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Math Help - Volume through integration

  1. #1
    Super Member angel.white's Avatar
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    Volume through integration

    Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

    x = 0, x = 9-y^2, about the line x = -1

    :/ Reviewing for a test next week and realize I've forgotten how to do these, tried two different ways with incorrect answers each time (I think my problem is by rotating it around x=-1 instead of an axis, because I just did two of these correctly which were rotated about an axis).
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  2. #2
    Moo
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    Hello,

    I'm not sure of my comprehension of the problem :s (and for the resolution )
    Find the domain for y, helping with x=9-y and the domain of x. I think that the latter is [-2,0] (symmetric to -1)

    The volume will be :

    \int_0^{2 \pi} dz \int_{y_1}^{y_2} dy \int_{-2}^0 dx
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  3. #3
    Super Member angel.white's Avatar
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    I appreciate it, but we haven't gotten that far yet. (I think we are introducing the z-axis after this test)

    Here is an example of one that I did earlier.
    Attached Thumbnails Attached Thumbnails Volume through integration-volume003.jpg  
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  4. #4
    Moo
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    I'm sorry, i didn't even know this method

    We studied volumes, but not that way :s
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    Quote Originally Posted by angel.white View Post
    Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

    x = 0, x = 9-y^2, about the line x = -1

    :/ Reviewing for a test next week and realize I've forgotten how to do these, tried two different ways with incorrect answers each time (I think my problem is by rotating it around x=-1 instead of an axis, because I just did two of these correctly which were rotated about an axis).
    I'd translate the region to the right by 1 unit (note that x = 9-y^2 --> x = 10 - y^2) so that the axis of revolution becomes the y-axis. Then use the shell method.
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  6. #6
    Super Member angel.white's Avatar
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    Quote Originally Posted by mr fantastic View Post
    I'd translate the region to the right by 1 unit (note that x = 9-y^2 --> x = 10 - y^2) so that the axis of revolution becomes the y-axis. Then use the shell method.
    I was able to get 10-y^2, but I don't think the shell method will work. My understanding is that you can use the shell method when the axis being rotated about is the same as the height of your function (ie a function of x being rotated around the y-axis). In this case, my height is x because this is a function of y, but my axis being rotated around is x=-1, which is a vertical axis. If I am not too confused, I would need to rotate around y=-1 or some parallel line in order to use the shell method.

    I went ahead and tried it anyway, and got:
    2\pi \int_{-3}^3 (y(10-y^2)) ~dy

    = 2\pi \int_{-3}^3 (10y-y^3)) ~dy

    = 2\pi \left[ 5y^2-\frac 14y^4\right]_{-3}^3

    = 2\pi \left[ 5(3)^2-\frac 14(3)^4-5(-3)^2 + \frac 14(-3)^4\right]

    = 0


    The correct answer should be \frac{1656\pi}5 if the back of the book is correct.
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  7. #7
    Super Member angel.white's Avatar
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    Okay, I figured it out.

    Cross-sectional area is \pi x^2 \Rightarrow \pi (0+1)^2, \pi (9-y^2+1)^2

    Then:
    V = \pi \int_{-3}^3 ( (10-y^2)^2 - (1)^2) ~dy

    V = \pi \int_{-3}^3 ( 99 -20y^2 + y^4) ~dy

    V = \pi \left(99y -\frac{20}3y^3 + \frac 15y^5\right)_{-3}^3

    V = \pi \left(99(3) -\frac{20}3(3)^3 + \frac 15(3)^5 -99(-3) + \frac {20}3(-3)^3 - \frac 15(-3)^5\right)

    V = \pi \left(594 -360 + \frac 15(486)\right)

    V = \frac \pi 5(1656)
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  8. #8
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    Here's the shell method since Angel white done the washer method.

    With shells, the cross-sections are parallel to the axis about which we ar rotating. Since x=-1 is vertical, the cross-sections will be 'stacked up' along the x-axis and we can integrate wrt x.

    4{\pi}\int_{0}^{9}(x+1)\sqrt{9-x}dx=\frac{1656\pi}{5}
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  9. #9
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    Quote Originally Posted by angel.white View Post
    Okay, I figured it out.

    Cross-sectional area is \pi x^2 \Rightarrow \pi (0+1)^2, \pi (9-y^2+1)^2

    Then:
    V = \pi \int_{-3}^3 ( (10-y^2)^2 - (1)^2) ~dy

    V = \pi \int_{-3}^3 ( 99 -20y^2 + y^4) ~dy

    V = \pi \left(99y -\frac{20}3y^3 + \frac 15y^5\right)_{-3}^3

    V = \pi \left(99(3) -\frac{20}3(3)^3 + \frac 15(3)^5 -99(-3) + \frac {20}3(-3)^3 - \frac 15(-3)^5\right)

    V = \pi \left(594 -360 + \frac 15(486)\right)

    V = \frac \pi 5(1656)
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  10. #10
    Super Member angel.white's Avatar
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    Thank you everyone for helping me out today ^_^

    These are things I should have kept up with, but my data structures and algorithms course has been soooooo exciting lately that Calc kind of fell by the wayside :/

    Good to be catching up on Calc, though (since I'm 2 weeks ahead in DS&A with nothing else to do)
    Quote Originally Posted by mr fantastic View Post
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