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Triple Integral in Rectangular Coordinates Converting to Spherical Coordinates

  1. Dec 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Given that:

    24o1zib.jpg

    Write an equivalent integral in spherical coordinates.

    2. Relevant equations

    bjj38n.jpg

    (Triple integral in spherical coordinates.)

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    2m7gn4g.gif
    2ik5iiw.gif
    27yu2qw.gif

    (Conversions from rectangular to spherical coordinates.)


    (What spherical coordinates entail)

    3. The attempt at a solution

    The region is going to be the region of the lower part of a sphere with radius a moved up a units on the z-axis that lies in the -x and -y plane. The limits of integration for theta should be 0 to pi/2. Need assistance with limits for phi and rho.
     
    Last edited by a moderator: Apr 18, 2017
  2. jcsd
  3. Dec 5, 2009 #2

    HallsofIvy

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    y ranges from -a to 0 so we are talking about the lower half plane in the xy- plane.

    For each y, x ranges from [itex]-\sqrt{a^2- y^2}[/itex] to 0. [itex]x= -\sqrt{a^2- y^2}[/itex] gives [itex]x^2+ y^2= a^2[/itex], the circle with center (0,0) and radius a. Since x is the negative square root and we already know that we are in the lower half plane (y< 0), we are looking at the portion of that circle in the third quadrant.

    For each (x,y), z ranges from [itex]a- \sqrt{a^2- x^2- y^2}[/itex] to a. [itex]z= a- \sqrt{a^2- x^2- y^2}[/itex] gives [itex]x^2+ y^2+ (z-a)^2= a^2[/itex], the sphere with center at (0,0,a) and radius a. Since z goes up to a, we are in the lower half sphere. That is, we are looking at the part of the sphere [itex]x^2+ y^2+ (z-a)^2= a^2[/itex] where x and y are negative and z< a.

    I would be inclined to introduce new coordinates: x'= x, y'= y, z'= z- a. In those coordinates, the region we are integrating over is the "seventh octant" in which x, y, and z are all negative. To get that sphere, let [itex]\rho[/itex] go from 0 to a, [itex]\theta[/itex] from [itex]\pi[/itex] to [itex]3\pi/2[/itex], and [itex]\phi[/itex] from [itex]\pi/2[/itex] to [itex]\pi[/itex]. The integrand is still "1" and the "differential of volume" is still [itex]\rho sin^2(\phi)d\phi d\theta d\rho[/itex]. In fact, because the integrand is "1", this integral is just the volume of the region: 1/8 of the volume of a sphere of radius a and so is [itex](1/8)(4/3)\pi a^3= (1/6)\pi a^3[/itex]. You can use that as a check.
     
  4. Dec 5, 2009 #3
    Shouldn't we be integrating the region of the sphere in the -x, -y, +z plane, since z goes from the bottom of the sphere, moved up a on the z-axis, to the plane z=a?
     
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