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Homework Help: Simple Double Integral

  1. Nov 28, 2009 #1
    1. The problem statement, all variables and given/known data

    I don't know what is going on on my brain. I am at a sage in a problem where I need to evaluate the double integral:


    where the surface is the is the portion of the plane x+y+x=1 that lies in the 1st octant.

    3. The attempt at a solution

    Forging ahead I arrive at:


    I know that I am correct up until this point. The text verifies up till this step. So clearly my trouble lies in evaluating this simple double integral.

    The text says the answer is 1/[itex]\sqrt3[/itex]. I keep getting:




    Before I go any further, is this last step ok? I am having trouble recalling whether it is ok or not to switch the order of integration from 'dxdy' to 'dydx' ?

    I can't think of a reason why it wouldn't be......but it has been awhile. From here, I keep getting [itex]\sqrt{3}/2[/itex] which does not match the text's answer.


    EDIT: I just tried it reversing the order which yields the same answer. I am convinced that my bounds are correct. Is the text just wrong?
    Last edited: Nov 28, 2009
  2. jcsd
  3. Nov 28, 2009 #2
    Wouldn't your y bounds have to be from 0 to 1-x?

    Edit: In the xy-plane, the "shadow" of the plane x+y+z = 1 will look like the triangle bounded by x=0, y=0 and x+y = 1.
    Thus, if you say x is bounded by 0 and 1, then when you hold an arbitrary x constant, y is bounded by 0 and 1-x.

    I worked out the integral with these new bounds and got the same answer as the text.
    Last edited: Nov 28, 2009
  4. Nov 28, 2009 #3
    Hmmm. You are conjuring up distant memories here.... but I am still failing to recall why we do this.

    Are the 'intercepts' of the plane with the x and y axes not fixed? Why do we need to write the y bounds in terms of x?

    If I was near a text, I would try to find the answer, but unfortunately I am out at a coffee shop right now. Delicious. :smile:

    Edit: I am examining you edit right now :smile:
  5. Nov 28, 2009 #4
    I am getting there, but I am still confused. I am still not understanding the answer to this "Are the 'intercepts' of the plane with the x and y axes not fixed?"

    What I was doing was letting x and z equal 0 to find the y-bounds and letting y and z equal 0 to find the x-bounds. I know that that is not right now.... but I am not seeing why.

    And why are the x-bounds from 0 to 1 correct? that is, why not choose the y-bounds to run from 0 to 1 and write the x-bounds in terms of y?

  6. Nov 28, 2009 #5
    The intercepts are indeed fixed. It would also be perfectly fine to let y go from 0 to 1 and then let x go from 0 to 1-y if we switch the dydx to dxdy.

    Maybe this will help: Draw out the region in the xy-plane. Now, we know the bounds for the outermost integral must be constants (or else we would end up with variables in our answer). So, let's choose x to go from 0 to 1. Now, to figure out the y bounds, we must fix an arbitrary x - say x is roughly 1/2. Now, draw a vertical line at x=1/2 and see what y is "cut off" by. You will notice that y is cut off by the line y=0 and the line y=1-x. So, we've found our bounds. The only time y goes from 0 to 1 is if x is 0. But if we chose x to be 0, we didn't pick a very arbitrary x ;)
  7. Nov 28, 2009 #6
    Oh.....I am so slow. :redface: The bounds that I gave are that of a rectangle :rofl:

    It is getting clearer now. thanks for your patience Dunkle :smile:
  8. Nov 28, 2009 #7
    I am a horrible person. I thought it would be a piece of cake after getting the bounds right. I can't figure out where I am messing up the procedure:


    [tex]=\sqrt{3}\int_{y=0}^{1-x}\left[x - xy\right]_0^1\,dy [/tex]



    From here, I can see that I am going to be left with 'x' in the answer..... where am I fidging this now?

    EDIT: I just found this on a website and also notice that Dunkle alluded to this point earlier: "In a double integral, the outer limits must be constant, but the inner limits can depend on the outer variable."

    Can someone give a 'precise' reason as to why this is the case? I know, now, that the integral simply cannot be evaluated properly if the outer bounds are not constants....but what is the 'mathematical' reasoning for this?
    Last edited: Nov 28, 2009
  9. Nov 28, 2009 #8
    Your integral should be either

    [tex]\sqrt{3}\int_0^1 \int_0^{1-x} 1-y \ dydx \ \ \ \ or \ \ \ \ \sqrt{3}\int_0^1 \int_0^{1-y} 1-y \ dxdy[/tex]

    I think the second is easier to integrate, so let's do it!

    [tex]\sqrt{3}\int_0^1 \int_0^{1-y} 1-y \ dxdy[/tex]

    [tex]= \sqrt{3}\int_0^1 \left[x-xy\right]_0^{1-y} \ dy[/tex]

    [tex]= \sqrt{3} \int_0^1 (1-y)-(1-y)y \ dy[/tex]

    [tex]= \sqrt{3} \int_0^1 y^{2}-2y+1 \ dy[/tex]

    [tex]= \sqrt{3} \left[\frac{1}{3}y^{3}-y^{2}+y\right]_0^1[/tex]

    [tex]= \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}[/tex]


    Exactly! You could even call it a square :wink:

    I don't have a "precise mathematical reason" for this, but you could think about it this way. Consider a region with area A in the xy-plane. If one were to integrate the function 1 over that region, the answer would be A (this follows from the definition of double integrals). If one did not have constant bounds on the outer integral, the answer would depend on either x or y. However, the area is always A and does not depend on x or y.
    Last edited: Nov 28, 2009
  10. Nov 28, 2009 #9
    That's a great explanation. Thank you Dunkle. Now I feel like before I continue with this vector calculus text, I need to do a serious review of my multivariable calculus.

    I hope PF is ready for an onslaught of questions :smile:
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