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Math Help - Triple integral question

  1. #1
    Junior Member
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    Triple integral question

    If f is a continuous and a is a positive number (think of a as a parameter), show that .

    Here, V = {}
    {}
    {}


    Hint: if you are good at visualization, you may be able to "see" that the same region can be written in a **different** way as

    V = {}
    {}
    {}
    (I have given you the limits for x.) If you get this right, the rest is easy.


    I am horrible at visualization. lol I have a sketch of this. I have that the top is z=a. The back is x=0. A side is x=y, and another side is y=z. I have no idea how to rewrite the region though. I think that it may be since z=a. But I can't figure out the "new" inequality for z. Furthermore, I really don't know how redefining this helps to prove the original statement.
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  2. #2
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    Hi. First consider the general expression when integrating with respect to dzdydx:  \int_0^a \int_{f_1(x)}^{f_2(x)} \int_{g_1(x,y)}^{g_2(x,y)} dzdydx. That's the usual one and is easiest to visualize. It's just integrating from the surface g_1(x,y) up to the surface g_2(x,y) between the curves f_1(x) and f_2(x) and going along the x-axis from zero to a. So now that you have

    \int_0^a\int_0^z\int_0^y dxdydz

    so that means we're integrating between the surfaces g_1(y,z)=0 to g_2(y,z)=y. That's going from the zy-plane to the red plane in the figure. And we're integrating from the lines f_1(z)=0 to f_2(z)=z which is that green triangle in the zy-plane, and finally we're going from 0 to a in the z direction or just from 0 to 1 in my figure. So it's that little red-purple-green wedge in there that we're integrating over. Now lets switch the order of integration to:

    \int\int\int dzdydx

    Now we want to integrate first in the z-direction which is from the purple surface up to the black surface so z goes from f_1(x,y)=y up to f_2(x,y)=a. Now in the y-direction, we go from the line that is g_1(x)=x to g_2(x)=a and finally in the x direction we go from 0 to a giving us:

    \int_0^a\int_x^a\int_y^a f(x)dzdydx

    and you can do that integration up to the dx part right? Also, this is a particular case of the general expression:

    \int_0^a\int_0^{x_1}\int_0^{x_2}\cdots \int_0^{x_{n-1}} f(x_n)dx_n dx_{n-1}\cdots d{x_1}=\frac{1}{(n-1)!}\int_0^x (x-t)^{n-1}f(x)dx

    which I'm pretty sure is derived via differentiation and integration directly.
    Attached Thumbnails Attached Thumbnails Triple integral question-int-volume.jpg  
    Last edited by shawsend; March 11th 2010 at 04:28 AM. Reason: corrected first and second integration limits for dxdydz
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  3. #3
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    Hi guys. Interesting problem for me. I'm pretty sure I made a mistake with the explanation of the integration limits for the dxdydz case up there so I corrected it. I think it's ok. It makes more sense perhaps if I wrote it as:

    \int_0^a \int_0^{y=h(z)=z} \int_0^{x=f(y,z)=z} dxdydz and also, it's easier to see this if I rotate the figure so that the x-direction is pointing upwards. See below. If you want Banana, use the following Mathematica code so that you can rotate it as you like so that you understand how the integration is being done:

    Code:
    p1 = Polygon[{{0, 0, 0}, {2, 2, 0}, {2, 2, 1}, {0, 0, 1}}]; 
    p2 = Polygon[{{0, 0, 0}, {0, 1, 1}, {1, 1, 1}, {0, 0, 0}}]; 
    p3 = Polygon[{{0, 0, 0}, {0, 1, 1}, {0, 0, 1}}]; 
    p4 = Polygon[{{0, 0, 1}, {1, 1, 1}, {0, 1, 1}}]; 
    Show[Graphics3D[{{Opacity[0.2], Red, p1}, {Opacity[0.2], Blue, p2}, 
        {Opacity[0.2], Green, p3}, {Opacity[0.5], Black, p4}}], Axes -> True, 
      PlotRange -> {{0, 2}, {0, 2}, {0, 2}}, AxesLabel -> {x, y, z}]
    Attached Thumbnails Attached Thumbnails Triple integral question-rotated-fig.jpg  
    Last edited by shawsend; March 11th 2010 at 05:01 AM.
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