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Math Help - Volumes of solid revolutions

  1. #1
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    Volumes of solid revolutions

    I'm fed up with solid revolutions and shapes. I've tried examples and it's not helping with the particular problem I'm trying to solve.

    Here it is:
    Use the washer method to compute the volume of the solid obtained by rotating the region bounded by the curves y = 1-X^(2), y = 0

    A (about the x axis)
    This is what I came up with, inner radius= 1-0x, outer radius = 1-(1-x^(2)) or x^(2) when calculated out.

    so... A(x) =pi (x^(2)+0)^(2)-(1-0x)^(2)) ==== pi integrated on bounds of -1 to 1(x^(4)-1)dx ===== turns into (1/5^(5)-1x) when I take the anti derivative. I put in 1 and -1 doing fundamental theorem of calc and end up with 5.026548.

    Is this even close?

    There is another problem that is on my sheet. The only variation is instead of y=0, it has y=-1 instead.

    If anyone could help me with solids of revolution it would be GREATLY appreciated. I passed calc I with a B, but I am having quite a lot of trouble right now with this class, I am trying my butt of and no matter what it feels like my efforts aren't good enough. I almost don't know what to do......
    Last edited by mr fantastic; July 10th 2011 at 04:43 PM.
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  2. #2
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    Re: I'm just incredibly frustrated right now with volumes of solid revolutions

    Quote Originally Posted by tastybrownies View Post
    Here it is:
    Use the washer method to compute the volume of the solid obtained by rotating the region bounded by the curves y = 1-X^(2), y = 0

    A (about the x axis) ...
    rotating the region between y = 1-x^2 and y = 0 about the x-axis would involve disks, not washers ... either you 're using the wrong technique, or something is missing from your problem statement.

    V = \pi \int_{-1}^1 (1-x^2)^2 \, dx

    rotating the same region about y = -1 would involve washers ... note the cylindrical "hole" formed by the rotation.

    V = \pi \int_{-1}^1 [(1-x^2) - (-1)]^2 - 1^2 \, dx
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  3. #3
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    Re: I'm just incredibly frustrated right now with volumes of solid revolutions

    Yeah, not sure why it says to use the washer method on the worksheet, maybe a mistake?

    Well, took your advice and used the other method, A(x) = pi x^(2) change in x for the disks.

    (-x^(2)+1)^(2) == x^(4)-2x^(2)+1, after integrating I ended up with

    pi * (1/5(x)^(5)-2/3x^(3) )+1x/ integrated from -1 to 1,

    I ended up with (8/5)- (-8/5) *pi = 16pi/15 or 3.35....

    For the other method, I did
    (x^(2)+1)^(2) - (2-0x)^(2)
    after multiplying and integrating this, i ended up with the equation,
    1/5x^(5) - 2/3x^(3) -3x/ integrated from -1 to 1,

    I ended up with -6.93*pi= -21.78. Not sure if this should automatically be an absolute value because its volume.

    Thank you again, I'm just trying to understand and wrap my mind around this!
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  4. #4
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    Re: I'm just incredibly frustrated right now with volumes of solid revolutions

    Quote Originally Posted by tastybrownies View Post
    Yeah, not sure why it says to use the washer method on the worksheet, maybe a mistake?

    Well, took your advice and used the other method, A(x) = pi x^(2) change in x for the disks.

    (-x^(2)+1)^(2) == x^(4)-2x^(2)+1, after integrating I ended up with

    pi * (1/5(x)^(5)-2/3x^(3) )+1x/ integrated from -1 to 1,

    I ended up with (8/5)- (-8/5) *pi = 16pi/15 or 3.35.... correct

    For the other method, I did
    (x^(2)+1)^(2) - (2-0x)^(2) ... no
    after multiplying and integrating this, i ended up with the equation,
    1/5x^(5) - 2/3x^(3) -3x/ integrated from -1 to 1,

    I ended up with -6.93*pi= -21.78. Not sure if this should automatically be an absolute value because its volume.

    Thank you again, I'm just trying to understand and wrap my mind around this!
    \pi \int_{-1}^1 [(1-x^2) - (-1)]^2 - 1^2 \, dx

    \pi \int_{-1}^1 (2-x^2)^2 - 1 \, dx

    \pi \int_{-1}^1 3 - 4x^2 + x^4 \, dx

    .. now try it again.

    also, understand that you should be making a graph sketch to determine which method is appropriate to determine the volume of revolution.
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  5. #5
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    Re: I'm just incredibly frustrated right now with volumes of solid revolutions

    Hey, thank you, I did try that second problem again and ended up with 56pi/15, hopefully this is correct. I am just having trouble visualizing these graphs in my mind, which my makes them hard to draw as well.

    Thank you again.
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