1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Disk Washer Method

  1. Jun 13, 2013 #1
    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. jcsd
  3. Jun 13, 2013 #2

    Mark44

    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.
     
  4. Jun 13, 2013 #3
    Would the correct way to set it up then be

    ∏ * ∫([1/sqrt(x)] + 1)^2) ?

    And follow up with:
    ∏* ∫[1/x + 2/sqrt(x) + 1]
    ∏ * ([ln x] + 4sqrt(x) + x) ?

    Thank you so much for all your help.
     
  5. Jun 13, 2013 #4

    Mark44

    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##.
     
  6. Jun 13, 2013 #5
    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.
     
  7. Jun 13, 2013 #6

    Mark44

    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.
     
  8. Jun 13, 2013 #7
    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?
     
  9. Jun 13, 2013 #8

    Mark44

    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?
     
  10. Jun 13, 2013 #9
    They are (1/sqrt(x) + 1) units apart?
     
  11. Jun 13, 2013 #10

    Mark44

    Staff: Mentor

    Yes, so this is the outer radius. Don't forget, it needs to be squared.
     
  12. Jun 13, 2013 #11
    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?
     
  13. Jun 13, 2013 #12

    Mark44

    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.
     
  14. Jun 13, 2013 #13
    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 -_-)
     
  15. Jun 14, 2013 #14

    Mark44

    Staff: Mentor

    Yes, that's what I get, too.
     
  16. Jun 14, 2013 #15
    Thank you very much sir!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Disk Washer Method
  1. Disk and washer method (Replies: 2)

Loading...