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Integrating with 2 variables

  1. Feb 8, 2008 #1
    I was trying my hand at integrals with 2 variables... So.. my first excercise was to find out the surface area of a sphere. So, the co-ordinate system is something like this:

    i] The whole co-ordinate system is mapped on the surface of a sphere.
    ii] The x-axis of the sphere is like the equator of the earth.
    iii] The y-axis of the sphere is like the prime meredian of the earth.

    So, for the area, I used:

    [tex]
    A = \int^{2\pi r}_{0}dx\int^{2\pi r}_{0}dy
    [/tex]

    giving me the area:

    [tex]
    A = 4 \pi^2 r^2
    [/tex]

    Which is ofcourse wrong.. because if i look from the first equation, it is more like i'm calculating the area of a square lamina having lengths [itex]2 \pi r[/itex] each.
     
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  3. Feb 8, 2008 #2

    Hootenanny

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    And that is exactly what you are doing. What you need to do, is figure out the equation of the projection of a sphere on the xy plane (imagine slicing a ball in half) and evaluate the double integral over that region (with correct limits).
     
    Last edited: Feb 8, 2008
  4. Feb 8, 2008 #3

    mathman

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    The angle around the equator goes from 0 to 2pi. However, the polar angle goes from -pi/2 to pi/2 and even more important its differential is cosydy, not dy.
     
  5. Feb 8, 2008 #4

    HallsofIvy

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    In general, the integral
    [tex]\int_a^b\int_c^d du dv[/tex]
    is the area of a rectangle in uv-space, not the surface of a sphere. For a problem like that, it is not enough to know how to do multiple integrals. You also have to know how to find the "differential of area" for a curved surface. The differential of area is "du dv" only for u and v being orthogonal coordinates in a plane.
     
  6. Feb 8, 2008 #5
    Thanks for replying. Can you please ellaborate on how the differential is cosy dy?
     
  7. Feb 9, 2008 #6
    Here's a way to do it.
    Lets take this surface and break it down.
    First we require a parametric equation for a sphere--> [tex]a^2=x^2+z^2+y^2[/tex]
    a is the radius of the sphere.
    We can then turn this into a vector equation representing the parametric surface
    [tex]\vec{r}=<x,y,(+/-)\sqrt(x^2+y^2-a^2)>[/tex]
    We can then take partial derivatives with respect to x and y.
    Then take the cross products of x with y, then the magnitude.
    Basically:

    [tex]\int\int|\vec{r}_x X\vec{r}_y|dxdy[/tex]
    Change to polar for convenience.
     
  8. Feb 9, 2008 #7

    HallsofIvy

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    Or, since this is a sphere, start from spherical coordinates:
    [tex]x= \rho cos(\theta) sin(\phi)[/tex]
    [tex]y= \rho sin(\theta) cos(\phi)[/tex]
    [tex]z= \rho cos(\phi)[/itex]

    On the surface of the sphere of radius R, [itex]\rho= R[/itex] so we have
    [tex]x= R cos(\theta) sin(\phi)[/tex]
    [tex]y= R sin(\theta) cos(\phi)[/tex]
    [tex]z= R cos(\phi)[/itex]
    in terms of the two parameters [itex]\theta[/itex] and [itex]\phi[/itex].

    Now, [itex]\vec{r}= R cos(\theta)sin(\phi)\vec{i}+ Rsin(\theta)\sin(\phi)\vec{j}+ R cos(\phi)\vec{k}[/itex]

    The cross product [itex]\left|\vec{r_\theta} \times \vec{r_\phi}\right|[/itex] is called the "fundamental vector product" of the surface and the "differential of surface area" is
    [itex]\left|\vec{r_\theta} \times \vec{r_\phi}\right|d\theta d\phi[/itex]
     
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