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Multiple Integrals

  1. Apr 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi, I've been trying this for days now and I really can't get it, so would appreciate some help please!



    Find the volume of the finite region between the two surfaces z=x^2 + 4y^2 and z=2x + 8y + 4


    2. Relevant equations



    3. The attempt at a solution


    I tried to do a double integral of (x^2 + 4y^2 - 2x - 8y - 4)dxdy. I thought the bounds of this on the x-y plane would be given by x^2 + 4y^2=2x + 8y + 4. However doing this gives a nasty quadratic for x in terms of y that makes things impossible. I'm also pretty sure I need to sub in something of the form x=a*r*cosP, y=b*r*sinP. The Jacobian for this would be abr.



    I got thie bound to be between x=1-[5+8y - 4y^2] and 1 + [5+8y - 4y^2] and y = -0.5 and 5/2



    I think this is wrong!



    I tried subbing in the sin and cos things also, but don't know how to work out the bounds.



    Please help!
     
  2. jcsd
  3. Apr 13, 2008 #2

    benorin

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    Think, what is the shadow of z=x^2 + 4y^2 in the xy-plane?
     
  4. Apr 14, 2008 #3

    tiny-tim

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    Hi joker_900! :smile:

    (btw, feel free to copy ² and anything else you like for future use. :smile:)

    Now, this isn't the best way of doing it … but when ever I get confused, I try to simplify the problem as much as possible … it may take a little longer … but that's better than being completely confused! :redface:

    I would make the substitution X = x, Y = 2y.

    Then we get a nice symmetric paraboloid, z = X² + Y², intersecting the plane z = 2X + 4Y + 4.

    And then I'd simplify again by putting X´ = X + 2Y, Y´= 2X - Y (so they're perpendicular, and we have 5z = X´² + Y´² intersecting z = 2X´ + 4).

    Does that help? :smile:
     
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