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Integrals with Respect to X and Y

  1. Mar 15, 2008 #1
    [SOLVED] Integrals with Respect to X and Y

    Hey folks, I've been working on this problem for too long.

    1. The problem statement, all variables and given/known data
    Sketch the area of the region bounded by the curves [tex]y^2+4x=0[/tex] and [tex]y=2x+4[/tex]. Set up two integrals, one with respect to x and one with respect to y, for finding the area of the region. Evaluate one of the integrals to find the area.

    2. Relevant equations
    [​IMG]

    I've got it drawn but I just can't get the integrals to match. The question says to only do one but I want to check my answer (and hopefully learn how to set these up). Here are my two latest attempts.

    3. The attempt at a solution
    [​IMG]

    [​IMG]
     
    Last edited: Mar 15, 2008
  2. jcsd
  3. Mar 15, 2008 #2

    HallsofIvy

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    One obvious point: the points of intersection have y= -4 and -1 but when y= -1, x is NOT 2!!!
     
  4. Mar 15, 2008 #3
    Thanks. That fixed the numerator, bringing me somewhat closer, but now the denominators are not matching. I have no idea what's going on here.
     
  5. Mar 15, 2008 #4
    You once again set the limits of integration incorrectly, look what happens if you proceede how u started.

    [tex]\int_{-4}^{-1}(2x+4)dx=(x^{2}+4x)|_{-4}^{-1}=(1-4)-(16-16)=-3[/tex]

    which doesn't actually make sense, since the area cannot be negative. Instead
    [tex]\int_{-4}^{-2}|(2x+4)|dx=-\int_{-4}^{-2}(2x+4)dx=-(x^{2}+4x)|_{-4}^{-2}=
    -[(4-8)-(16-16)]=5 [/tex]

    YOu have to be carefule, because as i explained in another post, the [tex]\int_a^bf(x)dx[/tex] only calculates the area that the curve of the function f(x) encloses with the x-axis, meaning that you have to set up a different integral with the limits of integration wherever the graph crosses the x-axis.

    Now which part of the region enclosed by those functions do you want to find by setting an integral with respect to y? In other words do you want to find the area that the two graphs enclose above x-axis or below x-axis, or do you want to do the whole thing with respect to y?
     
    Last edited: Mar 15, 2008
  6. Mar 15, 2008 #5
    The total enclosed area above and below the axis.
     
  7. Mar 15, 2008 #6
    I think you misunderstood me. I said that which part of that area would you like to find integrating with respect to y?
    Or did i misunderstand you?
     
  8. Mar 15, 2008 #7
    The whole thing wrt x and the whole thing wrt y.Thanks again.
     
    Last edited: Mar 15, 2008
  9. Mar 15, 2008 #8
    Since the line [tex]y=2x+4[/tex] crosses the x-axis at the point -2, whenever we need to find an area that this line encloses with the x-axis there will always be one limit of integration -2. SInce like i said also in the previous post :
    YOu have to be carefule, because the [tex]\int_a^bf(x)dx[/tex]
    only calculates the area that the curve of the function f(x) encloses with the x-axis, meaning that you have to set up a different integral with the limits of integration wherever the graph crosses the x-axis.



    So say we want to calculate the area that the line f(x)=2x+4 encloses with the x-axis from -8 to 4 then we have:

    [tex]A=\int_{-8}^{-2}|f(x)|dx+\int_{-2}^{4}f(x)dx[/tex] We have to proceede this way, among others, because the function f(x) from -8 to - is negative, notice it lies under the x-axis, so since we want the total area that f(x) encloses with the x-axis, we need to add the area that f(x) encloses from -8 to -2 and the area that f(x) encloses from -2 to 4. If we would not separate the integral this way, we would not end up adding these two areas but instead subtracting them, since like i mentioned f(x) from -8 to -4 has a neg. sign.

    Hope this helps!!!
     
  10. Mar 15, 2008 #9
    Suppose I transposed the functions so that the whole area was within a positive domain and range. Then my method would work, right? Of course that would be more trouble than just using the absolute values.

    If I understand what you're saying then these previous problems should also be wrong because the bounded area crosses the x axis and I have not used absolute values for the piece underneath. However, I was pretty sure they were correct, especially the triangle which I even checked with Heron's formula... hence my confusion.

    [​IMG]

    [​IMG]
     
    Last edited: Mar 15, 2008
  11. Mar 15, 2008 #10
    It does not have to be within a positive domain, only within a positive range. Let's reffere once more to the line you have y=2x+4, and say we want to find the area that this line encloses with the x-axis from -2 to 7. On this whole interval y is greater than or equal to 0( it is equal to zero only when x=-2). This way we would have:

    [tex]A=\int_{-2}^7(2x+4)dx[/tex]
     
  12. Mar 15, 2008 #11
    With transposing a function, do you mean like for ex, shifting it upwards, downwards and other such stuff? If so, then you cannot apply the exact procedure, since the transposing function is itself a function, so we have to treat it as a function also, unless it is symetric say around y-axis, and we only reflect it around it.
     
    Last edited: Mar 15, 2008
  13. Mar 15, 2008 #12
    Bump. So does that mean the questions I edited in above are also wrong?
     
  14. Mar 15, 2008 #13
    Let me check it, i did not notice that you edited your post. I'll see what i can do!
     
  15. Mar 15, 2008 #14
    well, i am wondering how come you got the right answer on the first one, because you did not set the first integral correctly. But the answer is correct, i checked it also integrating with respect to y. But i will see later, what happened there, cuz i need to run now.
     
  16. Mar 15, 2008 #15
    and for the second one, it looks fine to me. Moreover, you have taken in consideration the minus sign, but maybe you are just not aware of it, look at it closely. YOu could have also done the last one more quickly, and easier.
     
  17. Mar 15, 2008 #16
    Thank you again for taking the time to explain this. I will go back and see if I can get it worked out.
     
    Last edited: Mar 15, 2008
  18. Mar 17, 2008 #17
    After a few tries I was able to solve wrt using substitution. Thanks again.
     
  19. Mar 17, 2008 #18
    Like i said before you have taken into consideration the minus sign here. When u set up the first integral you did it like this,

    [tex]A=\int_{-2}^0[(x+2)-(x^{2}-4)]dx=\int_{-2}^0(x+2)dx-\int_{-2}^0(x^{2}-4)dx[/tex]
    Here at the very first sight it looks like you are subtracting the second integral from the first one, but indeed you are adding the area that the second integral calculates with the area that the first integral calculates, for the fact that the function [tex]f(x)=x^{2}-4[/tex] is negative from -2 to 0. So, this is actually what i meant when i said u have taken into consideration the minus sign. Because here you should be adding areas, instead of subtractig them. You have done the same thing with the second integral, which is correct.
     
  20. Mar 17, 2008 #19
    Yep, it all makes sense now. Thanks for the followup... and now on to the next concept I don't understand :bugeye:
     
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