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Prove cross-section of elliptic paraboloid is a ellipse

  1. May 19, 2012 #1
    1. The problem statement, all variables and given/known data
    a elliptic paraboloid is x^2/a^2+y^2/b^2<=(h-z)/h, 0<=z<=h, show that the horizontal cross-section at height z, is an ellipse


    2. Relevant equations



    3. The attempt at a solution
    i dont know how to prove this, i only know that the standard ellipse is x^2/a^2+y^2/b^2=1..
    could someone help me?
     
  2. jcsd
  3. May 20, 2012 #2

    tiny-tim

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    hi chris_usyd! :smile:

    (try using the X2 button just above the Reply box :wink:)
    what is the intersection of it with the plane z = C ? :wink:
     
  4. May 20, 2012 #3
    yes Tim, i think the intersection is the plane z=C, and c is (0,h);
    but how to rewrite this to standard ellipse form?
     
  5. May 20, 2012 #4

    tiny-tim

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    what do you mean? :confused:

    just write out the equation of the intersection
     
  6. May 20, 2012 #5
    z is between 0 and h....sorry my fault.

    i think the equation of the intersection @certain height 'z' is just x^2/a^2+y^2/b^2<=(h-z)/h..isn't it?
    but how to rewrite it as a standard ellipse form ( x^2/a^2+y^2/b^2=1)??
     
  7. May 20, 2012 #6

    tiny-tim

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    divide by (h-z)/h ?
     
  8. May 20, 2012 #7
    : )
    then the equation is : hx^2/((h-z)*a^2) + hy^2/((h-z)*b^2)=1
    this is an ellipse equation???
     
  9. May 20, 2012 #8
    if so, the area of the intersection @certain height is pi*((h-z)/h)*a*b, isnt it?

    since @ height z, when x=0, y=sqrt((h-z)/h)*b,when y=0, x=sqrt((h-z)/h)*a.
    to calculate the area, just simply pi*sqrt((h-z)/h)*sqrt((h-z)/a),which is pi*((h-z)/h)*a*b ????
     
  10. May 20, 2012 #9

    tiny-tim

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    (have a square-root: √ and a pi: π :wink:)
    (difficult to read, but …) yes :smile:
     
  11. May 20, 2012 #10
    nice. ^^
    thank you,T.
     
  12. May 22, 2012 #11
    Tim, same equation for the solid elliptic paraboloid, x^2/a^2+y^2/b^2<=(h-z)/h, 0<=z<=h.
    the question is : suppose a>=b, calculate the volume of that part of the paraboloid that lies above the disc x^2+y^2<=b^2.( use a suitable integral in polar coordinates.)
     
  13. May 22, 2012 #12
    how can i set up x and y in polar coordinates??
    we usually get a cylinder or whatever the intersection is a circle.
    in these cases, x=rcos(),y=rsin()...
     
  14. May 22, 2012 #13

    tiny-tim

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    let's see …

    the paraboloid sits on a base on the x-y plane which is the ellipse x²/a²+y²/b² = 1,

    and the ellipses get smaller up to height z = h, where they disappear

    and we want the volume of that inside the cylinder that just fits inside the base ellipse​

    ok, so slice it into horizontal "discs" of thickness dh at height h …

    each "disc" is actually the intersection of a circle (r = b) with an ellipse

    that's the area which you need the polar coordinates to find :wink:
     
  15. May 22, 2012 #14
    ok... i do this question like this.. tell me if i am right or not. : ))
    let x = r∙cos(θ),y = r∙sin(θ)
    dS = dxdy = r drdθ
    because
    x² + y² ≤ b² =>r²∙cos²(θ) + r²∙sin²(θ) ≤ b²=>r² ≤ b²
    and r is the distance to the origin, which can not be negative. So the range of integration in radial direction is:
    r:0->b
    θ:0->2pi
    the height above the disc is z = h∙(1 - (x²/a²) - y²/b²)
    => h∙(1 - (1/a²)∙r²∙cos²(θ) - (1/b²)∙r²∙sin²(θ))

    with the identities
    cos(2∙θ) = cos²(θ) - sin²(θ) = 2∙cos²(θ) - 1 = 1 - 2∙sin²(θ)
    =>
    cos²(θ) = (1/2)∙(cos(2∙θ) + 1)
    sin²(θ) = (1/2)∙(1 - cos(2∙θ))
    height z can be rewritten as:
    z = h∙(1 - (1/a²)∙r²∙(1/2)∙(cos(2∙θ) + 1) - (1/b²)∙r²∙(1/2)∙(1 - cos(2∙θ)))
    = (1/2)∙h∙(2 - ((1/a²) + (1/b²))∙r² - ((1/a²) - (1/b²))∙r²∙cos(2∙θ))

