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Find volume of circle. Cross-sections are squares. What am I doing wrong? :/

  1. Jan 21, 2013 #1
    Hi, everyone. I am just trying to do some practice problems on finding volume.

    So this is the one I'm working on:

    1. "Find the volume of the solid described below:

    The solid lies between the planes perpendicular to the x-axis at x=6 and x=-6. The cross-sections perpendicular to the axis on the interval -6≤x≤6 are squares whose bases run from the semicircle y= -√(36 - x^2) to the semicircle y= √(36 - x^2)."


    2. I first drew out the object, which is a sphere with a radius of 6.

    Every cross-section is a square, so each will have an area s^2. The area of each cross-section is 2(√(36 - x^2))^2.

    To find the volume, the dx represents the thickness of each square that fills the sphere.

    2∫-6 to 6 of (√(36 - x^2))^2

    = 2∫36 - x^2

    = 2(36x - (x^3)/3)dx |-6 to 6

    = Plug in values... Answer = 576units^3

    But this is counted as the wrong answer! What am I doing wrong here?! :/
    Thanks for your help! :)
     
  2. jcsd
  3. Jan 21, 2013 #2

    Dick

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    I think the area of each cross section should be (2√(36 - x^2))^2 not 2(√(36 - x^2))^2. See the difference?
     
  4. Jan 21, 2013 #3
    Oh, yes, I see what you mean. Since every cross-section has a side of 2√(36 - x^2), this whole thing is what needs to be squared.

    But- I did this in another problem, and it led me to the wrong answer! That is why I switched it to 2(√(36 - x^2))^2.

    The other problem was this:

    "Find the volume of the solid describe below:

    The solid lies between the planes perpendicular to the x-axis at x=0 and x=6 . The cross-sections perpendicular to the axis on the interval 0≤x≤6 are squares whose diagonals run from the parabola y= -√(x) to the parabola y= √(x)."

    So when I drew out the graph, it looked sort of like a parabola turned on its side.

    Since every cross-section is a square, the area = s^2.
    Following the should-be-correct-method, the area of each square = (2√(x))^2

    ∫0 to 6 of (2√(x))^2dx

    = ∫4xdx

    = 4(1/2*x^2) |0 to 6

    = 4(1/2(6)^2) - 0

    = 72
    This is counted as the wrong answer! The correct answer is 36!

    I get 36 only when I separate the 2 from the (√(x))^2. But we already went over how that would be incorrect.

    So what is wrong here...?
    Thanks so much for the help! :)
     
  5. Jan 21, 2013 #4

    Dick

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    In the first case they are giving you the side of the square s. Then the area is s^2. In the second case they are giving you the diagonal of the square. If the diagonal of the square is s, what's the area?
     
  6. Jan 21, 2013 #5
    Oh, right! Thanks!

    I thought that if the diagonals run from the parabola y=-√(x) to y=√(x),
    then each diagonal should equal 2*√(x)

    So since eace DIAGONAL = 2√(x), we have to figure out the length of each SIDE!

    2√(x)= side*√2

    [2√(x)]/√(2) = side

    ∫0 to 6 of [(2√(x))/√(2)]^2 dx

    = ∫4x/2 dx= ∫2x dx

    = (2x^2)/2 |0 to 6

    = (6)^2 - (0)^2

    = 36

    So, yay, I get the right answer. But am I thinking about the diagonals and sides correctly?

    Thank you! :)
     
  7. Jan 21, 2013 #6

    Dick

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    Yes you are. If they give you a side s then the area is s^2. If they give you a diagonal d, then s=d/sqrt(2) so the area is d^2/2.
     
  8. Jan 21, 2013 #7
    Oh, okay. Thank you! :D
     
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