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Evaluate double integral

  1. Dec 24, 2011 #1

    sharks

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    The problem statement, all variables and given/known data
    Evaluate the double integral:
    http://s1.ipicture.ru/uploads/20111224/K3MGcdS7.jpg

    The attempt at a solution

    Consider x as constant, first integrate w.r.t.y.

    I get
    [tex]\frac{xy^2}{2}[/tex]
    After evaluating the integral with limits y=0 to y=y, i get the same exact answer.

    Then, i integrate the above w.r.t.x

    I get
    [tex]\frac{x^2 y^2}{4}[/tex]
    Applying the limits x=0 to x=1

    I get the final answer as:
    [tex]\frac{y^2}{4}[/tex]

    However, the answer in my notes says: 1/8

    Also, i don't understand why i must always first integrate w.r.t.y then w.r.t.x and not vice versa?
     
  2. jcsd
  3. Dec 24, 2011 #2

    Mark44

    Staff: Mentor

    No. You need to integrate with respect to x first. That is the significance of dx coming before dy.
    It depends on the order of dx and dy in the iterated integral.
    In this integral you integrate with respect to x first, and then y.
    [tex]\int \int f(x,y) dx~dy[/tex]

    Here you integrate with respect to y first, and then x.
    [tex]\int \int f(x,y) dy~dx[/tex]
     
  4. Dec 24, 2011 #3

    sharks

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    Hi Mark44

    Using your method, i get the final answer:
    [tex]\frac{x^2}{4}[/tex]
    which is wrong.

    OK, for argument's sake (and to clear up my confusion), let's just consider another example:
    http://s1.ipicture.ru/uploads/20111224/VxWfVW7H.jpg

    From what you tell me, the answer should be: 2-2x
    Again, this is wrong. So, something about your method doesn't add up.
     
    Last edited: Dec 24, 2011
  5. Dec 24, 2011 #4

    ehild

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    How did you get that final answer? It has to be a number. Have you substituted the upper and lower bounds after integration with respect to x?

    ehild
     
  6. Dec 24, 2011 #5

    Curious3141

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    The answer should be numerical. First integrate wrt x, treating y as a constant. Then evaluate the definite integral between the bounds 0 and y, which means substituting y for x in that expression to get a form that only has y in it. Finally integrate that expression wrt y with the bounds 0,1.

    As it is written, the answer is 2 - 2x. This is because the inner bound with the sqrt(x) leaves the variable x in the integral after integrating wrt x and evaluating for the bounds, and it persists after the final integration wrt y.
     
  7. Dec 24, 2011 #6

    sharks

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    OK, i got the correct numerical answer for the first problem:
    http://s1.ipicture.ru/uploads/20111224/K3MGcdS7.jpg

    However, for the second problem:
    http://s1.ipicture.ru/uploads/20111224/VxWfVW7H.jpg
    The answer in my notes is numerical: 6/7
    The solution (according to my notes): The variable x is considered as constant and then integrate w.r.t.y (which is the other way round to what i've been led to believe in this thread!). I'm confused.
     
  8. Dec 24, 2011 #7

    gb7nash

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    The problem looks like it was copied or written wrong, then. It doesn't make sense to integrate with respect to x, and then plug in a function of x as a bound. My guess (and you'll have to check with your instructor) is that the second problem is supposed to be:

    [tex]\int_0^1 \int_\sqrt{x}^1 12xy^2 dy dx[/tex]
     
  9. Dec 24, 2011 #8
    I think you have a problem with the geometry so I believe you should focus on it for now rather than the algebra: Suppose I said I want the volume underneath the function [itex]f(x,y)=12 x^2[/itex] that lies between the curves [itex]y_1(x)=\sqrt{x}[/itex] and [itex]y_2(x)=1[/itex] from the points x=0 to x=1. Now forget about the integrals for now. Just draw that, nicely. Label everything. Now go back to your text and study how the double integral is defined as either:

    [tex]\int_{x_1}^{x_2}\int_{y_1(x)}^{y_2(x)} f(x,y)dydx[/tex]

    or if instead I have the functions in terms of y as [itex]x_1(y)[/itex] and [itex]x_2(y)[/itex] from the points [itex]y_1[/itex] to [itex]y_2[/itex]:

    [tex]\int_{y_1}^{y_2}\int_{x_1(y)}^{x_2(y)} f(x,y)dxdy[/tex]

    and then adapt your problem to one of those.
     
