Chapter 5: Applications of Integration
Section 5.2: Volume of a Solid of Revolution
If A is the plane region bounded by the x-axis and the graphs of y=x2 and x=1, use the method of shells to calculate the volume of the solid of revolution formed when A is rotated about the line x=2.
The height of the generic cylindrical shell for this solid is Lx= x2. This shell is formed from a rectangular slab dx thick, and of height Lx and width 2 π ρ, where ρ=2− x.
Thus, the volume of the slab from which the shell is constructed is
2 π ρ L dx=2 π x22−x.
The total volume is then 2 π ∫01x2 2−x ⅆx = 5⁢π6
Figure 5.2.8(a) shows the
tutor defaulting to the shell method when finding the volume of the solid of revolution formed by rotating the plane region A is about a vertical axis.
In the section "Line of Revolution" select "Vertical" and change the default 0 to 2 as the distance of the line of rotation from the coordinate axis. Click "Plot Options" and choose both Constrained Scaling and framed axes. Click the "Display" button to update the figure and to obtain the appropriate calculation.
Figure 5.2.8(b) shows the solid segmented into a stack of concentric shells. The VolumeOfRevolution command at the bottom of the tutor, or the tutor itself can be used to generate a similar figure. (The actual code used to generate Figure 5.2.8(b) is hidden behind the table cell in which the figure resides.)
Figure 5.2.8(a) Volume of Revolution tutor
Student:-Calculus1:-VolumeOfRevolution(x^2, 0 .. 1, 'axis' = vertical, 'distancefromaxis' = 2, 'showvolume' = false, 'showsum' = true, 'showregion' = false, 'method' = midpoint, 'partition' = 6, 'output' = plot, axes = none, sumvolumeoptions = [color = pink, transparency = 0, lightmodel = none, style = patch], caption = "", orientation = [-120, 75, 0], glossiness = 0, scaling = constrained, showfunction = false, showrotationline = false);
Figure 5.2.8(b) Solid segmented into shells
Calculation of the volume
Expression palette: Definite-integral template
Context Panel: Evaluate and Display Inline
2 π ∫01x2 2−x ⅆx = 5⁢π6
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