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Math Help - Volumes of rotation- Shell method, multiple problems.

  1. #1
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    Volumes of rotation- Shell method, multiple problems.

    I just learned the shell method for volumes of rotation today and I'm swinging and missing at all these problems except a few, so I can't understand what I'm getting wrong. Attached are the two scans of my work, since I can't type in my work very quickly/well on the computer since I'm not extremely tech savvy but here it is.

    Volumes of rotation- Shell method, multiple problems.-002.jpgVolumes of rotation- Shell method, multiple problems.-001.jpg
    Last edited by mr fantastic; January 3rd 2011 at 09:10 PM. Reason: Title.
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    Quote Originally Posted by cubejunkies View Post
    I just learned the shell method for volumes of rotation today and I'm swinging and missing at all these problems except a few, so I can't understand what I'm getting wrong. Attached are the two scans of my work, since I can't type in my work very quickly/well on the computer since I'm not extremely tech savvy but here it is.

    Click image for larger version. 

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    Problem 1.

    Washer method:
    The cross-sectional area is the difference in area of 2 discs,
    one with radius 2,
    the other with radius \sqrt{x}


    \displaystyle\ V=\int_{0}^4{\pi}{\left(2^2-\sqrt{x}^2\right)}dx={\pi}\int_{0}^4{(4-x)}dx

    \displaystyle\ =4{\pi}(4-0)-\frac{{\pi}\left(4^2-0\right)}{2}
    Last edited by mr fantastic; January 3rd 2011 at 09:11 PM.
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    But we're supposed to use the shell method which is 2pi int((shell radius)(shell height) dx) or dy, and of course on the correct interval with respect to the variable which the volume is being integrated in terms of
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    y = \sqrt{x}

    y^2 = x

    height = y^2

    radius = y , not (2-y)

    \displaystyle V = 2\pi \int_0^2 y^3 \, dy
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    Quote Originally Posted by cubejunkies View Post
    But we're supposed to use the shell method which is 2pi int((shell radius)(shell height) dx) or dy, and of course on the correct interval with respect to the variable which the volume is being integrated in terms of
    Shell method:

    Integrate over the y-axis as the shells are horizontal.

    \displaystyle\int_{0}^2{2{\pi}rh}\;dy=2{\pi}\int_{  0}^2{f(x)x}\;dy=2{\pi}\int_{0}^2{(y)y^2}dy
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    for problem #2 ... why just the region in quad I ? the region between y = x^2 and y = 1 is in quads I and II, which probably explains why you are getting 1/2 the correct volume.
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    for problem #3 ...

    region in quad I defined by x = 12(y^2 - y^3) and the y-axis ...


    rotated about y = k \le 0

    \displaystyle V = 2\pi \int_0^1 (y - k)[12(y^2-y^3)] \, dy


    rotated about y = k > 1

    \displaystyle V = 2\pi \int_0^1 (k-y)[12(y^2-y^3)] \, dy
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    Quote Originally Posted by skeeter View Post
    for problem #3 ...

    region in quad I defined by x = 12(y^2 - y^3) and the y-axis ...


    rotated about y = k \le 0

    \displaystyle V = 2\pi \int_0^1 (y - k)[12(y^2-y^3)] \, dy


    rotated about y = k > 1

    \displaystyle V = 2\pi \int_0^1 (k-y)[12(y^2-y^3)] \, dy
    THANK YOU BOTH SO MUCH!!!!!!! now the last thing im having problems with is really understanding what youre doing for the third problem, it makes sense to me, but could you just explain your thinking process to arrive upon where you did? just merely so i can get ones like this in the future

    THANK YOU GUYS SO SO MUCH!!!
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    note the diagram ...

    red thick line is a representative shell of thickness dy to be rotated.

    blue dashed horizontal lines are the axes of rotation (one greater than y = 1 , one less than y = 0)

    green dotted lines are the radii lengths.
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    oh i get it now! its all so much clearer for me, thank you so so much!
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