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Application of integration: Volume

  1. Dec 4, 2005 #1
    Find the volume of the right cone of height 9 whose base is an ellipse. Major axis 12 and minor axis 6.
     
  2. jcsd
  3. Dec 4, 2005 #2

    arildno

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    use the formula for the volume of a cone.
     
  4. Dec 4, 2005 #3

    HallsofIvy

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    I didn't know there was a formula for the volume of a cone with a non-circular base!

    Hmm- I guess there will be once we do the integral!

    nyfan0729: At height "y" above the base of the cone, a cross section will be an ellipse with major axis of length (2/3)(9- y) and minor axis of length (1/3)(9-y): that would have area
    [tex]\frac{2\pi}{9}(9-y)^2[/tex]

    It should be easy to integrate that from y= 0 to 9.
     
  5. Dec 4, 2005 #4

    arildno

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    If it is done correctly, it is one third of the product of the base area and the height (a right cone is, of course, assumed):

    Let (x,y) be an arbitrary point in the base area A lying in the plane z=0, and for simplicity, let the position of the apex be (0,0,h), where h is the height,

    Thus, the cone may be represented as:
    [tex]\vec{F}(x,y,t)=((x,y,0)-(0,0,h))t+(0,0,h)=(xt,yt,h(1-t)), (x,y)\in{A}, 0\leq{t}\leq{1}[/tex]
    The Jacobian is [itex]ht^{2}[/tex], from which the volume formula follows:
    [tex]V=\int{dV}=\iint_{A}\int_{0}^{1}t^{2}hdtdA=\frac{h}{3}\iint_{A}dA=\frac{Ah}{3}[/tex]
     
  6. Dec 4, 2005 #5

    Tide

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    Arildno,

    I believe the result is even more general than that. You don't have to assume a "right cone." As long as the horizontal cross sections of the cone are self-similiar then you'll get the same result.
     
  7. Dec 5, 2005 #6

    arildno

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    I assumed a "right cone" since that was assumed in the exercise.
    You are right, of course..:wink:
     
  8. Dec 5, 2005 #7
    But it's obviously half the volume of a right circular cone of diameter 12 because it's just compressed to half size in the direction of the minor axis, so [tex]\frac{\pi 6^2}{2.3}[/tex]
     
    Last edited: Dec 5, 2005
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