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Find the volume of a solid bounded by different planes

  1. Jan 21, 2014 #1
    1. The problem statement, all variables and given/known data

    It asks to find the volume of the solid given these planes:

    z = x
    y = x
    x + y = 2
    z = 0

    It also asks to find the volume using 2 iterated integrals with different orders of x and y integration.

    2. Relevant equations



    3. The attempt at a solution

    I found the volume of the solid which is 1/3 by setting up the volume as
    ∫(x=1 --> x=0) ∫(y=2-x -->y=0) [x]dydx

    which then gave me 1/3.

    However, when I tried to do it by integrating with respect to x first, I get a different answer with variables of y.

    ∫(y=2-x -->y=0)∫(x=1 --> x=0) [x]dxdy

    first integration wrt 'x': (x^2)/2 from (x=1 --> x=0) = 1/2

    This would leave ∫(y=2-x -->y=0) [1/2]dy = y/2 (y=2-x -->y=0) => This gives simply variables.

    What am I doing wrong?
     
  2. jcsd
  3. Jan 21, 2014 #2

    LCKurtz

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    First you have to describe the region more carefully. Give us an exact statement of the problem. I assume it wants a region in the first octant but you haven't said that. Even so, there are still two regions that fit your description. Exact statement please.
     
  4. Jan 22, 2014 #3

    haruspex

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    It looks to me as though there is only one region bounded by those four planes.
    You haven't allowed x to exceed 1. It can.
     
  5. Jan 22, 2014 #4
    I'm sorry, it is in the 1st octant. How can x be extended? I found that value ( x =1) from the point of intersection from y = x and x + y = 2.
     
  6. Jan 22, 2014 #5

    LCKurtz

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    I agree haruspex; I had my picture wrong.

    coolusername, my question now for you is have you drawn a careful picture of the region? And, unfortunately, it is a bit tricky to draw so you can see what you are doing. You need to do that to have any hope of setting up the integrals correctly. Neither of your two integrals is set up correctly. If you use a dydx integral you will need to break up the problem into two pieces.

    In your second integral, the inner limits need to be x as a function of y. Again, you need a good picture.
     
  7. Jan 22, 2014 #6

    haruspex

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    The point (1.5, 0.2, 0.1) is also in the region; y < x, y < 2-x, 0 < z < x.
    I suggest cutting it into two volumes with the plane x = 1.
     
  8. Jan 22, 2014 #7

    LCKurtz

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    I had a little time to kill so I drew a picture for you:

    region3d.jpg

    The vertical sides are the xz plane (yellow), the plane x = y (green) and the plane x+y=2 (cross hatched). The celing is the plane z=x (blue) and the floor is z = 0 (brown). You need to look in the xy plane for your limits.
     
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