Problem 22 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when \(R\) is revolved about the \(x\) -axis. $$y=x^{3}, y=1, \text { and } x=0$$

Step-by-Step Solution

Answer

Answer: The volume of the solid generated is $$\frac{3\pi}{14}$$.

Step 1: Determine the limits of integration

We need to determine the bounds of the region R. To do this, we need to find the x-coordinate where the curves y = x^3 and y = 1 intersect. We can find this by setting the two functions equal: 1 = x^3 Solve for x (since x can't be negative as y = x^3 >= 0 for x >= 0): x = 1 So the region R is bounded by the curves y = x^3, y = 1, and x = 0 and by x = 1.

Step 2: Set up the shell method integral

Now that we have the limits of integration, we can set up the integral to find the volume of the solid generated when the region is revolved around the x-axis: $$ V = 2\pi \int_0^1 y(1-x^3)dx $$ The integral represents the sum of the volumes of the cylindrical shells formed by revolving the region around the x-axis from 0 to 1.

Step 3: Solve the integral

We will now evaluate the integral to find the volume: $$ V = 2\pi \int_0^1 y(1-x^3)dx = 2\pi \int_0^1 (x^3)(1-x^3)dx $$ To do this, we will first expand the integral: $$ V = 2\pi \int_0^1 (x^3-x^6)dx $$ Now, compute the integral: $$ V = 2\pi \left[ \frac{1}{4}x^4 - \frac{1}{7}x^7 \right]_0^1 $$ Evaluate the expression at the limits of integration: $$ V = 2\pi \left( \left[ \frac{1}{4}(1)^4 - \frac{1}{7}(1)^7 \right] - \left[ \frac{1}{4}(0)^4 - \frac{1}{7}(0)^7 \right] \right) $$ $$ V = 2\pi \left( \frac{1}{4} - \frac{1}{7} \right) $$

Step 4: Find the volume

Finally, calculate the volume by multiplying the result by 2π: $$ V = 2\pi \left( \frac{3}{28} \right) = \frac{6\pi}{28} = \frac{3\pi}{14} $$ The volume of the solid generated when the region R is revolved around the x-axis is $$\frac{3\pi}{14}$$.

Key Concepts

Volume of Revolution

The volume of revolution is a concept in calculus for finding the volume of a three-dimensional object generated by rotating a two-dimensional region around an axis. This is particularly useful for creating objects with symmetrical shapes, like cylinders, cones, or spheres. When a region is revolved around an axis, it sweeps out a solid shape. Calculating the volume of this solid can be done using various techniques, such as the disk, washer, or shell methods.

These methods use different approaches to break down the solid into simpler shapes whose volumes we can easily calculate and sum up. The disk method involves slicing the solid perpendicular to the axis of rotation, forming disk-shaped slices. Similarly, the shell method involves slicing the solid with shapes that resemble cylindrical shells. The choice of method often depends on the axis of rotation and the functions that bound the region.

Integration

Integration is a fundamental concept in calculus used to find areas, volumes, central points, and many other useful things. In the context of finding the volume of a revolution, integration allows us to sum up infinite small parts to find the entire volume.

When using the disk method, we set up an integral that accounts for the sum of cylindrical disks that form as the region revolves around the axis. The integral will typically look something like \[ \int_{a}^{b} \pi [f(x)]^2 \, dx \] where \(f(x)\) is the function defining the curve, and \([a, b]\) are the bounds for the integration. This formula results from the area of a circle \(\pi r^2\), where \(r = f(x)\), and the integration adds up all these infinitesimally thin disks along the axis of revolution. The result gives us the total volume of the solid.

Cylindrical Shells

The cylindrical shells method represents another way to compute the volume of a solid of revolution. When using this method, we imagine that the solid is composed of many thin, hollow cylinders rather than disks. Each shell has a small thickness, defined by \(dx\) or \(dy\), depending on the axis of revolution.

To apply the method, use the formula: \[ V = 2\pi \int_{a}^{b} (radius)(height) \, dx \]where "radius" is the distance from the axis of rotation to the middle of the shell (often represented as \(x\) or \(y\)), and "height" is determined by the function being revolved. The \(2\pi\) factor accounts for the circular aspect of each shell.

This method is particularly beneficial when the region is better suited for integration parallel to the axis of rotation, making it easier to set up and solve the integral for specific problems, such as those involving more complex functions or asymmetric regions.