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Double integral of volume bounded by plane and paraboloid

  1. Sep 26, 2007 #1
    Evaluate the volume of the solid bounded by the plane z=x and the paraboloid z = x^2 + y^2


    I have tried to graph this, and they don't bound anything? have i graphed it wrong. and is there a way to do these problems where you don't need to draw the graph.
     
  2. jcsd
  3. Sep 27, 2007 #2

    HallsofIvy

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    They surely do bound something! If the 3-d graph bothers you assume y= 0. Then your graphs are z= x and z= x^2, which intersect at (0,0) and (0,1).
    Letting z= x in z= x^2+ y^2, you get x^2- x+ y^2= x^2- x+ 1/4 + y^2= 1/4 so
    (x-1/2)^2+ y^2= 1/4. In the xy-plane, that's a circle with center at (1/2, 0) and tangent to the y-axis. That's the projection of the actual intersection in the xy-plane.
     
    Last edited: Sep 27, 2007
  4. Sep 27, 2007 #3

    EnumaElish

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    Graphs can be very useful in this type of problem.

    z = x is like a ramp that runs the length of the y axis.

    What does the parabola equation describe? Can you visualize it?
     
  5. Sep 27, 2007 #4

    HallsofIvy

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    This was also posted in the homework forum. I'm merging it into that thread.
     
  6. Sep 27, 2007 #5
    am i able to change to polar coordinates with the shifted circle? or should i keep it in x y form

    let me know if this is correct:
    [tex]\int^{1/2}_{0}\int^{\sqrt{1/4-(x-1/2)^{2}}}_{-\sqrt{1/4-(x-1/2)^{2}}} x^{2} + y^{2} dy dx[/tex]

    I think i want to convert to polar form.. but i don't know what to do with the 1/4 inside the ( ) with the x when converting.
    also, is my double integral even correct?
     
  7. Sep 27, 2007 #6

    HallsofIvy

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    If it were me, I would convert to polar coordinates. If [itex](x- 1/2)^2+ y^2= 1/4[/itex], then [itex]x^2+ y^2- x+ 1/4= 1/4[/itex] or [itex]x^2+ y^2- x= 0[/itex] so, in polar coordinates [itex]r^2- rcos(\theta)= 0[/itex] and, finally, [itex]r= cos(\theta)[/itex]. Looks nice!
     
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