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Fux integral

  1. Apr 29, 2006 #1
    Surface area integral

    sorry, this is not about flux integration... but surface area! sorry about the title!

    Find the surface area of the part of the sphere [tex]x^2+y^2+z^2=36 [/tex] that lies above the cone [tex]z=\sqrt{x^2+y^2}[/tex]

    [tex]z=\sqrt{36-x^2-y^2} [/tex]

    [tex]A(S)=\int\int_D \sqrt{1+\left( \frac{\partial z}{\partial x} \right) ^2 + \left( \frac{\partial z}{\partial y} \right) ^2 } dA[/tex]

    [tex] \frac{\partial z}{\partial x} =\frac{1}{2}\left( 36-x^2-y^2 \right) ^{\frac{3}{2}} \left( -2x \right) [/tex]

    [tex] \frac{\partial z}{\partial y}=\frac{1}{2}\left( 36-x^2-y^2 \right) ^{\frac{3}{2}} \left( -2y \right) [/tex]

    [tex]A(S)=\int\int_D \sqrt{1+\left( \frac{1}{2}\left( 36-x^2-y^2 \right) ^{\frac{3}{2}} \left( -2x \right) \right) ^2 + \left( \frac{1}{2}\left( 36-x^2-y^2 \right) ^{\frac{3}{2}} \left( -2y \right) \right) ^2 } dA[/tex]

    is this correct so far? how would I find the ends of integration for a cone?

    I've taken the liberty of changing the title of this thread. Since it wasn't even "flux integration" it was bothering me!
    Last edited by a moderator: May 1, 2006
  2. jcsd
  3. Apr 29, 2006 #2
    It would be much easier if you convert to polar coordinates



    and [tex]z=r[/tex]

    Remember to use the Jacobian when you're changing coordinates.
  4. Apr 29, 2006 #3
    oh wow, okay, i got it in polar form...

    [tex]\int \int \sqrt{1+r^2 (36-r)^3}[/tex]

    [tex]\int \int \sqrt{1+46656x^2-3888x^4+108x^6-x^8}[/tex]

    but what would be the ends of integration for a cone?
    Last edited: Apr 29, 2006
  5. Apr 29, 2006 #4


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    Science Advisor

    First find the intersection between the cone and the sphere. This will give you a condition that you can turn into your limits of integration.
  6. Apr 29, 2006 #5
    how would I do that?

    here's my guess:

    the cone has equation:

    the sphere has equation:
    [tex]x^2+y^2+z^2=36 [/tex]

    putting them together:
    [tex]x^2+y^2=18 [/tex]

    so my ends would be:

    [tex]\int _0 ^{2\pi} \int_0 ^{\sqrt{18}[/tex]

    also, when i change from [tex]A(S)=\int\int_D \sqrt{1+\left( \frac{1}{2}\left( 36-x^2-y^2 \right) ^{\frac{3}{2}} \left( -2x \right) \right) ^2 + \left( \frac{1}{2}\left( 36-x^2-y^2 \right) ^{\frac{3}{2}} \left( -2y \right) \right) ^2 } dxdy[/tex] to the polar form [tex]\int \int \sqrt{1+r^2 (36-r)^3}dA[/tex] do I have to add the variable r at the end such that [tex]\int \int \sqrt{1+r^2 (36-r)^3}rdrd \theta[/tex]?
    Last edited: Apr 29, 2006
  7. May 1, 2006 #6


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    Homework Helper
    Gold Member

    It should be
    [tex]\int \int \sqrt{1+r^2 (36-r^2)^3} rdrd \theta[/tex]

    Apart from that, I think everything else is correct.
  8. Oct 28, 2007 #7
    how the heck do you integrate that?
  9. Oct 28, 2007 #8
    may I suggest this?
    A(S) = \int\int_S dA = \int_u \int_v \left| { \frac{\partial \mathop r\limits^ \to}{\partial u} \times \frac{\partial \mathop r\limits^ \to}{\partial v} } \right| du dv
    now considering spherical coordinates, and representing the integral surface with respect to spherical coordinates
    S : x = 6 \sin \theta \cos \phi , y = 6 \sin \theta \sin \phi , z = 6 \cos \theta
    [tex] 0 \leq \theta \leq \frac{\pi}{4}, 0 \leq \phi \leq 2 \pi [/tex]
    A(S) = \int_u \int_v \left| { \frac{\partial \mathop r\limits^ \to}{\partial u} \times \frac{\partial \mathop r\limits^ \to}{\partial v} } \right| du dv
    = \int_\theta \int_\phi \left| { \frac{\partial \mathop r\limits^ \to}{\partial \theta} \times \frac{\partial \mathop r\limits^ \to}{\partial \phi} } \right| d \theta d \phi
    = \int_\theta \int_\phi 36 \left| { \sin \theta } \right| d \theta d \phi
    = 36 \int_\theta \left| { \sin \theta } \right| d \theta \int_\phi d \phi
    =72 \pi \left( { 1 - \frac{1}{\sqrt{2} } \right)
  10. Oct 28, 2007 #9
    What is that technique called?
    haven't seen that before
  11. Oct 28, 2007 #10
    Multiple integrals, usually taught in a third calculus course. Typically comes with a few conversion ideas, like polar coordinates in this example. Each integral and its limits correspond to a variable being integrated.
  12. Oct 28, 2007 #11
    cool, I guess we'll learn that in my next course
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