    Hence,
    ..... b.. 2∙π
    V = ∫... ∫ (1/2)∙h∙(2 - ((1/a²) + (1/b²))∙r² - ((1/a²) - (1/b²))∙r²∙cos(2∙θ))∙r dθdr
    ..... 0.. 0
    .. b
    = ∫ (1/2)∙h∙( [2 - ((1/a²) + (1/b²))∙r²]∙(2∙π - 0)- (1/2)∙((1/a²) - (1/b²))∙r²∙(sin(4∙π) - sin(0))∙r dr
    . 0
    .. b
    = ∫ (1/2)∙h∙(2 - ((1/a²) + (1/b²))∙r²)∙2∙π∙r dr
    . 0
    .. b
    = ∫ h∙π∙( 2∙r - ((1/a²) + (1/b²))∙r³ dr
    . 0
    = h∙π∙{ (b² - 0²) - (1/4)∙((1/a²) + (1/b²))∙(b⁴ - o⁴) }
    = h∙π∙( b² - (1/4)∙(b⁴/a²) - (1/4)∙b²
    = (1/4)∙h∙π∙b²∙(3 - (b²/a²))
    ?????
    right???
     
  16. May 22, 2012 #15

    tiny-tim

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    sorry, i'm not following what you're doing :redface:

    you seem to be calculating ∫∫ zr drdθ :confused:

    can you please explain in words how you're slicing the volume, and what you're integrating over?
     
  17. May 22, 2012 #16
    OK.Tim, this is what i learn from my textbook.
    If f(x,y)>=0 for all (x,y) in some region R, then z= f(x,y) represents a surface sitting above the xy-plane and over R. the double integral f(x,y) dxdy can then be interpreted as the colume of the solid under the surface z=f(x,y),over R, since the volume of a small strip sitting over deltA=delt(x)*delt(y) is approximately f(x,y)*deltA
     
  18. May 22, 2012 #17
    : ((
    i am confused as well, but this is what is written in my textbook-'course notes for Math2061", University of Sydney, school of mathematics and statistics.
    basically i think it is just using double integral to find the volume...
    Tim, does that make sense?
     
  19. May 22, 2012 #18
    oh about the r drdθ..
    that is deltA=r drdθ..
     
  20. May 22, 2012 #19

    tiny-tim

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    ah, now i see :smile:

    yes, that is a correct way of finding the volume …

    you slice the volume into square vertical columns of height z = f(x,y) and base dxdy (and therefore volume x dxdy), and then sum all the individual volumes

    so the volume is ∫∫ f(x,y) dxdy, = ∫∫ f(x,y)r drdθ ​

    but it's not the only way … you can slice the volume other ways, which may be easier

    in this case, since the previous parts of the question are all about the horizontal cross-sections,

    i'm guessing that they intended you to use horizontal slices, find the area of the intersection of that circle-and-ellipse (the circle part is easy, the ellipse part fairly easy), and then integrate that area over z

    (your solution looks ok, but i haven't checked right through it)
     
  21. May 22, 2012 #20
    thanks Tim, in part a) of this question, i found the semi-axes of this elliptic paraboloid are b*sqrt((h-z)/h) and a*sqrt((h-z)/h)
    therefore the intersection of the ellipse at height z is gonna be [a*sqrt((h-z)/h)]*[b*sqrt((h-z)/h) ]*pi, isn't? which is just pi*(h-z)/h*a*b.
    then how am i gonna use polar coordinates??? because a and b are constant numbers...

    : )) tim its 11pm in sydney now, time for bed...
    if u reply my post, i might not be able to read it tonight, but i will check tomorrow moring.. thanks for your help, have a good day!!
     
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