y=1/x , y=0 , x=1 , x=2 | Find the volume using the shell method?

I want to know how to set up the integral? Thanks

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- Jun 6th 2005, 05:17 PMthefx45Shell Method
y=1/x , y=0 , x=1 , x=2 | Find the volume using the shell method?

I want to know how to set up the integral? Thanks - Jun 7th 2005, 01:34 AMticbol
Volume? Shell method?

I'd say you want to find the volume generated if the area bounded by the given 4 lines/curve is rotated about the y-axis.

The rotation could be about the y-axis, or about the x=1 line, or about the x=2 line, or about any vertical line. I just feel it is about the (x=0)- or y-axis.

The volume could be for the area rotated about the x-axis, or about any horizontal line, but if by Shell method, then this method is less appropriate than by Disc method.

(Next time, please complete your questions so that we can avoid guessing.)

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Imagine, or draw on paper the figure.

The area is bounded by the two vertical lines (x=1, x=2), by the horizontal line y=0, and by the right hyperbola y = 1/x. The area is under the y = 1/x curve.

That area is to be rotated about the y-axis.

By shell method,

We get an infinitesimal shell.

The height is [y of hyperbola] minus [y of y=0], or it is (1/x -0) = 1/x.

The radius is x away from the y-axis.

The thickness is dx

So, the volume is

dV = (circumference)*(height)*(thickness)

dV = (2pi*x)(1/x)(dx)

dV = 2pi dx

The boundaries of dx are x=1 and x=2, or, the interval of dx is from x=1 to x=2, so the volume of the figure is:

V = INT.(1->2)[2pi dx]

V = (2pi)*INT.(1->2)[dx]

V = (2pi)*[x](1->2)

V = (2pi)[2 -1]

V = (2pi)[1]

V = 2pi cu.units ----answer.