The area between the curve y = 1/x, the y-axis and the lines y = 1 and y = 2 is rotated about the y-axis. Find the volume of the solid of revolution formed.

Our response:

We arrived at the correct answer of `1/2` `pi` using disks.

However we are told that it is better to use shells if the defined variable (in this exercise, y) and the axis of revolution are the same.

Using shells, our answer is 2 `pi`.

We are trying to figure out why are we not getting `1/2` `pi` as our answer.

We drew vertical shells each with radius = x, height = y = 1/x, and thickness = dx.

Using the formula V=2π∫(radius)(height)(thickness), we substituted the values of our

shell with the limits of integration.

The x's were cancelled in the integrand leaving dx. Integrating that leaves us with x.

So 2`pi`[2 - 1] = 2 `pi` is our result.

X

Exercise 2:
The area between the curve y = 1/x, the y-axis and the lines y = 1 and y = 2 is rotated about the y-axis. Find the volume of the solid of revolution formed.
Our response:
We arrived at the correct answer of `1/2` `pi` using disks.
However we are told that it is better to use shells if the defined variable (in this exercise, y) and the axis of revolution are the same.
Using shells, our answer is 2 `pi`.
We are trying to figure out why are we not getting `1/2` `pi` as our answer.

Relevant page
<a href="http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-volumes-2009-1.pdf">http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-volumes-2009-1.pdf</a>
What I've done so far
We drew vertical shells each with radius = x, height = y = 1/x, and thickness = dx.
Using the formula V=2π∫(radius)(height)(thickness), we substituted the values of our
shell with the limits of integration.
The x's were cancelled in the integrand leaving dx. Integrating that leaves us with x.
So 2`pi`[2 - 1] = 2 `pi` is our result.

@Phinah: You've almost got it, but the problem is in your expression for height.

What you have found is the volume for the area bounded by the curve and `y=0` and `y=2`, since you've said "height `= 1/x`". But that's not the height of the shell when the lowest possible `y`-value is `1.`

Do you think you can pick it up from there?

X

@Phinah: You've almost got it, but the problem is in your expression for height.
What you have found is the volume for the area bounded by the curve and `y=0` and `y=2`, since you've said "height `= 1/x`". But that's not the height of the shell when the lowest possible `y`-value is `1.`
Do you think you can pick it up from there?

OK, but the `x`-value limits we're integrating from aren't `1` to `2` - those are the `y`-value limits. What are the corresponding `x`-values?

Once you've integrated that, we still won't be at the final answer, because the volume you've found up to this point is just the volume involved with the curved part of the shape. Does that give enough hint about what the missing piece is?

X

OK, but the `x`-value limits we're integrating from aren't `1` to `2` - those are the `y`-value limits. What are the corresponding `x`-values?
Once you've integrated that, we still won't be at the final answer, because the volume you've found up to this point is just the volume involved with the curved part of the shape. Does that give enough hint about what the missing piece is?

If I am going to use shells and have 'y' limits of integration then I would revolve a horizontal strip around the x-axis and use 2π y f(y) dy. Substituting gives us an integrand of 2πy(1/y) dy from y=1 to y=2 which leads to 2π dy. Integrating: 2π y = 2π[2-1] = 2π.

X

If I am going to use shells and have 'y' limits of integration then I would revolve a horizontal strip around the x-axis and use 2π y f(y) dy. Substituting gives us an integrand of 2πy(1/y) dy from y=1 to y=2 which leads to 2π dy. Integrating: 2π y = 2π[2-1] = 2π.

Okay going back to you mentioning the corresponding x-values and then the missing piece.

Converting y-limits to x-limits: when y =2, x=1/2; when y=1, x=1. The greater y-limit should lead to the greater x-limit I thought. But it does not. Substituting and integrating: 2π ∫ x(`1/x` - 1) dx from x=1 to x=`1/2`

X

Okay going back to you mentioning the corresponding x-values and then the missing piece.
Converting y-limits to x-limits: when y =2, x=1/2; when y=1, x=1. The greater y-limit should lead to the greater x-limit I thought. But it does not. Substituting and integrating: 2π ∫ x(`1/x` - 1) dx from x=1 to x=`1/2`

Yes, the first part is now correct. Let me just formalize it:

In general, shell method for revolving around the `y`-axis is

`"Vol"=2pi int_a^bxf(x)dx`

In this case, for the curved part of the shape, we will have

`"Vol"=2pi int_(1/2)^1x(1/x-1)dx`

`=2pi int_(1/2)^1(1-x)dx`

`=2pi[x-x^2/2]_(1/2)^1`

`=2pi(1/8)`

`=pi/4`

The bit that's not right yet is the part "the half of the shell on the left of the axis of revolution". The formula takes care of the whole revolution of the vertical segments (forming a cylindrical shell).

So we still have a missing part.

HINT: Currently, our shape has a round hole in it (from `0` to `1`). What is the volume of that hole?

X

Yes, the first part is now correct. Let me just formalize it:
In general, shell method for revolving around the `y`-axis is
`"Vol"=2pi int_a^bxf(x)dx`
In this case, for the curved part of the shape, we will have
`"Vol"=2pi int_(1/2)^1x(1/x-1)dx`
`=2pi int_(1/2)^1(1-x)dx`
`=2pi[x-x^2/2]_(1/2)^1`
`=2pi(1/8)`
`=pi/4`
The bit that's not right yet is the part "the half of the shell on the left of the axis of revolution". The formula takes care of the whole revolution of the vertical segments (forming a cylindrical shell).
So we still have a missing part.
HINT: Currently, our shape has a round hole in it (from `0` to `1`). What is the volume of that hole?