int_0^2 x^3 dx\) uses disks because it applies \(\pi x^2\), illustrating the radius \(x = f(x)\), revolved around the x-axis.

Integral Bounds

Integral bounds are the limits that define the section of the function to be revolved to create the solid.

They are crucial because they indicate where to start and stop the integration process, thus accounting for the entire area necessary for volume calculation.

• The lower bound (\(a\)) is the point where the region beings, in this exercise, it is \(x=0\).
• The upper bound (\(b\)) is where the region ends, here \(x=2\).

These bounds are plugged directly into the integral, defining the range over which you sum the disk volumes. Properly understanding and applying these limits is key in setting up and successfully solving the integral.

Revolution Axis

The revolution axis is the line around which the region is revolved to form the solid.

Understanding this axis is vital; it dictates the direction and shape of the solid generated.

• The axis in this case is the x-axis as indicated by the integral form. This allows the function \(f(x) = x^3\) to be rotated horizontally.
• When revolving around the x-axis, regions above the x-axis create solid forms below.

Selecting the correct axis ensures the solid's volume is calculated accurately via the Disc Method.