1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor

    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: 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

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?