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Upper and lower bounds of a triple integral

  1. Apr 15, 2016 #1
    1. The problem statement, all variables and given/known data

    Let ##T \subset R^3## be a set delimited by the coordinate planes and the surfaces ##y = \sqrt{x}## and ##z = 1-y## in the first octant.

    Write the intgeral
    [tex]
    \iiint_T f(x,y,z)dV
    [/tex]
    as iterated integrals in at least 3 different ways.

    2. Relevant equations

    [tex]
    \iiint_T f(x,y,z)dV = \int_a^b \int_{y_1 (x)}^{y_2 (x)} \int_{z_1 (x,y)}^{z_2 (x,y)} f(x,y,z) dx dy dz
    [/tex]

    3. The attempt at a solution

    My interpretation of this problem is that they want me to "simply" find out the upper and lower bounds of integration for 3 different cases, since there is no specific function to integrate.

    This is a bit difficult to explain, since I can't show you a picture of the volume because I can't seem to be able to draw the ##y = \sqrt{x}## correctly in Matlab using this code:

    syms x y z

    f(x,z) = sqrt(x);
    g(x,y) = 1 - y;

    hold on
    xlabel('x')
    ylabel('y')
    zlabel('z')
    axis equal

    ezsurf(f,0:1);
    ezsurf(g,0:1)

    Anyways, I'm having trouble seeing what the upper bounds are for the innermost iteration, whichever way I try and integrate this thing. For example, if I were to integrate first with respect to y, then x and finally z, I would write:
    [tex]
    \int_0^1 \int_{0}^{(-z+1)^2} \int_{0}^{?} f(x,y,z) dy dx dz
    [/tex]
    Just looking at the horrible picture that I drew on paper (which kind of looks like a tetrahedron cut off by the ##\sqrt{x}##-surface) is not helping. Does anybody have any hints on where to start. My brain is telling me that ##?## should be a plane of some sort, but the fact that it's cut off by the ##\sqrt{x}## is really throwing me off.

    EDIT: The shape is simple (as in x-, y- and z-simple) in all 3 dimensions, so I was able to print out the projections to the different coordinate planes:

    y(x):

    H5_4yx.JPG

    z(y)

    H5_4zy.JPG


    x(z):

    H5_4zx.JPG
     
    Last edited: Apr 15, 2016
  2. jcsd
  3. Apr 15, 2016 #2
    After thinking about this for a bit I think this might be the correct answer for the case ##dy dx dz##:
    [tex]
    \int_0^1 \int_{0}^{(1-z)^2} \int_{\sqrt{x}}^{1} f(x,y,z) dy dx dz
    [/tex]
    I'm saying this, because
    \begin{cases}

    y = \sqrt{x} \iff x = y^2\\
    y+z = 1 \iff z = 1- \sqrt{x} \iff x = (1-z)^2
    \end{cases}

    How does this look?
     
    Last edited: Apr 15, 2016
  4. Apr 15, 2016 #3
    Using the same logic as above for two other cases:

    ##dxdydz##:

    [tex]\int_{0}^{1} \int_{0}^{1-z} \int_{0}^{y^2}f(x,y,z) dxdydz[/tex]
    and ##dzdxdy##:
    [tex]\int_{0}^{1} \int_{0}^{y^2} \int_{0}^{1-y} f(x,y,z) dzdxdy[/tex]
     
    Last edited: Apr 15, 2016
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