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Triple Integral restricted to first octant.

  1. Nov 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Evaluate the triple integral of function 14xz bounded between z=y^2 and z=8-2x^2-y^2 in the first octant.


    3. The attempt at a solution
    So the first octant would mean the bottom parameter on all my integral will be zero since (x,y,z)>0. Then I set the equations equal to each other and got 4=x^2+y^2. It appears they intersect in a circle of radius two. I'm having a bit of trouble applying this knowledge to set up the limits of integration. Do I integrate with respect to Z first between 0 and x^2+y^2 and then evaluate dy and dx between zero and tw0?
     
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  3. Nov 2, 2013 #2

    vanhees71

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    Setting the two equations for [itex]z[/itex] equal is good. This gives you the projection of the curve, where the two surfaces intersect, to the [itex]xy[/itex] plane. You find that this is a circle around the origin. Thus, since in the first octand [itex]x,y[/itex] must run over the corresponding quarter-circle, and [itex]z[/itex] between the given boundaries.

    Hint: It helps to draw the boundary surfaces to get a picture. Mathematica does a nice job, but I guess you can also do that with a free program like gnuplot.
     
  4. Nov 2, 2013 #3
    okay cool, so I did dzdxdy, dz was bounded by y^2 at the bottom and 8-2x^2-2y^2 at the top, and dy and dx were both from 0 to 2. Is that the correct way to set this up? If it is, screw me because I did the nasty integral by hand and rechecked my work a few times and got the same answer yet my online homework program is saying it's wrong
     
  5. Nov 2, 2013 #4

    LCKurtz

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    You have determined that the xy region is a circle of radius two. When you let both x and y go from 0 to 2, aren't you describing a square? You need to fix that.
     
  6. Nov 2, 2013 #5
    what if I make x go from 0 to (4-y^2)^(1/2) and then y go from 0 to 2? is that right?
     
  7. Nov 2, 2013 #6

    LCKurtz

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    Yes. I'm assuming you have the correct integrand, but I don't know since you never showed your integral. You might find it easier to calculate using polar coordinates.
     
  8. Nov 2, 2013 #7
    is there a way to make wolfram alpha evaluate this X_X seems kind of nasty to do by hand... i'm integrating (14xy)dzdxdy
     
  9. Nov 3, 2013 #8
    in polar coordinates, would dz go from (rsin(theta))^2 to 8-2r^2? and d(theta) go from 0 to pi/2? and dr go from 0 to 8?
     
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