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Triple Integral ( IS THIS RIGHT?)

  1. Nov 8, 2009 #1

    een

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    Triple Integral ( IS THIS RIGHT??)

    1. The problem statement, all variables and given/known data
    Let R be the solid enclosed by the planes x=0, y=0, z=2 and the surface x^2+y^2, where
    x[tex]\geq[/tex]0, y[tex]\geq[/tex]0. Compute[tex]\int\int\int[/tex]xdxdydz

    2. Relevant equations



    3. The attempt at a solution
    I did [tex]\int[/tex]0-1[tex]\int[/tex]0-1[tex]\int[/tex](x^2+y^2)-2 xdzdydx

    [tex]\int[/tex]0-1[tex]\int[/tex]0-12x-x^3-xy^2 dydx
    [tex]\int[/tex]0-1[2xy-x^3y-xy^3/3]0-1 dx
    [tex]\int[/tex]0-12x-x^3-x/3dx = 1-1/4-1/6 = 7/12
    Can someone tell me if this is the correct answer
     
  2. jcsd
  3. Nov 8, 2009 #2

    lanedance

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    Homework Helper

    Re: Triple Integral ( IS THIS RIGHT??)

    pretty hard to read, below might help your formatting...
    [tex]\int_{0}^1dx\int_{0}^1dy[/tex]
     
  4. Nov 9, 2009 #3
    Re: Triple Integral ( IS THIS RIGHT??)

    you got x and y's upper limits wrong.
    you are calculating the volume between these two surfaces (z(x,y) = x2 + y2, z < 2), see attachment.
     

    Attached Files:

    Last edited: Nov 9, 2009
  5. Nov 9, 2009 #4

    een

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    Re: Triple Integral ( IS THIS RIGHT??)

    okay.. this is what i got...
    the lower limit of x is 0 the upper limit is sqrt(2)
    same for y
    the lower limit of z is x^2+y^2 and the upper limit is 2...
    is that right?
     
  6. Nov 10, 2009 #5
    Re: Triple Integral ( IS THIS RIGHT??)

    yes, my result is 4/3.
     
  7. Nov 10, 2009 #6

    een

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    Re: Triple Integral ( IS THIS RIGHT??)

    i am getting sqrt(2)/3... what am i doing wrong?... i used the limits that you told me and im pretty sure that i integrated correctly... can you show me step by step
     
  8. Nov 11, 2009 #7

    Mark44

    Staff: Mentor

    Re: Triple Integral ( IS THIS RIGHT??)

    I have switched the order of integration, leaving the integrand unchanged. I think these are the correct limits of integration.
    [tex]\int_{x = 0}^{\sqrt{2}} \int_{y = 0}^{\sqrt{2 - x^2}} \int_{z = x^2 + y^2}^{2} x~ dz dy dx [/tex]

    The projection of this region onto the x-y plane is the disk x^2 + y^2 <= 2, a disk of radius sqrt(2) centered at the origin. Geometrically, we have a stack of blocks dx by dy that extend from the paraboloid up to the plane z = 2. Then we let the stack sweep out along the y-axis from y = 0 to y = sqrt(2 - x^2) (i.e., out to the boundary of the disk), then finally let this wall of blocks sweep from x = 0 to x = sqrt(2).
     
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