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Multiple Integration Limits

  1. Mar 24, 2009 #1
    I'm throwing in the towel - I need some explanation help / clarification -

    I'm brushing up on my polar / cylindrical coordinates and integration with polar / cylindrical coordinates.

    In working through some examples - the book seems to write the limits as if it is straightforward or I should already know how to generate the limits - it just states them and moves on. (I can do the integration once I have the limits.)

    With working backwards from the limits given I agree with the limits.

    For example - an example states, "calculate the volume of the ellipsoid
    [tex] 4x^2 + 4y^2 + z^2 = 4 [/tex]

    The limits given are:

    [tex] 0<= z <= 2\sqrt{(1-x^2 - y^2)} [/tex]

    I agree with this: solve for z

    next

    [tex] -\sqrt{(1-x^2)} <= y <= \sqrt{(1-x^2 )} [/tex]

    Not as sure about this - do we set z= 0, do we ignore z, if so why?

    [tex] -1 <= x <= 1 [/tex]

    Just like the limits on y, do I ignore z and y or set them equal to 0 , if so why?

    Can you explain how to acquire these limits?

    Thanks
    Sparky
     
  2. jcsd
  3. Mar 24, 2009 #2

    HallsofIvy

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    Because, having integrated with respect to z, you are now working in the xy-plane. It is not so much "setting z=0" as "projecting the figure onto the xy-plane". For an ellipse, projecting onto the xy-plane gives the outline of the ellipse which is the same as taking z= 0. If this were, say, an cone such as [math]z= \sqrt{x^2+ y^2}[/math] with z from 0 to 2, although taking z=0 gives the single point (0,0), projecting the entire ellipse to the xy-plane gives the circle [math]x^2+ y^2= 4[/math].

    Because projecting every point in the ellipse onto the x-axis, you get every value of x from -1 to 1.

    I prefer to think in the other direction: from the outside in: In order that triple integral be a number rather than a function of one or more variables, the limits of the outer integral must be a constant. Yet the integral must "cover" the entire figure. If that outer integral is with respect to, say, x, the limits must be the smallest and largest possible values of x in that figure.

    In this case, the ellipse [tex] 4x^2 + 4y^2 + z^2 = 4 [/tex], which we could also write as [tex]\frac{x^2}{1}+ \frac{y^2}{1}+ \frac{z^2}{4}= 1[/tex], x ranges from -1 to 1, y from -1 to 1 and z from -2 to 2. If we decide to integrate so that the x-integral is the "outer" integral, its limits will be -1 to 1. If we decide to integrate so that the y-integral is the "outer" integral, its limits will be -1 to 1. If we decide to integrate so that the z-integral is the "outer" integral, its limits will be -2 to 2.

    If, having set up the "outer integral" with respect to x, the limits of the next integral can depend on x but not on y or z. If we decide to do that integral with respect to y, we need to use the least and greatest values of y for a given x. Again, it is not just "taking z= 0" but projecting the figure onto the xy-plane. In this case the "widest" part of the ellipse happens to be when z= 0:
    [tex]4x^2+ 4y^2= 4[/tex]
    so the maximum and minimum y are [math]\pm\sqrt{1- x^2}[/math].

    For given x and y, the boundary is given by solving [tex]4x^2+ 4y^2+ z^2= 4[/tex] solving that for z gives [tex]z= \pm\sqrt{4- 4x^2- 4y^2}[/tex].

    Lets look at the problem I mentioned before: find the volume of the cone [tex]z= \sqrt{x^2+ y^2}[/tex] between z= 0 and z= 2. The "broadest" part of that is at the top:z= 2 where [tex]2= \sqrt{x^2+ y^2}[/tex] or [tex]x^2+ y^2= 4[/tex], the circle centered on the origin with radius 2. every point in the cone projects down to a point in that circle. Projecting that circle to the x-axis gives the interval from x=-2 to x= 2.

    The limits on the "outer" integral, with respect to x would be -2 and 2. On that circle, drawing a vertical line for a given x, y ranges from [tex]-\sqrt{4- x^2}[/tex] to [tex]\sqrt{4- x^2}[/tex]. Those are the limits of integration on the "dy" integral.

    Finally, the part of the cone that we want is above [tex]z= \sqrt{x^2+ y^2}[/tex] and below z= 2. The integral for the volume of the cone is
    [tex]\int_{x= -2}^2\int_{y= -\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{\sqrt{x^2+ y^2}^2 dzdydx[/tex]
     
  4. Mar 27, 2009 #3
    HallsofIvy

    Thank you so much!!

    your explanation is very helpful!!
     
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