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Triple Integral question

  1. Apr 22, 2012 #1
    I need some help in this

    The question states find the volume for the region bounded below by z = 3 - 2y and above by z = x^2 + y^2 using cylindrical coordinates .

    Now i tried to do it but i got stuck with the part where i find the radius . i found
    r^2 = 3-2r*sin(zeta) :( its hard to factorize and use this for the boundary

    i already know that z will range from r^2 to 3-2r*sin(zeta) and zeta range from zero to 2 pie

    so how can this question be solved ??? :)

    btw the correct answer is 8pie
     
  2. jcsd
  3. Apr 22, 2012 #2

    HallsofIvy

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    First, you have "below" and "above" reversed. The plane z= 3- 2y is above the paraboloid [itex]z= x^2+ y^2[/itex] in the bounded region.

    They intersect where [itex]z= x^2+ y^2= 3- 3y[/itex] which is equivalent to [itex]x^2+ y^2+ 3y= 3[/itex]. [itex]x^2+ y^2+ 3y+ 9/4= x^2+ (y+3/2)^2= 3+ 9/4= 21/4[/itex]. That is a circle with center at (0, -3/2) which is why ordinary polar coordinates do not give a simple equation. Either
    1: integrate with x from [itex]-\sqrt{21}/2[/itex] to [itex]\sqrt{21}/2[/itex] and, for each x, y from [itex]-3/2- \sqrt{21/4- x^2}[/itex] to [itex]-3/2+ \sqrt{21/4- x^2}[/itex].

    2: Let [itex]x= rcos(\theta)[/itex] and [itex]y= -3/2+ rsin(\theta)[/itex], shifting the origin to (0, -3/2).
     
  4. Apr 22, 2012 #3
    Oh i c :D !!! Thanks alot . I wasn't aware of shifting the origin :) !!

    Thanks :D !
     
  5. Apr 23, 2012 #4
    LOL 8pie
     
  6. Apr 23, 2012 #5

    HallsofIvy

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    For desert?
     
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