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Washer and Shell Methods

  1. Jun 15, 2013 #1
    1. The problem statement, all variables and given/known data
    The region in the first quadrant bounded by the lines y=4 and x=2 and the curve x^2=4y is revolved about the y-axis. Find the volume:
    a) by the method of washers
    b) by the method of shells

    2. Relevant equations

    3. The attempt at a solution

    I drew the region, and I can picture what the solid volume would look like. I think the limits of the integral are 0 and 4, if I'm not mistaken. Also, for the washer method, I need the inner and outer radius. I think the outer radius is x^2/4, but I'm not sure what the inner radius would be. As for the shell method, I think the radius would be 4 and the height, I'm not so sure about. Any help is appreciated. Thank you so much.
  2. jcsd
  3. Jun 15, 2013 #2
    Here is a graph of the region in question. Note the region in question is the region above the parabola (and below y = 4), not the region below the parabola (and above y = 0).

    You correctly identified the limits of integration for the washer method; however, the upper and lower limits of integration for the shell method are x = 0 and x = 2, respectively. I suggest you review the formulas associated with the washer and shell methods.

    For the washer method, the radius [itex]R(y)[/itex] is given by

    [tex]R(y) =
    \left \{
    2\sqrt{y} , & 0 < y < 1\\
    2 , & 1 < y < 4

    Because the radius is a piece-wise function, two integrals must be evaluated to find the volume.

    For the shell method, the radius [itex]p(x)[/itex] is given by [itex]p(x) = x[/itex] and the height [itex]h(x)[/itex] is given by [itex]h(x) = 4 - \dfrac{x^2}{4}[/itex].

    After you review the washer and shell methods, you should be able to progress from here.
  4. Jun 15, 2013 #3
    I do not understand how you determined there to be two integrals for the radius for the washer method. If you could please explain that to me, I'd appreciate it. Thank you.
  5. Jun 15, 2013 #4
    By the washer method, the formula for the volume [itex]V[/itex] is given by

    [tex]V = \pi \int_0^4 [R(y)]^2 \, dy[/tex]

    Keeping in mind [itex]R(y)[/itex] is a piecewise-defined function with two subfunctions, can you evaluate [itex]\int_0^4 [R(y)]^2 \, dy[/itex] with only one integral?
  6. Jun 16, 2013 #5
    No you would need two integrals, one with a limit from 0 to 1, and then the other integral with a limit from 1 to 4.

    For the first integral:

    [itex] ∫_0^1 4y dy [/itex]
    [itex] 2y^2 | _0^1 [/itex]
    [itex] 2∏ [/itex]

    Second Integral:

    [itex] ∫_0^1 4 dy [/itex]
    [itex] 4y | _1^4 [/itex]

    12∏ + 2∏= 14∏

    The same answer as I got for the shell method. I understand how you got the 2√y for the function of the first integral, but I do not get how you got the '2' for the second integral. Can you please explain that to me? Thank you so much.
  7. Jun 16, 2013 #6
    I attached a graphic to aid our discussion. The black line at [itex]y = 1[/itex] appropriately divides the graph into two regions, one below the line and one above the line. In each region, an arbitrary radius is represented by a red rectangle. In the lower region where [itex]0 < y < 1[/itex], the radius [itex]R(y)[/itex] is given by [itex]R(y) = 2\sqrt{y}[/itex] as you already know. In the upper region where [itex]1 < y < 4[/itex], what is the radius [itex]R(y)[/itex]? From the attached graphic, it should be clear [itex]R(y)[/itex] is constant in this region.

    Attached Files:

  8. Jun 16, 2013 #7
    Thank you so much. The graph really helped me understand it better.
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