Question

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?

Answer 1

`V = 208/3pi`

Explanation:

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`