How do you find the volume of the solid generated by revolving the region bounded by the graphs #y=x, y=0, y=4, x=6#, about the line x=6?

Here is the region described:

We can find this area in one of two ways. Method 1 uses intuitive geometric properties to find the volume, while Method 2 uses the Disk Method to find the volume.

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Method 1

Rotating this shape about #x=6# will produce a conical frutsum.

In other words, it will be the volume of the cone with base radius #6# and height #6# (as defined by the lines #y=x# and #x=6#) MINUS the volume of the cone with base radius #2# and height #2# (as defined by the area between #y=x, y=4, and x=6#).

Therefore, as described above, the volume is:

#V = 1/3pi(6)^2(6) - 1/3pi(2)^2(2)#

#V = 1/3pi(216)-1/3pi(8)#

#V = 208/3pi#

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Method 2

Since the axis of rotation is vertical, we will integrate with respect to #y#.

The Disk Method formula, therefore, is:

#V = int_a^bpi(x-"axis")^2dy#

#V = piint_0^4(x-6)^2dy#

And since we can make the substitution #y=x#,

#V = piint_0^4(y-6)^2dy#

#V = pi*[(y-6)^3/3]_0^4#

#V = pi[(-2)^3/3 - (-6)^3/3]#

#V = pi[-8/3 + 216/3]#

#V = 208/3pi#