1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:

    [tex]\int\int_S(x+z)\,dS[/tex]

    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:

    [tex]\sqrt{3}\int\int_R(1-y)\,dx\,dy[/tex]

    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:

    [tex]\sqrt{3}\int\int_R(1-y)\,dx\,dy[/tex]

    [tex]=\sqrt{3}\int_{x=0}^1\int_{y=0}^1(1-y)\,dx\,dy[/tex]

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

    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.

    Thanks!

    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?

    thanks,
    Casey
     
  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(\int_{x=0}^1(1-y)\,dx\right)\,dy[/tex]

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

    [tex]=\sqrt{3}\int_{y=0}^{1-x}(1-y)\,dy[/tex]

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

    Edit:

    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:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook