1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    (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


    User Avatar
    Science Advisor
    Homework Helper

    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
    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:
    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))

    ..... 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))
  16. May 22, 2012 #15


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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!!
  22. May 22, 2012 #21


    User Avatar
    Science Advisor
    Homework Helper

    you need to find the four values of θ where the circle and ellipse meet

    over two sections, that's just the area of a sector of a circle, 1/2 r21 - θ2)

    over the other two sections, it's the area of a sector of an ellipse, which is … ? :wink:

    sleep tight! :zzz:​
  23. May 22, 2012 #22
    I'm confused about how you got the area?

    PS I'm doing this same course and assignment too
  24. May 22, 2012 #23


    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi TDR14! Welcome to PF! :smile:
    There's various ways of doing it, but the simplest is to note that an ellipse is a squashed circle, by a factor b/a along the minor axis,

    so the area is πa2*b/a, = πab :wink:

    (technically, one proves this with the substitution y' = ay/b o:))
  25. May 22, 2012 #24
    Re: Welcome to PF!

    ok, I know that.

    But how do you put it in terms of x,y,z etc?

    I got A=∏*(x/sqrt((h-z)/h))*(y/sqrt((h-z)/h))

    Then I just follow chris_usyd's way of integrating and so on and so forth???
  26. May 23, 2012 #25


    User Avatar
    Science Advisor
    Homework Helper

    i don't understand :confused:

    A does not depend on x or y, only on z, see chris's formula …
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook