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Math Help - Triple Integral Help

  1. #1
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    Triple Integral Help

    Given function f(x,y,z)=y
    Region bounded by x+y+z=2, x^2+z^2=1, and y=0

    Integrate given function over indicated region.
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  2. #2
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    Quote Originally Posted by veritaserum2002 View Post
    Given function f(x,y,z)=y
    Region bounded by x+y+z=2, x^2+z^2=1, and y=0

    Integrate given function over indicated region.
    Lets re-arrange, so in effect what we have is a plane and a cylinder of radius 1

    z=2-x-y
    z= \sqrt{1-x^2}

    To find our domain in the XY plane let these equal eachother such that

    2-x-y= \sqrt {1-x^2}

     y= 2-x- \sqrt {1-x^2}

    Now, what we are looking for is

    \iiint ydV

    We know our Z bounds (between the 2 equations we formatted). But what about X and Y. Well we might want to evaluate Y last because we have a function Y as our density.

    So let us find when x=0

     y= 2- \sqrt {1}

    So now we know Y runs from 0 to the above. Find your X bounds, and which Z is bigger (either the cylinder or the plane) and put in your bounds accordingly. We are done!
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  3. #3
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    From that, I got that the limits of integration are:
    z= [-sqrt(1-x^2) , sqrt(1-x^2)]
    y= [0 , 2-x-sqrt(1-x^2)]
    x= [-1,1]

    The limits on z are from the cylinder equation, bounded on the y axis by the intersection of the cylinder and plane. Does this sound right? Then the limits on x are just from "end to end" of the cylinder.
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  4. #4
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by veritaserum2002 View Post
    From that, I got that the limits of integration are:
    z= [-sqrt(1-x^2) , sqrt(1-x^2)]
    y= [0 , 2-x-sqrt(1-x^2)]
    x= [-1,1]

    The limits on z are from the cylinder equation, bounded on the y axis by the intersection of the cylinder and plane. Does this sound right? Then the limits on x are just from "end to end" of the cylinder.
    Typically, when we have an integral such as

    \iint ydV

    It is easiar to evaluate y in terms of constant bounds (i.e integrate y last) but you don't have to. In this case it might be easiar to find the bounds by integrating x last. So let us do that.

    <br />
y= 2-x- \sqrt {1-x^2}<br />

    Set y=0 and find your bounds for X. Then y of course will be from

    <br />
0 \le y \le 2-x- \sqrt {1-x^2}<br />

    In the z plane you are bounded by both of the equations we derived. So in fact, our bounds would look something like

    \int_{ \sqrt{1-x^2}}^{2-x-y} dz

    Of course since I havent actually figured out the bounds for x, i am only guessing that

     \sqrt{1-x^2} \le 2-x-y for region R

    Edit- it is 3am and this is the most I can muster for tonight, I will actually compute this tomorrow is somebody else hasent done so already.
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  5. #5
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    Thanks much for your help, I think I may be able to run with it now!
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