# Disk Washer Method

1. Jun 13, 2013

### Justabeginner

1. The problem statement, all variables and given/known data
Find the volume of the solid generated by revolving the region bounded by y=1/sqrt(x) and y=0 for 1≤x≤2 about the line y=-1

2. Relevant equations
Disk Method? ∏ * ∫[f(x)]^2 dx

3. The attempt at a solution
A and B (lower and upper limits will be 1 and 2)?
∏ * ∫(1/sqrt(x)^2 dx
∏ * ∫1/x dx
∏ * ln x
Since limits are 2 and 1, my final answer is: ∏ * ln 2

Is this right? Thanks.

2. Jun 13, 2013

### Staff: Mentor

This formula is applicable only if the rotation is around the x-axis.

The formula would be better written as ∏ * ∫[radius]2dx
No. Your integrand doesn't reflect the fact that the region is being rotated around the line y = -1. You can't use a disk in this case - you need to use a washer if you use vertical strips, or you could use shells if you go with horizontal strips.

3. Jun 13, 2013

### Justabeginner

Would the correct way to set it up then be

∏ * ∫([1/sqrt(x)] + 1)^2) ?

∏* ∫[1/x + 2/sqrt(x) + 1]
∏ * ([ln x] + 4sqrt(x) + x) ?

Thank you so much for all your help.

4. Jun 13, 2013

### Staff: Mentor

No, this isn't set up correctly. You're using disks when you should be using washers.

For a washer, the volume is $\pi [(R_{outer})^2 - (R_{inner})^2]Δx$.

5. Jun 13, 2013

### Justabeginner

But I just don't understand, even after drawing this diagram, how there's an inner AND an outer radius? If you could explain that to me, I'd appreciate it.

6. Jun 13, 2013

### Staff: Mentor

The region to be rotated is roughly trapezoidal in shape. Its bottom edge is the x-axis, the left and right edges are the lines x = 1 and x = 2, and the top edge is the curve y = 1/√x.

When the region is rotated around the line y = -1, you get a kind of ring-shaped solid that is hollow. The inner radius is 1, and the outer radius extends from the line y = -1 to the curve.

7. Jun 13, 2013

### Justabeginner

Thank you! I am horrible at visualizing things in 3D and your description made it so much clearer.

But how would I determine the length of the outer radius? I know that the points on the curve are (1, 1) and (2, sqrt(2)/2)- how would that help me in this problem, or would it not be of any use?

8. Jun 13, 2013

### Staff: Mentor

Points on the curve are of the form (x, 1/√x). The upper end of the outer radius is at this point, and the lower end is at (x, -1). How far apart are these points?

9. Jun 13, 2013

### Justabeginner

They are (1/sqrt(x) + 1) units apart?

10. Jun 13, 2013

### Staff: Mentor

Yes, so this is the outer radius. Don't forget, it needs to be squared.

11. Jun 13, 2013

### Justabeginner

So what I get now is:
∏ *∫{[(1/sqrt(x) + 1)^2] - (1^2)} * Δx (Isn't delta x = 1 since x=1 and x=2 are bounding the region?)
∏ * ∫(1/x) + 2/sqrt(x) + 1 - (1)
∏ * ∫(1/x) + 2/sqrt(x)
∏ * (ln x + 4sqrt(x))
And since x=1 and x=2 are the left and right bounds, I'd plug that in to solve for the final volume?

∏* [(ln 2 + 4*sqrt(2) - (ln 1 + 4 sqrt(1)]
∏ * [ln 2 + 4sqrt(2) - 4]

Is this correct?

12. Jun 13, 2013

### Staff: Mentor

No, that's not what Δx means. It corresponds loosely with dx in the integral.

I don't have time to look at this, right now, but will later on if nobody else jumps in.

13. Jun 13, 2013

### Justabeginner

Okay thank you so much! (Yes I just realized, delta x can also be used for dx- I don't know why I even said that -_-)

14. Jun 14, 2013

### Staff: Mentor

Yes, that's what I get, too.

15. Jun 14, 2013

### Justabeginner

Thank you very much sir!