Calculus: finding volume of revolution: the disc method
The tutor gives an example of the disc method for finding volume of revolution.
Usually, the disc method is preferred when the graph is revolved about the x axis.
Example: Find the volume generated, x=7 to x=10, when y=-(x-9)2+5 is revolved about the x axis.
Solution:
Here we have an ideal case for the disc method. The radius of each disc is the height of the graph above the x axis; each disc’s area, therefore, is π(height)2. Of course, the height is y, or -(x-9)2+5. The “thickness” of the disc is dx. Using the concept that
volume=(area)(thickness)
we construct the integral:
V=∫710π(-(x-9)2+5)2dx
which becomes, from expanding the outer square,
V=∫710π((x-9)4-10(x-9)2+25)dx
We can factor out π, then integrate term by term:
π|107 (x-9)5/5 – 10(x-9)3/3 + 25x
Plugging in for the limits of integration we get
π((10-9)5/5-10(10-9)3/3+25(10)
-[(7-9)5/5-10(7-9)3/3+25(7)]
then
π(1/5-10/3+250 -[-32/5+80/3+175])
=π(33/5-90/3+75)=π(33/5+45)=π(258/5)
Apparently the volume arising from the revolution described above is 258π/5.
I’ll be discussing other methods of finding volumes of revolution in future posts:)
Source:
Larson, Roland and Robert Hostetler. Calculus, third edition. Toronto: D C Heath and Company, 1989.
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
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