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=6y=x,y=0,y=4,x=6, about the line x=6?

1 Answer
May 30, 2017

V = 208/3piV=2083π

Explanation:

Here is the region described:
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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=6x=6 will produce a conical frutsum.

In other words, it will be the volume of the cone with base radius 66 and height 66 (as defined by the lines y=xy=x and x=6x=6) MINUS the volume of the cone with base radius 22 and height 22 (as defined by the area between y=x, y=4, and x=6y=x,y=4,andx=6).

Therefore, as described above, the volume is:

V = 1/3pi(6)^2(6) - 1/3pi(2)^2(2)V=13π(6)2(6)13π(2)2(2)

V = 1/3pi(216)-1/3pi(8)V=13π(216)13π(8)

V = 208/3piV=2083π

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Method 2

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

The Disk Method formula, therefore, is:

V = int_a^bpi(x-"axis")^2dyV=baπ(xaxis)2dy

V = piint_0^4(x-6)^2dyV=π40(x6)2dy

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

V = piint_0^4(y-6)^2dyV=π40(y6)2dy

V = pi*[(y-6)^3/3]_0^4V=π[(y6)33]40

V = pi[(-2)^3/3 - (-6)^3/3]V=π[(2)33(6)33]

V = pi[-8/3 + 216/3]V=π[83+2163]

V = 208/3piV=2083π