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Double integration

  1. Feb 26, 2005 #1
    could someone please check my work on the following problem:

    Find the volume under [itex]z= 2x + y^2[/itex] and above the region bounded by [itex]x= y^2[/itex] and [itex]x= y^3[/itex].

    endpoints of region: (0,0) and (1,1)
    [tex] V = \int_{0}^{1}\int_{0}^{1} (2x + y^2) \,dx\,dy = \int_{0}^{1} [x^2 + xy^2]|_{0}^{1}\,dy = (x^2 y + \frac{1}{3}xy^2)|_{(0,0)}^{(1,1)} = \frac{4}{3} [/tex]
     
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  3. Feb 26, 2005 #2

    dextercioby

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    Shouldn't the points be in 3D...?And the integral be a triple one...?

    Daniel.
     
  4. Feb 26, 2005 #3
    you set your limits point incorrectly, they do not only depend on the end points, (0,0) and (1,1), but also depend on the curve x=y^2 and x=y^3

     
  5. Feb 26, 2005 #4

    dextercioby

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    Shouldn't the volume of a domain D from R^{3} have the following form:
    [tex] V_{D}=\iiint_{D} dx \ dy \ dz [/tex]

    Daniel.
     
    Last edited: Feb 27, 2005
  6. Feb 26, 2005 #5
    So it should be:
    [tex]V = \int_{0}^{1}\int_{y^3}^{y^2} (2x + y^2) \,dx\,dy = \int_{0}^{1} [x^2 + xy^2]|_{y^3}^{y^2}\,dy = \int_{0}^{1} [(y^4 - y^4) - (y^6 - y^5)]\,dy = \int_{0}^{1} y^5 - y^6 \,dy = [\frac{1}{6}y^6 - \frac{1}{7}y^7]|_{0}^{1} = \fract{1}{6} - \frac{1}{7} = \frac{1}{42} ?[/tex]

    Somewhat confused.
     
  7. Feb 27, 2005 #6

    dextercioby

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    Are u sure that the values of the square (y²) are less than the values of the cube (y³) ...?If so,then the first integral is correct.Unfortunately,the 3-rd integral is incorrect,and hence everything that follows.

    Daniel.

    P.S.Pay attention to the signs.
     
  8. Feb 27, 2005 #7

    dextercioby

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    What...? :surprised :yuck: What's your definition for "ABOVE"...?

    To the OP:disregard Marlon's post.Your integration (the one you started with was good,the arithmetics that followed was sloppy).

    Daniel.
     
  9. Feb 27, 2005 #8
    [tex]V = \int_{0}^{1}\int_{y^3}^{y^2} (2x + y^2) \,dx\,dy [/tex]

    this is what it should be ... I was wrong. Sorry bout that. I mixed the two functions

    marlon
     
  10. Mar 1, 2005 #9
    okay i fixed the math in
    [tex]V = \int_{0}^{1}\int_{y^3}^{y^2} (2x + y^2) \,dx\,dy[/tex]

    and got 4/3. I think this is right.
     
  11. Mar 1, 2005 #10

    dextercioby

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    Nope,My Maple says it's wrong.It should come down to [tex] \frac{19}{210} [/tex]

    Daniel.
     
  12. Mar 1, 2005 #11

    Galileo

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    Check this step again.
     
  13. Mar 1, 2005 #12
    OMG stupid arithmetic

    [tex] \int_{0}^{1} [x^2 + xy^2]|_{y^3}^{y^2}\,dy = \int_{0}^{1} [(y^4 + y^4) - (y^6 + y^5)]\,dy = (\frac{2}{5}y^5 - \frac{1}{7}y^7 - \frac{1}{6}y^6)|_{0}^{1} = \frac{19}{210} [/tex]

    Finally. Thanks guys.
     
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