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Finding volume of a nose cone with a given r with integration

  1. Oct 14, 2014 #1

    LBK

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    I'm still confused on some of these volume problems, so please bear with me :)
    1. The problem statement, all variables and given/known data
    Find the volume of a reentry spacecraft nose cone that has a cross-section radius of (1/4)x2 taken x feet from the nose and perpendicular to the axis of sym. We are given that the length of the cone is 10 ft.

    2. Relevant equations
    Well, I'm not sure if a "nose cone" is a standard cone or is this like a frustrum?
    I see a formula for 2pi*Int(0, 10) x*(1/4)x2dx but is that all there is to it?

    3. The attempt at a solution
    Using the equation above I would get 2pi/4[(1/2)x^2*(x^3)/3] over interval 0 to 10
    so: 2pi/4[50*1000/3]-0=2pi/4*50,000/3=100,000pi/12=25,000pi/3 This just seems way too huge.
     
  2. jcsd
  3. Oct 14, 2014 #2

    SteamKing

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    The shape is not a cone with straight sides like this:

    Cone_%28geometry%29.png
    The radius of the cone, however, is given a a function of the distance x measured from the apex. The radius is 0 at the apex and (1/4)*102, or 25 feet at the base.

    I'm not sure if you have set up the integral expression for the volume correctly. After all, the element of volume dV = A(x) dx, where A(x) is the cross-sectional area of the cone at distance x from the apex.

    If you are trying to use one of the Theorems of Pappus, it doesn't appear that you have applied it correctly to this problem.
     
  4. Oct 14, 2014 #3

    LCKurtz

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    It looks like you are trying to use a formula for the volume of a solid of revolution but you are getting mixed up. You have to decide whether to use shells or disks and be consistent with the formulas. If you are going to use ##y## limits of ##0## to ##10##, then you need the formula for disks using a ##dy## integral. If you are going to use a ##dx## integral you would use the shells method$$
    V =\int_a^b \pi x^2(y_{upper}-y_{lower})dx$$ You have elements of both in your attempt.

    [Edit] Hold on, I may have misinterpreted something....

    [Edit, added] Yes, I did, but you are still mixing up the two methods. You want to revolve ##y=\frac 1 4 x^2## about the ##x## axis. Us the disk method.
     
    Last edited: Oct 14, 2014
  5. Oct 14, 2014 #4

    LBK

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    OK, I'm still reading up, and while there's not a direct example, if I can use disk method, I get
    A(x)=pi(f(x))2 from which I get V=INT A(x)dx
    So V=INT[0,10]*pi((1/4)x2)2dx integrating this I would get pi/16*(x3/3)| 0 to10
    =pi/16(1000/3)-0=1000pi/48 =125pi/6
    Am I warm now? :D Oh, wait I still have a problem with dy vs. dx? The actual text of my question said the radius was measured at x feet from the tip, so I took it to be like a cone with axis of symmetry being horizontal.
     
    Last edited: Oct 14, 2014
  6. Oct 14, 2014 #5

    SteamKing

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    If r(x) = (1/4)x2, then what is [r(x)]2? Check your algebra here before integrating.
     
  7. Oct 14, 2014 #6

    LBK

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    Oh good grief, thank you!
     
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