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Double integral to find volume between two surfaces

  1. Oct 30, 2015 #1
    1. The problem statement, all variables and given/known data
    set up and evaluate a double integral to find the volume of the solid bounded by the graphs of the equations
    y = 4 - x^2
    z= 4 - x^2
    first octant

    3. The attempt at a solution
    I am fairly confident in my ability to evaluate double integrals , but I am having a problem figuring out how to set this one up. In the example in my book they give two equations for z and equate them to find the region in the xy plane from this region they find the limits of integration.

    I think for this one I have to rewrite the first equation in terms of z?
    so z = f(x,y) = y + x^2 - 4 = 0
    and then I set this equal to the other equation for z

    y + x^2 -4 = 4 - x^2
    y -4 = 4 - 2x^2
    y = -2x^2 + 8

    I am not sure if anything up to here is correct and I don't know where to go from here. can someone give me a hint to help me get started ?
     
  2. jcsd
  3. Oct 30, 2015 #2

    LCKurtz

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    No, you are making it too complicated. In the xy plane plot the first quadrant portion of ##y=4-x^2##. That region is the base of your volume and you can find the xy limits there. The height of the solid is ##z=4-x^2##. Just do a double integral in xy.
     
  4. Oct 30, 2015 #3
    Oh I think I get it now. This is the integral I came up with:

    [itex] \int_0^4 \int_0^{(4-y)^\frac{1}{2}} (4-x^2) dxdy [/itex]
     
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