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Double Integral problem

  1. Sep 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the volume under the surface z = y(x+2) and over the area bounded by y+x = 1, y = 1 and y = sqrt(x)

    2. Relevant equations



    3. The attempt at a solution

    Based on the geometry of the bounds, I broke this integral into two parts. I first found the intersection of the sqrt(x) and y = 1-x. This is at x = [itex] \dfrac {1} {2}\left( 3-\sqrt {5}\right) [/itex]

    Then my first integral was this:

    [tex]\int _{x=0}^{x=\dfrac {1} {2}\left( 3-\sqrt {5}\right) }\int _{y=1-x}^{y=1}y\left( x+2\right) dydx[/tex]

    and my second integral is this:

    [tex]\int _{x=\dfrac {1} {2}\left( 3-\sqrt {5}\right) }^{x=1}\int _{y=\sqrt {x}}^{y=1}y\left( x+2\right) dydx [/tex]

    However when I solve this integral, I don't get the correct answer (which should be 9/8 according to the solution).

    Where am I going wrong?

    Thanks!
     
  2. jcsd
  3. Sep 2, 2014 #2

    Zondrina

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    By plotting the region it looks like you should be using four double integrals.
     
    Last edited: Sep 2, 2014
  4. Sep 2, 2014 #3

    Ray Vickson

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    I assume you want to bound the volume below by the plane z = 0.

    What was your final answer? I do not get the value 9/8, either. In fact, since 9/8 > 1 it cannot be the correct answer, for even getting a crude upper bound gives an answer < 1. Below, let
    [tex] a = \frac{3}{2}-\frac{\sqrt{5}}{2} \doteq 0.381966012 [/tex]
    be x-value of the intersection between the square root and the slanted line.

    The base of the first integral is the right-triangle with sides ##a## and ##a##, so has area ##A_1 = a^2/2 \doteq 0.07294901715.## A simple upper bound on ##z## is ##z \leq 3## for all feasible ##x,y##, so the first integral, ##I_1##, has the simple upper bound ##I_1 \leq U_1 = 3 A_1 \doteq 0.2188470514##.

    The base of the second integral is a "triangle" with a curved hypotenuse, so has area less than a true triangle with the same base and height; that is we have:
    [tex] \text{area}_2 < A_2 \equiv a (1-a)/2 \doteq 0.1180339888 [/tex]
    Thus, an upper bound on the second integral, ##I_2##, is ##I_2 \leq U_2 = 3A_2 \doteq 0.3541019664##. An upper bound on the whole double integral is ##U = U_1 + U_2 \doteq 0.5729490180.## This is a lot smaller than the claimed value of 9/8 = 1.125.
     
    Last edited: Sep 2, 2014
  5. Sep 2, 2014 #4

    LCKurtz

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    Also, I would point out that, given the shape of that region, a dxdy order of integration allows setting up only one double integral.
     
  6. Sep 2, 2014 #5
    @Zondrina: I don't think there are 4 double integrals because this area is bounded by y = 1 and comes to a point at the intersection of y = 1 - x and y = sqrt(x)

    @Ray Vickson: yeah I get a number much smaller than 1. I calculated the integrals on wolfram alpha and I ended up getting something like .38


    @LCKurtz: I'm not sure I understand what you mean. Care to explain more?
     
  7. Sep 3, 2014 #6

    LCKurtz

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    @LCKurtz: I'm not sure I understand what you mean. Care to explain more?

    I mean put the proper limits on$$
    \iint y(x+2)~dxdy$$Just one double integral is necessary.
     
  8. Sep 3, 2014 #7

    Zondrina

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    Yes I didn't notice this when I was sleepy. The region looks like it's divided up into three pieces, not four (somehow my brain convinced me there was a line at ##x=1## last night); The left triangle like piece, the upper triangular like piece and the bottom piece.

    The limits for ##x## appear to be different in each region, as are the limits for ##y##. If I'm seeing this now, it looks like you would need five integrals. Not quite sure how you would do this with just one integral.
     

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  9. Sep 3, 2014 #8

    Ray Vickson

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    Intuitively, you have chosen to split up the base area into thin vertical slices parallel to the y-axis. Essentially, the suggestion of LCKurtz is that you split it up instead into thin horizontal slices parallel to the x-axis.
     
  10. Sep 3, 2014 #9

    LCKurtz

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    attachment.php?attachmentid=72664&d=1409750071.png

    @Zondrina: The only region that is described in the original post is the upper one, and it takes one double integral if done as a dxdy integral.
     
  11. Sep 3, 2014 #10

    Zondrina

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    I see all the lines only have one region where they all form a closed boundary.

    Thank you for clarifying that.
     
    Last edited: Sep 3, 2014
  12. Sep 3, 2014 #11
    Hey, thanks so much! Makes sense now.

    Thanks!
     
  13. Sep 3, 2014 #12

    Ray Vickson

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    Which response are you replying to? Please use the "quote" button when responding; it keeps the thread under control.
     
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