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Homework Help: Spherical Coordinates Question

  1. Oct 1, 2006 #1
    Hi, I am having trouble starting off this question. Could someone help me start off? Thanks in advance...

    Use spherical coordinates to evaluate
    ∫∫∫н (x² + y²) dV,
    where H is the hemispherical region that lies above the x-y plane and below the sphere x² + y² + z² = 1.
     
  2. jcsd
  3. Oct 1, 2006 #2

    HallsofIvy

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    In spherical coordinates, [itex]x=\rho cos(\theta)sin(\phi)[/itex], [itex]y= \rho sin(\theta)sin(\theta)[/itex], z= [itex]\rho cos(\phi)[/itex] so
    [itex]x^2+ y^2= \rho^2 sin^2(\theta)[/itex], [itex]dV= \rho^2 sin(\phi)d\rho d\theta d\phi[/itex], and the limits of integration are [itex]\rho[/itex] from 0 to 1, [itex]\theta[/itex] from 0 to [itex]2\pi[/itex], and [itex]\phi[/itex] from 0 to [itex]\pi/2[/itex].
     
    Last edited by a moderator: Oct 2, 2006
  4. Oct 1, 2006 #3
    Thanks HallsofIvy for the reply. So is this basically is this just finding a volume to the hemisphere that's below the x-y plane? Sorry with the questions, I"m new to this... :tongue2:
     
  5. Oct 2, 2006 #4

    HallsofIvy

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    No, it's not. For one thing, you said it was the portion of the sphere above the x-y plane, not below. More importantly, it's not finding the volume because that would be just [itex]\int \int \int dV[/itex], not [itex]\int \int \int (x^2+ y^2) dV[/itex]. You are integrating the function x2+ y2 over the unit hemisphere above the xy-plane.
     
  6. Oct 3, 2006 #5
    Clarification, from a casual observer

    Is sine supposed to be squared in dV? If not, why?

    Also, what's the difference between "evaluating a hemisphere" as stated in the original problem and "finding volume"? When I took VectorCalc, this kind of stuff used to stump me all the time.

    And lastly, how do you know what the limits are for each variable, e.g. 0 to 1, 0 to 2pi, etc?
     
  7. Oct 4, 2006 #6

    HallsofIvy

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    the sin2 was my error. I have edited it.

    The problem did not say "evaluate a hemisphere" (I don't know what that could mean). It said "evaluate the integral" on the hemisphere.
    Surely you know what it means to "evaluate an integral".
     
  8. Apr 18, 2008 #7
    since x=psin(thie)cos(theta) and y= psin(thie)sin(theta) squaring both quantanties would give you p^2sin^2(thie)cos^2(theta) + psin^2sin(thie)sin^2(theta) and using an identity give yous p^2sin^2(thie) + p^2sin^2(thie) since the theta's would equal one. Then adding these together this would give you 2p^2sin^2(thie) wouldn't it. then multiplying by p^2sin(thie) give you the thing you integrate first by rho, then theta, and finaly thie right. This is problem18 on p887 of James Stewart Calculus Concepts and Contexts 3
     
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