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Calc. Double Integrals: Setting Limits

  1. Apr 2, 2005 #1
    Here is the question:

    Let D be the region bounded by the y axis and the parabola x = -4y^2 + 3. Compute 'double integral: (x^3)y dxdy'

    I'm having a hard time setting limits to the double integral. I got y = [-sqrt(3/4), +sqrt(3/4)] for one of the limits, but I don't know how to set the x limits for integration. Also, can someone show me an easy way to quickly find the limits of integration. Thanks.
     
  2. jcsd
  3. Apr 2, 2005 #2
    For double integrals you always want the limits of the outer integral to be constants. Occasionally both integrals will have constant limits.

    [tex] \int \int_D x^3y dx dy[/tex]

    The region D is contained by x = 0 and x = 3-4y^2. Since your outer integral is dy, your y limits will be constant, and x limits will be variable. Your limits alon the x axis are given by x = 0, and x = 3-4y^2, and:

    [tex] \int \int_0^{3-4y^2} x^3y dx dy [/tex]

    Plot the graph of your domain to determine the range of y. You will see that the maximum y value occurs on the y axis, at x = 0.

    Solving 0 = 3-4y^2, you find y = +/- sqrt(3/4).


    The easiest way to find limits of integration is to sketch the region of integration.

    The final integral becomes [tex] \int_{-\sqrt{3/4}}^{\sqrt{3/4}} \int_0^{3-4y^2} x^3y dx dy [/tex]
     
    Last edited: Apr 2, 2005
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