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Volume Using Polar Coordinates

  1. Nov 15, 2008 #1
    1. The problem statement, all variables and given/known data
    Use polar coordinates to find the volume bounded by the paraboloids z=3x2+3y2 and z=4-x2-y2


    2. Relevant equations



    3. The attempt at a solution
    Somehow, through random guessing, I managed to get the right answer, it's just that I don't understand how I got it. Also, because the z is involved, I actually used cylindrical coordinates, but would that still be considered the same thing as polar coordinates? So anyway, I changed the two paraboloid equations to z=3r2 and z=4-r2. Then setting these two equations equal to each other (since they are both equal to z), I solved for r and got the limits of -1,1. For the limits of theta, I just happened to take it from 0 to 2pi. Lastly for the z limits, I just tried from 4-r2 to 3r2, so that gave me the equation:
    [tex]\int^{2\pi}_0\int^1_{-1}\int^{3r^2}_{4-r^2}dzrdrd\theta[/tex]
    However, solving this equation didn't give me the right answer, so I changed the limits of r to 0 to 1, and switched the z-limits around, so now it is:
    [tex]\int^{2\pi}_0\int^1_0\int^{4-r^2}_{3r^2}dzrdrd\theta[/tex]
    And solving for this, gave me the right answer of 2pi. The problem is that I don't understand the real logic behind what I did.
    So in summary, what I didn't understand was how to establish the limits for theta, r, and z.
     
    Last edited: Nov 15, 2008
  2. jcsd
  3. Nov 16, 2008 #2

    HallsofIvy

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    Those two paraboloids intersect when z= 3x2+ 3y2= 4- x2- y2 or 4x2+ 4y2= 4 which reduces to x2+ y2= 1. That projects onto the xy-plane as the unit circle. It shouldn't be hard to see that the volume you want is above that circle. To cover that circle, [itex]\theta[/itex] goes from 0 to [itex]2\pi[/itex] and r goes from 0 to 1. Those are your limits of integration.

    For each (x, y), the upper boundary is z= 4- x2- y2 and the lower boundary is z= 3x2+ 3y2. The volume of a "thin" rectangular solid used to construct the Riemann sums for this volume would be [(4- x2- y2)- (3x2+ 3y2)]dxdy= (4- 4x2- 4y2)dxdy= 4(1- x2- y2)dxdy and that, in polar coordinates, is 4(1- r2)rdrd[itex]\theta[/itex].

    Yes, "cylindrical coordinates" is just polar coordinates in the xy-plane.

    No, you didn't. Or you shouldn't. r cannot be negative. r must be between 0 and 1.

    Did you think at all about the geometry of the situation? This has circular symmetry. To cover the entire volume you have to cover the entire circle: 0 to [itex]2\pi[/itex].

    Surely you recognized that z= 4- r2 is ABOVE z= 3r2 for r<= 1?

     
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