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Another coordinate conversions.

  1. Nov 14, 2009 #1
    1. The problem statement, all variables and given/known data

    The actual question is to evaluate the integral. All I need help on is the setting up part.

    Instead of making a thread for each, I will post 3 integral question with my attempts.

    Just tell me if you see something wrong from rectangular to spherical conversion. Thanks.

    1) Evaluate the integral
    [tex]\int\int\int_{E} e^\sqrt(x^2+y^2+z^2) dv[/tex]

    where E is enclosed by the sphere x^2+y^2+z^2 = 9 in the first octant


    This is what I got in spherical coordinate :

    [tex]\int^{\pi/2}_{0}[/tex][tex]\int^{\pi/2}_{0}[/tex][tex]\int^{3}_{0} e^\rho\rho^2 *d\rho*sin(\phi)d\phi*d\theta[/tex]



    Now for the seconds one :

    Evaluate the integral :

    [tex]\int\int\int_{E}x^2 dV[/tex]

    where E is bounded by the x-z plane and the hemisphere y = sqrt(9-x^2-z^2) and y = sqrt(16 - x^2 - z^2)

    Here is my integral setup in spherical coordinates :

    [tex]\int^{\pi}_{0}[/tex][tex]\int^{\pi/2}_{0}[/tex][tex]\int^{4}_{3} (p^2*sin^2(\phi)*cos^2(\theta) )* p^2d\phi * sin(\phi) d\phi * d\theta[/tex]

    And for the last one :

    Evaluate the integral by converting it into spherical coordinate :


    In rectangular coordinate :

    [tex]\int^{1}_{0}[/tex][tex]\int^{\sqrt(1-x^2}_{0}[/tex][tex]\int^{\sqrt(2-x^2+y^2)}_{\sqrt(x^2+y^2)} xz *dzdydx[/tex]


    Here is my partial attempt to convert it into spherical coordinate, but some are missing because I am not sure what it is supposed to be.

    [tex]\int^{2*pi}_{0}[/tex] [tex]\int^{?}_{?}[/tex] [tex]\int^{2}_{?} xz \rho^2 d\rho sin(\phi)d\phi d\theta[/tex]

    where x = p sin(phi) cos(theta) and z = p*cos(phi)

    Thanks for any help.
     
  2. jcsd
  3. Nov 15, 2009 #2
    If anyone could help on any of the 3 problems, checking if my conversion is correct, then
    I would really appreciate.
     
  4. Nov 16, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, that looks good.



    Since z includes both positive and negative values, [itex]\phi[/itex] will have to go from 0 to [itex]\pi[/itex].

    Since x ranges from 0 to 1, we are in the right half space. That means that [itex]\theta[/itex] goes from [itex]-\pi/2[/itex] to [itex]\pi/2[/itex], not 0 to [itex]2\pi[/itex]. For each x, y ranges from 0 to [itex]\sqrt{1-x^2}[/itex] or the circle [itex]x^2+ y^2= 1[/itex]. Finally, for each (x,y), z ranges between the cone [itex]z^2= x^2+ y^2[/itex] and the sphere [itex]z^2= 2- x^2-y^2[/itex] (or [itex]x^2+ y^2+ z^2= 2[/itex] which just happen to have the circle [itex]x^2+ y^2= 1[/itex], z= 1 as intersection. In spherical coordinates that cones is [itex]\rho^2cos^2(\phi)= \rho^2 sin^2(\phi)[/itex] or [itex]tan^2(\phi)= 1[/itex], [itex]\phi= \pi/4[/itex] and sphere is [itex]\rho^2 cos^2(\phi)= 2- \rho^2 sin^2(\phi)[/itex] or [itex]\rho^2= 2[/itex], [itex]\rho= \sqrt{2}[/itex]

    [itex]\rho[/itex] goes from 0 to [itex]\sqrt{2}[/itex] (the distance from (0,0,0) to each point on the circle of intersection), [itex]\theta[/itex] goes from [itex]-\pi/2[/itex] to [itex]\pi/2[/itex] and [itex]\phi[/itex] goes from 0 to [itex]\pi/4[/itex].
     
    Last edited: Nov 16, 2009
  5. Nov 16, 2009 #4
    There is no way to give reps or something similar, because you earned a million of them.
    Thanks a lot, as usual.
     
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