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Math Help - Integration first converting to spherical polars

  1. #1
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    Integration first converting to spherical polars

    I'm trying to work through a question, but I cannot get my result to look anything like the model answers.

    The question is to integrate

    z/((x^2+y^2)^(1/2)) over the volume of a region bounded by the surfaces z=0 and z=(9-x^2-y^2)^(1/2).

    I know I need to use spherical polar co-ordinates to do this, but I can't seem to get the substitutions right.

    The model answer says that it becomes r^2*cos(theta) integrated over the obvious region, but my attempt to get it is as such:

    z = r*cos(theta)
    r = (x^2+y^2)^(1/2) (in 2 dimensions, ignoring z)

    So,

    the integral is

    integral( (r*cos(theta) / r) * r^2 * sin(theta) drd(theta)d(phi) )

    Of course this gives me r^2*cos(theta)*sin(theta).

    What am I missing?
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  2. #2
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    Quote Originally Posted by StarWrecker View Post
    I'm trying to work through a question, but I cannot get my result to look anything like the model answers.

    The question is to integrate

    z/((x^2+y^2)^(1/2)) over the volume of a region bounded by the surfaces z=0 and z=(9-x^2-y^2)^(1/2).

    I know I need to use spherical polar co-ordinates to do this, but I can't seem to get the substitutions right.

    The model answer says that it becomes r^2*cos(theta) integrated over the obvious region, but my attempt to get it is as such:

    z = r*cos(theta)
    r = (x^2+y^2)^(1/2) (in 2 dimensions, ignoring z)

    So,

    the integral is

    integral( (r*cos(theta) / r) * r^2 * sin(theta) drd(theta)d(phi) )

    Of course this gives me r^2*cos(theta)*sin(theta).

    What am I missing?
    Why did you choose to ignore z?

    In spherical coordinates  x = r \cos(\varphi) \sin (\theta) , and  y = r \sin(\varphi) \sin(\theta) . You should use this definition to calculate  (x^2 + y^2)^{\frac{1}{2}}, and you should get  (x^2 + y^2)^{\frac{1}{2}} = r \sin(\theta)

    Giving you an integrand of  \frac{r \cos(\theta)}{r \sin(\theta} r^2 \sin(\theta) = r^2 \cos(\theta)
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  3. #3
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    That's one reason why it is more common to use " \rho" as the radial variable in spherical coordinates (and they are NOT normally referred to as "spherical polar coordinates), to avoid confusing 3d and 2d coordinates.

    In spherical coordinates x= \rho cos(\phi) sin(\theta) and y= \rho sin(\phi) sin(\theta). Then x^2+ y^2= \rho^2 cos^2(\phi)sin^2(\theta)+ \rho^2 sin^2(\phi)sin^2(\theta) = \rho^2 sin^2(\theta)(cos^2(\phi)+ sin^2(\phi))= \rho^2 sin^2(\phi) so that (x^2+ y^2)^{1/2}= \rho sin(\phi)
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