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Plane in the space

  1. Apr 7, 2010 #1
    Hi, I got following problem. It might contain little language mistakes, since I translated it from danish.

    Let P describe that plane in the space [tex]R^{3}[/tex], that is given by the equation [tex]6x+3y+z=6[/tex]

    (i) State the coordinates for the planes crossing points with three coordinate axis, and sketch that triangle, which verticies are these three crossing points.
    State moreover an equation for plane’s intersection line with xy-plane

    (ii) Calculate triple integral
    [tex]\iiint\limits_{0} y dV[/tex]
    where K is that pyramid, that is restricted by the plane P and three coordinate planes.
  2. jcsd
  3. Apr 7, 2010 #2
    i.) to find the intersection with the z axis, set y=x=0; for y intersection, set x=z=0..and so on.
    ii.) you have the function P = 6x + 3y + z - 6. Note that dV = dxdydz, so pick one (lets say dz) and integrate with respect to it to get integral(P, z) = 3xz + 3yz/2 + (z^2)/2 - 3z. Now, you do the same thing with respect to x and then y (in whichever order you please).
    Finally, if you're looking for the area beneath the pyramid, you simply use the axis intercepts as your limits of integration. Note that if you're not finding the definite integral, then you need to ignore the limits of integration and just add a constant at the end of your final answer.
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