    Last edited: Dec 24, 2011
  10. Dec 24, 2011 #9

    Mark44

    Staff: Mentor

    I agree.
     
  11. Dec 24, 2011 #10

    SammyS

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    Of course, evaluating this does not give the result 6/7.
     
  12. Dec 24, 2011 #11

    sharks

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    OK, thank you all for your help. I think it is a typo error then. I will email my instructor.

    Best wishes for the holidays!:smile:
     
  13. Dec 24, 2011 #12
    Actually, I think this was an intentional error. If you draw the region, you'll find that you can change the integral to

    [STRIKE]∫010y^2 12xy2 dx dy

    Evaluation of this integral DOES result in 6/7.[/STRIKE]
     
    Last edited: Dec 24, 2011
  14. Dec 24, 2011 #13

    sharks

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    I'm struggling to understand the basic concepts here. Without a solid foundation, i can't proceed with further sections in double integrals and i have to study triple integrals after that. Can someone please give me a definitive answer?
     
  15. Dec 24, 2011 #14
    Sorry. I made a mistake earlier. It should be

    010√x 12xy2 dy dx

    which results in 8/7.

    Whenever you have a double integral, you should just treat it as an integral within an integral.

    01√x1 12xy2 dx dy

    = ∫01 (∫√x1 12xy2 dx) dy

    As with a regular integral:

    ab f'(x,y) dx

    a represents the lower bound and b represents the upper bound with respect to x. In the case of general regions, instead of having a line as the lower bound, the lower bound is, instead, a curve.

    Whenever there is a multiple integrals over a general region, you want all the variables to disappear by the end of the integration. In other words, when f(x) is one of the bounds on an integral, there had better be a dx on one of the outer layer of the nested integral.

    In your case, the bounds with respect to x is √x to 1. The only outer layer is an integral with respect to y. This is a problem, since integrating outside with respect to y does not get rid of the function of x in the x-bound. The only choice here is to change how the region of integration is defined.

    If you draw the region on a piece of paper, you'll find that the region of integration is the inner section of the upper region of a horizontal parabola, with vertex at (0,0), cut off at the line x=1. One easy way rewrite the bounds is to have the y run from 0 to √x. Then the x bounds run from 0 to 1. This means that you'll be integrating with respect to y on the inner layer of the integral, and you'll be integrating with respect to x on the outer layer.

    010√x 12xy2 dy dx

    = ∫01(∫0√x 12xy2 dy) dx

    = ∫01 (4xy3 |y=0y=√x) dx

    = ∫01 (4xx3/2) dx

    = ∫01 4x5/2 dx

    = 4 (2/7) x7/2 |01

    = 8/7

    Keep in mind that the integral you wrote down:

    http://s1.ipicture.ru/uploads/20111224/VxWfVW7H.jpg

    doesn't actually make sense by itself. My interpretation is just one way to crunch numerical results out of the nonsensical integral.
     
    Last edited: Dec 24, 2011
  16. Dec 25, 2011 #15

    sharks

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    OK, i'm beginning to see a clearer pathway of reasoning. However, the first problem appears to be a trick question, since it has 2 possible answers??
     
  17. Dec 26, 2011 #16

    Mark44

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    No, there is just one answer -- 1/8, the same as in your notes.

    After the first (inner) integration, you should get y3/2. After integrating that with respect to y, you should get y4/8, which you evaluate at 1 and at 0, resulting in the final answer of 1/8.


     
  18. Dec 26, 2011 #17

    sharks

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    OK, i got it. Thanks to all.
     
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