Question

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the \(y\) -axis. $$ y=2 x, \quad y=4, \quad x=0 $$

Short answer

The volume of the solid generated by revolving the plane region about the y-axis is \(\frac{32\pi}{3}\) cubic units.

Step-by-step solution

Understanding the Region

The given equations \(y=2x\), \(y=4\), and \(x=0\) form a triangular region in the x-y plane. The equation \(y=2x\) is a line passing through the origin. The equation \(y=4\) is a horizontal line crossing the y-axis at \(y=4\). The equation \(x=0\) is the y-axis itself.

Setting Up the Integral

For shells centered at the y-axis, the volume of the solid is given by the integral \[ V = 2\pi \int_a^b{x(y)h(y)dy}\] Where \[x(y)\] is the distance from the shell to the rotation axis and \[h(y)\] is the height of the shell. Here, \[x(y) = y/2\] from \(y=2x\) and \[h(y) = y/2 - 0\] is the x-range (from x = 0 to x = y/2) for the shell. The y-range is from 0 to 4 (for the region between \(y=2x\) and \(y=4\)), so our integral becomes \[V = 2\pi \int_0^4 {(y/2)(y/2)dy} = \pi \int_0^4 {y^2 dy}\]

Evaluating the Integral

The antiderivative of \(y^2\) is \(\frac{1}{3}y^3\), so evaluating the integral gives \[V = \pi \left[\frac{1}{3}y^3 \right]_0^4 = \pi \left[\frac{1}{3}(4)^3 - \frac{1}{3}(0)^3\right] = \frac{32\pi}{3}\]

Key concepts

Volume by Shell Method

The shell method is a powerful technique in calculus for finding the volume of a solid of revolution. When a region in the x-y plane is revolved around a line (usually one of the axes), it creates a three-dimensional object. To determine the volume using the shell method, we imagine slicing the solid into cylindrical shells.

The volume of each infinitesimally thin shell can be computed by the formula:
\[\text{Volume of a shell} = 2\pi \times (\text{radius}) \times (\text{height}) \times (\text{thickness})\]
By integrating this expression over the entire range of the solid, we sum the volumes of all the shells, therefore obtaining the total volume of the solid. In our example exercise, this meant integrating with respect to y, since we revolved the region around the y-axis to form the solid.

Solid of Revolution

A solid of revolution is a three-dimensional object obtained by rotating a two-dimensional shape around an axis. This method of creating solids can be visualized in numerous cases such as pottery making or metalworking. In mathematics, we use this principle to determine the volume of the shape through integration.

In the case of our exercise, rotating the triangular region formed by the equations \(y=2x\), \(y=4\), and \(x=0\) around the y-axis generates a cone. To find the volume, we must consider the change in radius and height across the entire region being rotated. These changes are reflected in the integral used to calculate the volume.

Setting Up Integrals

\(x(y) = y/2\), as derived from the equation of the line \(y=2x\). The height of a shell is also a function of y and is determined by the horizontal distance the region spans in the x-direction, in this case simply \(y/2\). After establishing these functions, we set the limits from 0 to 4—the range of y in the enclosed area—to complete the integral setup for the solid's volume.

Volume of a Solid

Calculating the volume of a solid is a common problem in calculus, which can often be solved by integrating a cross-sectional area along the axis of the solid. The volume represents how much three-dimensional space the solid occupies.

Using integral calculus, we can sum an infinite number of infinitesimally thin cross-sections of the solid to find its volume. In the context of the shell method, these cross-sections are cylindrical shells, and the integral expresses the continuous sum of their volumes, as seen in our example where the final volume of the cone-like solid was established to be \(\frac{32\pi}{3}\). This result is found by evaluating the integral that represents the accumulated volume of all shells making up the solid.