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Use - disks, washers or shells? integrate volume of revolving solids

  1. Mar 30, 2013 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 30, 2013 #2
    Disks and washers are essentially equivalent - it's analogous to finding area versus finding the area between two curves. So just thinking disks verses washers: it depends on the shape of the relation (not necessarily functions!), what is the best method to use. If you have a relation that's also a function and even 1-1, then it probably doesn't matter. If you have a function that's not 1-1, try [itex]dx[/itex] disks. If you have [itex]x[/itex] as a function of [itex]y[/itex], but it's not 1-1, try shells with a [itex]dy[/itex]. You don't want the infinitesimal shape to intersect the relation except at the endpoints, or you'll have to break up your integral into multiple pieces, which ends up being more work.
     
  4. Apr 3, 2013 #3
    How do I set up the integral for the curve y = x, x = 2 - y^2, and y = 0 to find the volume
    when it is revolved

    around x axis
    around y axis
    about the line x = -1
    y = 1

    I am quite confused. Do I change x = 2 - y^2 to y = Squareroot (2 - x), when required. If I do so, does the limit of integral also change or does it remain the same.
     
  5. Apr 3, 2013 #4
    You will need to find the intersections of all three defining curves. Where does that happen?

    For the around the x axis case, I'd recommend shells using a dy.

    For the around the y axis case, I'd recommend washers using a dy. Use the same for about the line x = -1.

    For around the line y=1, I'd recommend shells using a dy.

    Can you see why I picked these?
     
  6. Apr 4, 2013 #5

    Mark44

    Staff: Mentor

    The following is copied from a PM sent by Rattanjeet.

     
  7. Apr 4, 2013 #6
    Thanks for your suggestion.

    Intersections are (0,0), (1,1) and (2,0)

    And my working of this question is as follows:

    a) Volume when the region is rotated about x-axis:

    Method used: shell method

    V=2∏∫[c-d] y g(y) dy
    V=2∏{∫[0-1] y (y) dy - ∫[1-2] y(2-y^2)} dy
    V=2∏{∫[0-1] y^2 dy - ∫[1-2] (2y-y^3)} dy

    b) Volume when the region is rotated about y-axis:

    Method Used: Washer Method

    V = ∏∫[c-d] f(y)^2 - g(y)^2 dy
    V= ∏{∫[0-1] (2-y^2)^2dy - ∫[1-2] (y)^2 dy }

    c) Volume when the region is rotated about x = -2:

    Method used: Wahser method

    V = ∏∫[c-d] f(y)^2 - g(y)^2 dy
    V= ∏{∫[0-1] (2-y^2 + 2)^2dy - ∫[1-2] (y+2)^2 dy }
    V= ∏{∫[0-1] (4-y^2)^2dy - ∫[1-2] (y+2)^2 dy }

    d) Volume when the region is rotated about y = 1:

    Method Used: Washer Method

    V = ∏∫[c-d] f(y)^2 - g(y)^2 dy
    V = ∏∫[0-1] (1 - y)^2dy - ∏∫[0-1](1-(2-y^2))^2 dy
    V = ∏∫[0-1] (1 - y)^2dy - ∏∫[0-1](y^2-1))^2 dy

    Is my approach correct?

    Rattanjeet

     
  8. Apr 4, 2013 #7

    Mark44

    Staff: Mentor

    Let's take a look at the first two parts of your question. Generally it is better to ask a single question in a thread.
    These are correct.
    Your setup in the 2nd line above is incorrect, and will give you the negative of the volume. For the length of the shell, you want xparabola - xline. You have it the other way around, which produces a negative value.

    In the shell method you are taking thin horizontal strips that extend from the line x = y to the parabola x = 2 - y2. The length of a strip is 2 - y2 - y. The width of each strip is Δy, a small subinterval in the interval [0, 1] on the y-axis.

    ΔV, the volume of a typical volume element, is given by ΔV ≈ 2##\pi## [radius] * [length of strip] * Δy

    This problem could also be done using washers, but you would need two integrals.
    This is wrong as well. The typical volume element here is
    ΔV ≈ ##\pi## [xouter2 - xinner2] * Δy
    = ##\pi## [(2 - y2)2 - y2]Δy

    You don't need two integrals.
     
  9. Apr 4, 2013 #8
    Thanks Mark,

    But I have to ask you how to remove a thread that I posted just today. To this very question I got another reply saying that my answer was correct and I should put it on thread. I did that. Since the answer was wrong, I would like to withdraw it so that others who read it are not misled.

    Please guide.

    Thanks once again.
    Rattanjeet
     
  10. Apr 4, 2013 #9

    Mark44

    Staff: Mentor

    You have a total of 4 posts, and they're all in this thread. I deleted one of your posts, as it seemed to be a duplicate of what is already in this thread.
     
  11. Apr 4, 2013 #10

    Mark44

    Staff: Mentor

    I should also mention that it's really important to draw a sketch of the region you're working with, and another sketch of the solid you get by revolving the region. It is very difficult to get these problems right if you don't have a reasonably good picture.
     
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