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Volume of the resulting solid

  1. Oct 25, 2004 #1
    I am having a lot of trouble with these last 3 problems out of 10 that I have done relating to volumes. I have tried just about every method for each of these and I am just not getting the right answer (according to this online math program I am using).

    2. Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis:
    y = x^2, y = 1; about y = 2.

    The resulting solid for me looks like a volcano. So I used the Shell method like so:

    A(x) = length * width
    A(x) = 2*pi*(x)*x^2 = 2*pi*x^3

    V = integral b/w 0 and 1 of A(x) ...

    Not getting the right answer here either...

    3. A ball of radius 11 has a round hole of radius 4 drilled through its center. Find the volume of the resulting solid.

    I tried using vertical washers like so:

    A(x) = pi*11^2 - pi*4^2

    V = integral b/w 1 and 11 of A(x) = ...

    Incorrect here as well. Not sure how else to approach this one.

    Thanks for help on any of these. At least I got the other 7 problems done on my own. :P
    Last edited: Oct 26, 2004
  2. jcsd
  3. Oct 26, 2004 #2


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    [itex]40x-8x^2[/itex] is a parabola with it's maximum at x=2.5. So the region bounded by this parabola and the line y=0 is like a hill.
    When you rotate it around the y-axis you get something like a donut, but with the upper part like a parabola and the bottom part removed, so it's flat on the bottom side.
    Using horizontal disks is right, but you have to integrate from y=0 to y=50. (From bottom to top).
    Each horizontal disk is like a ring. Try setting up the integral from here.
  4. Oct 26, 2004 #3
    Thanks, I eventually figured it out.

    Still stuck on the other two problems...
  5. Oct 26, 2004 #4


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    In the second, I would suggest that you split the volume up in the following manner:

    Inner Volume: Radius 2-1=1
    The volum of this cylinder is subtracted from:

    Outer Volume: Radius:[tex]2-x^{2}[/tex]

    Hence, your volume should be:
  6. Oct 26, 2004 #5
    I ended up getting 5.864306287 as the answer but the online homework program says it's still incorrect.

    I agree that the small radius is 1. How did you come up with 2 - x^2 as the big radius though? I don't get how to determine what the outer radius is in this case. Thanks for any help.

    Any ideas on the last problem?
  7. Oct 26, 2004 #6
    OK, got an idea for the last problem but still not getting the right answer.

    OK so if I place a sphere of radius 11 in the center of the x-y plane, then the top right quadrant curve of that sphere will equal (r^2 - x^2)^(1/2) correct? So then I can do the volume like so (using vertical washers):

    V = 2pi * Integral b/w 0 and 11 of (11^2 - x^2) - (4^2 - x^2) dx

    But it isn't right according to this program.
  8. Oct 26, 2004 #7


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    You are to rotate it around y=2, right?
    The distance from y=2 to the parabola is 2-x^2
  9. Oct 26, 2004 #8
    Yeah I suppose so. Grrr this stupid program. It must be a bug or something...
  10. Oct 26, 2004 #9


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    For the last problem, you want to rotate the area bounded by [itex]y=\sqrt{11^2-x^2}[/itex] and [itex]y=2[/itex] about the x-axis.

    The answer I get is [itex]\frac{4268\pi}{3}\approx 4469.44[/itex]
  11. Oct 26, 2004 #10
    Wouldn't it be y = 4 and not y = 2 for the other one?

    I tried your answer and it did not work while I tried my answer using y = 4 and of course it says incorrect.

    I'm running out of ideas as to what I could have done wrong for both of these problems.
  12. Oct 26, 2004 #11
    Anyone? These problems are really frustating me...
  13. Oct 27, 2004 #12


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    Yes it is. Sorry for that. It should be 4506.84 (if I did it right this time).
    Does that confirm?

    Also, the integral Arildno gave should run from -1 to 1, so the answer will be doubled.
  14. Oct 27, 2004 #13


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    That's where my mistake was!
    Thx, Galileo, sorry Caldus..:redface:
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