# 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=∫_{7}^{10}π(-(x-9)^{2}+5)^{2}dx

which becomes, from expanding the outer square,

V=∫_{7}^{10}π((x-9)^{4}-10(x-9)^{2}+25)dx

We can factor out π, then integrate term by term:

π|^{10}_{7} (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.