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Surface area of between 2 cones

  1. Jul 12, 2014 #1
    find the area of the cylinder x^2+z^2=a^2 that is inside the cylinder x^2+y^2=a^2.

    my attempt:
    parameterise x^2+z^2=a^2 as a vector r(x,y) = (x,y,(a^2-x^2)^1/2).
    using the formula given here : http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceArea.aspx, I found the surface area = the double integral of a(a^2-x^2)^-1/2 dxdy, over the circle x^2+y^2=a^2 on the x-y plane.
    change variables into polar coordinates, so we obtain the double integral of a(a^2-(r cos (u))^2)^-1/2 rdrdu, for 0<r<a, 0<u<2pi.
    Solving this i get zero, which doesnt seem right.

    where did i go wrong?
     
    Last edited: Jul 12, 2014
  2. jcsd
  3. Jul 12, 2014 #2

    HallsofIvy

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    This is wrong. The area of one cylinder inside the other does NOT cover that circle.

     
  4. Jul 12, 2014 #3
    thanks, HallsofIvy. but i dont see why it isnt integrated over a circle, since x^2+y^2=a^2 is cylinder centred at origin, surely if the other cylinder goes through it, the cross section (on the x-y plane) must be a circle?
     
  5. Jul 12, 2014 #4

    LCKurtz

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    I just deleted a post as soon as I made it as I'm not sure it was correct. More later.
     
  6. Jul 12, 2014 #5

    LCKurtz

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    OK, I'm back with a little more time. Sandylam966, I think you are correct about the domain being a circular disk and your integral is set up correctly. But there is no way it should work out to be zero because the integrand is positive on the region. So you obviously have something wrong in your working of the integral.

    Also, I'm wondering where this problem came from, because the integral you have apparently is not a simple integral. Maple gives an answer in terms of elliptic functions.
     
  7. Jul 12, 2014 #6
    thanks, LCKurtz.

    here's my working:
    the surface area A = the double integral of a(a^2-x^2)^-1/2 dxdy, over the circle x^2+y^2=a^2 on the x-y plane,
    so A = double integral of a(a^2-(r cos (u))^2)^-1/2 rdrdu, for 0<r<a, 0<u<2pi
    first we solve the dr part, so
    A = integral (wrt to u only now) of a[(-1/(cos u)^2)(a^2-(r cos (u))^2)^1/2] du, the integrand evaluated for r from 0 to a, then
    A = integral of a(-1/(cos u)^2)[(a^2(1-cos (u))^2)^1/2 - a] du
    = integral of a(-1/(cos u)^2)(a sin u -a) du
    = integral of a^2 ((-tan u)(sec u) + sec u)^2) du
    = a^2 [-sec u + tan u] evaulated for u from 0 to 2pi, which is clearly zero.

    so i did something wrong here right?
     
  8. Jul 13, 2014 #7
    I am having some trouble understanding this thread. Are we not trying to find the surface area of the shape that comes about from two cylinders intersecting at right angles?
     
    Last edited: Jul 13, 2014
  9. Jul 13, 2014 #8
    hi Quesadilla, yes thats what i'm trying to do, but i think i made some mistake in my working above.
     
  10. Jul 13, 2014 #9
    I see no mistakes, BUT a(a^2-(r cos (u))^2)^-1/2 r is never negative in your domain, so if you get zero the mistake is in the computation of the integral which you did not give us.

    EDIT: sorry did not see post #6 / sorry LCKurtz did not see your post... time to wake up
     
  11. Jul 13, 2014 #10
    (1-cos(u)^2)^1/2=|sin(u)|, not sin(u), then two times the integral from -pi/2, pi/2
     
  12. Jul 13, 2014 #11
    oh I see thanks!
     
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