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Homework Help: Calculating volumes by shell and slicing

  1. Aug 19, 2011 #1
    1. The problem statement, all variables and given/known data

    The area under the graph of the function y = cos inverse x on the interval [0; 1] is rotated
    about the x-axis to form a solid of revolution.
    (a) Write down the volume V of the solid as a de nite integral with respect
    to x according to the disc/slicing method. Do NOT attempt to evaluate this
    (b) Write down the volume V of the solid as a de nite integral with
    respect to y according to the shell method.
    (c) Using the antiderivative,
    Integral y cos y dy = y sin y + cos y + C;
    or otherwise, find the volume V of the solid as an exact real number.

    2. Relevant equations

    V=2*PI integral x*f(x) dx

    3. The attempt at a solution

    (a)V=integral 1 to 0 Pi*cos^-2x^2 dx

    (b)V=2PI integral 1 to 0 ycosy dy

    (c) i solved it and i got 2PI(PI/2 -1)
  2. jcsd
  3. Aug 19, 2011 #2


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    Your function is [tex]y = \cos^{-1} x .[/tex] Be aware that this is inverse functional notation, so [tex] \cos^{-1} x \neq \frac{1}{\cos x} .[/tex] So when you write the integral for the "disc/washer method", it should be

    [tex]V = \pi \int_{0}^{1}(\cos^{-1}x)^{2} dx , [/tex]

    since you are squaring an inverse cosine function.

    On part (b), watch out when working with y as the variable of integration. What does the graph of inverse cosine x from 0 ≤ x ≤ 1 look like? When you "turn it sideways" to make the slices parallel to the x-axis for the shells, is there only one set of boundaries for the "lower" and "upper ends" of the shells?
  4. Aug 20, 2011 #3
    the y values should be from 0 to PI/2 therfore how would i take a slice of the shell, would i just integrate ycosy dy :P
  5. Aug 20, 2011 #4


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    You really should make a graph of the region that is being revolved about the x-axis: the curve x = cos y is only a boundary for part of it if you integrate in the y-direction (as you will have to for shells).

    What ARE the boundaries of the region and what will act as the "upper" and "lower curves" if you are integrating with respect to y? (Using y cos y is only a part of this.)
  6. Aug 20, 2011 #5


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    (Forget that last note: now that I look at this again this morning, I see that the grapher I used took y = cos-1 x and plotted y = sec x anyway. )

    The curve does close properly from (0, pi/2) to (1, 0). You will need to change your limits of integration.

    So using shells, you should have [tex] 2\pi \int_{0}^{\pi/2} y cos y dy .[/tex]

    Sorry for my earlier comment.

    Your answer for part (c) appears to be correct. So you integrated correctly, but wrote the limits wrong in part (b).
    Last edited: Aug 20, 2011
  7. Aug 21, 2011 #6
    Hi, I was just passing by and I don't follow this integrations..
    Sorry but dynamicsolo can you please explain in details how to get to the answers for (b) and (c)?
  8. Aug 21, 2011 #7
    hmmm its a method of integration to calculate volumes by the shell method
  9. Aug 21, 2011 #8


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    He's asking exactly how you set up the shells; we're discussing this by PM.
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