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Difficulty in finding upper limit of x

  1. Apr 21, 2012 #1
    I've come across the following question:

    ∫∫∫6xy dxdydz
    where R lies under the planez=x+y+1 and above the region in the xy-plane bounded by y=0,y=√x and x=1.

    Now the limits would be:

    √x→y→0 ; x+(√x)+1→z→2 and ?→x→1

    Now I can't get the upper limit of x as a number (I've set up the x-integral as the outermost integral in my triple integratrion so it has to have numeric values for upper and lower limits).

    Could someone tell me HOW to find the limit and NOT WHAT IT IS.
     
  2. jcsd
  3. Apr 21, 2012 #2

    arildno

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    First, FORGET about the plane. (take care of that LATER!!)

    Try to visualize:
    What CLOSED region in the xy-plane are bounded by the three curves you've given there?
    What is the minimum x-value, what is the maximum x-value?
     
  4. Apr 21, 2012 #3
    Um. I tried visualizing it and I even drew a graph of the three curves in Matlab. Now what I found was that the curves intersect at (0,1),(0,0) and (1,1). Attached is a screenshot of the graph I plotted in matlab. Now these coordinates of intersections mean that the lower limit of x is 0. However when I evaluated the answer it came out to be wrong. What am I doing wrong? Is my lower limit of x even right?

    Again don't tell me WHAT the limit is, just tell me WHERE and HOW am I going wrong?
     

    Attached Files:

  5. Apr 21, 2012 #4

    HallsofIvy

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    Your screen shot shows a single line. The graph of [itex]y= \sqrt{x}[/itex] is a parabola. The top plane, z= x+y+ 1 crosses all three boundaries, y= 0, x= 1, and [itex]y= \sqrt{x}[/itex] above the xy-plane, so, as far as the x and y bounds are concerned, you only need to look at x and y.

    Now, looking only a the xy-plane and the region bounded by y= 0, x= 1 and [itex]x= \sqrt{y}[/itex], what are the smallest and largest values of x? For each x what are the smallest and largest values of y (the answer depends on x, of course).

    Then, in three dimensions, for each point, z goes from 0 to z= x+y+ 1.
     
  6. Apr 22, 2012 #5
    I don't get the following:

    1) How does the top plane, z= x+y+ 1 'cross' all three boundaries, y= 0, x= 1, and y=√x above the xy-plane.

    2) 'Now, looking only a the xy-plane and the region bounded by y= 0, x= 1 and x=√y , what are the smallest and largest values of x? For each x what are the smallest and largest values of y (the answer depends on x, of course).'

    The smallest value of x would be 0. Right, since x=sqrt(y) and is bounded above by 1 so since the lowest value of y is 0 so x=sqrt(0)=0 thus the lower limit of x is 0.

    Geomterically this also makes sense. I plotted the three curves on a piece of paper. The three points at which the 3 curves intersected each other were (0,0), (1,0), (1,1) so
    largest value of x=1; smallest values of x=0
    " " " y=1; " " " y=0

    3) 'Then, in three dimensions, for each point, z goes from 0 to z= x+y+ 1. '


    Now I agree wtih your upper limit of z, but don't get your lower limit of z. Should'nt it be 1.
    x and y can't be negative. Their lowest possible value is zero (at this point we're not tking into account that we've determined their lower limits to be 0). So:
    z=x+y+1=0+0 +1=1

    So how can it be 0?
     
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