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Set Up Double Integral to find Vol. Solid, Bounded by Graphs

  1. Apr 26, 2012 #1
    1. The problem statement, all variables and given/known data
    ...Bounded by graphs of equations:
    z=xy,
    z=0,
    y=x,
    x=1

    I don't know what z=xy is. The rest of boundaries are clear.
    I assume that when y=1 and x=1, z=1. But, is this a z=1 plane?
    Check my figure attached.

    Thank you.
    2. Relevant equations

    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Apr 26, 2012 #2

    sharks

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    You can use a 3D plotter for the graph of z=xy. I have attached the graph.

    Now, you can better visualize the required volume. It's an object with a triangular base and the top surface is z=xy.

    [tex]\int^1_0 \int^y_1 xy\,.dxdy[/tex]
     

    Attached Files:

  4. Apr 26, 2012 #3
    Yeah, I plotted it in Matlab as well, but I still couldn't draw it.
    So, how can I build my double integral?

    I found that if I horizontally sliced my the triangular base I get:
    0<=y<= x
    y<=x<=1
    Now, I have no clue how to involve the upper bound. As sharks said, "the top surface".

    Would it be ## \int _{y}^{1}\int _{0}^{x}xy\left( dydx\right) ## ?
     
  5. Apr 26, 2012 #4

    LCKurtz

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    Are you missing a ##y=0## boundary? And, assuming that, Sharks has his limits wrong.
     
  6. Apr 26, 2012 #5

    LCKurtz

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    No. The outer integral can never have a variable limit if the result is to be a number.
     
  7. Apr 26, 2012 #6

    sharks

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    You are obviously correct. :redface:

    knowLittle, a boundary (plane) is missing at the back of your drawing. If it's not been given, then ignore my suggestion.
     
  8. Apr 26, 2012 #7
    Sorry about that LC, It's restricted to 1st octant.
    The full problem is:
    Set up a double integral to find the volume of the solid bounded by the graphs of the equations.
    z=xy,
    z=0,
    y=x,
    x=1,
    1st Octant.
     
  9. Apr 26, 2012 #8
    So, I guess that the equation would be:
    ## \int _{0}^{1}\int _{y}^{1}xydxdy##
     
  10. Apr 26, 2012 #9

    LCKurtz

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    Yes. It's even slightly less work to do dydx.
     
  11. Apr 26, 2012 #10
    Thanks :) I'll try dydx.
     
  12. Apr 26, 2012 #11

    sharks

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    Is that equation the same as what i suggested earlier?
    [tex]\int^1_0 \int^y_1 xy\,.dxdy[/tex]
    The limits for when x varies are reversed. Does that mean, the volume will have an opposite sign?

    EDIT: Indeed, it will have an opposite sign.
     
    Last edited: Apr 26, 2012
  13. Apr 26, 2012 #12
    It gives the opposite volume, but I'm not sure if it's always valid.

    I have another question from the same problem:
    Part B:
    z= x+y,
    x^2+y^2=4,
    y>=0

    From the equation of the circle and Y>=0, the region limits are:
    - (4-y^2)^(1/2) <=x <= (4-y^2) ^(1/2)
    0<=y<=2

    The volume equation:
    ## \int _{0}^{2}\int _{-\sqrt {4-y^{2}}}^{\sqrt {4-y^{2}}}\left( x+y\right) dxdy
    ##
    And, I got -(4/3) 2^(1/2) , but I don't think it's correct.
     
    Last edited: Apr 26, 2012
  14. Apr 26, 2012 #13
    I tried a polar approach:

    ##\int _{0}^{\pi }\int _{0}^{2}\left( r\cos \theta + r\sin \theta \right) rdrd\theta##

    Is this correct?
    My result is :
    (8/3) (sin (pi) - cos(pi) +1)
     
  15. Apr 26, 2012 #14

    sharks

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    Your limits for [itex]\theta[/itex] are not correct. I assume that part (b) requires the volume from the 1st octant?
     
  16. Apr 26, 2012 #15
    They do not mention 1st octant in part B. However, 1st octant is mentioned in part A and part C.

    What's wrong with my limits for theta?
     
  17. Apr 26, 2012 #16

    sharks

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    If the volume is in the first octant, then θ varies between 0 and ∏/2
     
  18. Apr 26, 2012 #17
    It's not in the first octant.
     
  19. Apr 26, 2012 #18
    Check figure.
     

    Attached Files:

  20. Apr 26, 2012 #19

    LCKurtz

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    When you set up an area or volume integral, you must always do your integrations in the positive directions of the variables.
     
  21. Apr 26, 2012 #20
    Thank you, LC. Could you help me with part B of my problem?
     
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