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Which Method Should We Use?

We have studied several methods for finding the volume of a solid of revolution, but how do we know which method to use? It often comes down to a choice of which integral is easiest to evaluate. Figure 6.34 describes the different approaches for solids of revolution around the \(x -\text{axis} .\) It’s up to you to develop the analogous table for solids of revolution around the \(y -\text{axis} .\)

This figure is a table comparing the different methods for finding volumes of solids of revolution. The columns in the table are labeled “comparison”, “disk method”, “washer method”, and “shell method”. The rows are labeled “volume formula”, “solid”, “interval to partition”, “rectangles”, “typical region”, and “rectangle”. In the disk method column, the formula is given as the definite integral from a to b of pi times [f(x)]^2. The solid has no cavity in the center, the partition is [a,b], rectangles are vertical, and the typical region is a shaded region above the x-axis and below the curve of f(x). In the washer method column, the formula is given as the definite integral from a to b of pi times [f(x)]^2-[g(x)]^2. The solid has a cavity in the center, the partition is [a,b], rectangles are vertical, and the typical region is a shaded region above the curve of g(x) and below the curve of f(x). In the shell method column, the formula is given as the definite integral from c to d of 2pi times yg(y). The solid is with or without a cavity in the center, the partition is [c,d] rectangles are horizontal, and the typical region is a shaded region above the x-axis and below the curve of g(y).
Figure 6.34

Let’s take a look at a couple of additional problems and decide on the best approach to take for solving them.

Example 6.17

Selecting the Best Method

For each of the following problems, select the best method to find the volume of a solid of revolution generated by revolving the given region around the \(x -\text{axis} ,\) and set up the integral to find the volume (do not evaluate the integral).

  1. The region bounded by the graphs of \(y = x ,\) \(y = 2 - x ,\) and the \(x -\text{axis} .\)
  2. The region bounded by the graphs of \(y = 4 x - x^{2}\) and the \(x -\text{axis} .\)

Solution

  1. First, sketch the region and the solid of revolution as shown.
    This figure has two graphs. The first graph is labeled “a” and has two lines y=x and y=2-x drawn in the first quadrant. The lines intersect at (1,1) and form a triangle above the x-axis. The region that is the triangle is shaded. The second graph is labeled “b” and is the same graphs as “a”. The shaded triangular region in “a” has been rotated around the x-axis to form a solid on the second graph.
    Figure 6.35 (a) The region \(R\) bounded by two lines and the \(x -\text{axis} .\) (b) The solid of revolution generated by revolving \(R\) about the \(x -\text{axis} .\)

    Looking at the region, if we want to integrate with respect to \(x ,\) we would have to break the integral into two pieces, because we have different functions bounding the region over \(\left[\right. 0 , 1 \left]\right.\) and \(\left[\right. 1 , 2 \left]\right. .\) In this case, using the disk method, we would have
    \[V = \int_{0}^{1} \left(\right. \pi x^{2} \left.\right) d x + \int_{1}^{2} \left(\right. \pi \left(\left(\right. 2 - x \left.\right)\right)^{2} \left.\right) d x .\]

     If we used the shell method instead, we would use functions of \(y\) to represent the curves, producing
    \[\begin{aligned} V & = \int_{0}^{1} \left(\right. 2 \pi y \left[\right. \left(\right. 2 - y \left.\right) - y \left]\right. \left.\right) d y \\ & = \int_{0}^{1} \left(\right. 2 \pi y \left[\right. 2 - 2 y \left]\right. \left.\right) d y . \end{aligned}\]

     Neither of these integrals is particularly onerous, but since the shell method requires only one integral, and the integrand requires less simplification, we should probably go with the shell method in this case.
  2. First, sketch the region and the solid of revolution as shown.
    This figure has two graphs. The first graph is labeled “a” and is the curve y=4x-x^2. It is an upside down parabola intersecting the x-axis at the origin and at x=4. The region above the x-axis and below the curve is shaded and labeled “R”. The second graph labeled “b” is the same as in “a”. On this graph the shaded region “R” has been rotated around the x-axis to form a solid.
    Figure 6.36 (a) The region \(R\) between the curve and the \(x -\text{axis} .\) (b) The solid of revolution generated by revolving \(R\) about the \(x -\text{axis} .\)

    Looking at the region, it would be problematic to define a horizontal rectangle; the region is bounded on the left and right by the same function. Therefore, we can dismiss the method of shells. The solid has no cavity in the middle, so we can use the method of disks. Then
    \[V = \int_{0}^{4} \pi \left(\right. 4 x - x^{2} \left.\right)^{2} d x .\]

Checkpoint 6.17

Select the best method to find the volume of a solid of revolution generated by revolving the given region around the \(x -\text{axis} ,\) and set up the integral to find the volume (do not evaluate the integral): the region bounded by the graphs of \(y = 2 - x^{2}\) and \(y = x^{2} .\)

This lesson is part of:

Applications of Integration

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