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Double Integals and Volume

  1. Oct 21, 2014 #1
    1. The problem statement, all variables and given/known data
    Evaluate:
    [itex] \int_{0}^{a} \int_{0}^{b} e^{max(b^{2}x^{2}, a^{2}y^{2})} dy \hspace{1 mm} dx [/itex]
    Where a and b are positive.

    2. Relevant equations


    3. The attempt at a solution
    I'm having some trouble getting started with this problem, mostly because I am not very familiar with the notation. I am assuming that the "max" portion of the question means "the set of points of the argument for which the function attains its max value"? That is from the wikipedia page for "Arg max". If that is correct then I am looking for the volume of the region where the given argument of e takes on a maximum value, no?
     
  2. jcsd
  3. Oct 21, 2014 #2
    Beats me! I would never write anything quite that way, so I have no clue what it means. Best to ask the original author what he had in mind.
     
  4. Oct 21, 2014 #3
    Unfortunately that's not possible. Anyway, thanks for your time :)
     
  5. Oct 22, 2014 #4
    Anyone have any ideas? I know it looks like I haven't really made an effort here, but that's honestly because I don't know where to start :/
     
  6. Oct 22, 2014 #5
    Start with determining the regions where ##b^2x^2>a^2y^2## and ##a^2y^2>b^2x^2##, can you see on what region you are integrating?
     
  7. Oct 22, 2014 #6
    so I would need to separate double integrals, divided by a line right? I just don't see how to obtain that line. Does that make any sense?
     
  8. Oct 22, 2014 #7

    Mark44

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    I think it's probably simpler than this. Max{b2x2, a2y2} chooses the larger of the two numbers.

    The region over which integration takes place is the rectangle whose corners are at (0, 0), (a, 0), (a, b), and (0, b). If b2x2 > a2y2, then bx > ay (since x and y are nonnegative and it's given that a and b are positive). In this case, what does the integrand simplify to? What part of the rectangle does bx > ay represent?

    On the other hand, if b2x2 < a2y2, then bx < ay. Same questions as before.
     
  9. Oct 22, 2014 #8

    Orodruin

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    I would start with a variable substitution, the integral is going to look much more symmetric after that... Then you can split it into two regions which will be equivalent and give the same contribution so you only need to compute one integral.
     
  10. Oct 22, 2014 #9
    So could I use bx = ay as a line segment that separates two non-intersecting regions and then set up two separate double integrals? One that looks like:
    [itex] \int_{0}^{a} \int_{0}^{\frac{bx}{a}} [/itex]
    And another:
    [itex] \int_{0}^{b} \int_{0}^{\frac{ay}{b}} [/itex]
     
  11. Oct 22, 2014 #10
    Yes.
     
  12. Oct 22, 2014 #11

    Mark44

    Staff: Mentor

    I hadn't thought about it quite like this, but I think that something like this will work, provided you have the right limits of integration.

    BTW, I've always found it easier to keep track of things in iterated integrals by writing in the variable in the limits of integration. Like this:
    $$ \int_{x = 0}^{a} \int_{y = 0}^{b} \dots dy dx$$
     
  13. Oct 22, 2014 #12
    Oh, I thought that is what you were hinting at by writing bx>ay and ay>bx.....Also, that is a good suggestion, I've found myself getting confused sometimes because I can't remember which limit belongs to which variable...
     
  14. Oct 22, 2014 #13
    Thank you!
     
  15. Oct 22, 2014 #14
    I'm afraid I'm not entirely sure what you mean...
    Edit: you just mean a simple u-sub?
     
  16. Oct 22, 2014 #15

    Orodruin

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    There is a trivial substitution which will make your integral area as well as the integrand symmetric in the integration variables. If you do it, both of the split integrals will be the same and you can multiply by two instead of doing two (admittedly very similar) integrals.
     
  17. Oct 22, 2014 #16
    Awesome, thanks for the help..
     
  18. Oct 22, 2014 #17
    Ok, I ended up solving using the integrals I posted earlier (unfortunately, I couldn't figure out the sub Orodruin referred to, perhaps he'd be so kind as to tell me). Here is my solution:
    [itex] \int_{0}^{a} \int_{0}^{\frac{bx}{a}} e^{b^{2}x^{2}} dy \hspace{1 mm} dx + \int_{0}^{b} \int_{0}^{\frac{ay}{b}} e^{a^{2}y^{2}} dx \hspace{1 mm} dy [/itex]
    [itex] = \int_{0}^{a} \frac{bx}{a} e^{b^{2}x^{2}} \,dx + \int_{0}^{b} \frac{ay}{b} e^{a^{2}y^{2}}\,dy [/itex]
    [itex] = \frac{b}{a} \int_{0}^{a} xe^{b^{2}x^{2}}\,dx + \frac{a}{b} \int_{0}^{b} ye^{a^{2}y^{2}}\,dy [/itex]
    Here, I performed a u-sub on each integral, for the first integral I did:
    [itex] u = b^{2}x^{2} \hspace{2 mm} du = 2b^{2}x dx [/itex]
    For the second integral:
    [itex] u = a^{2}y^{2} \hspace{2 mm} du = 2a^{2}y dy [/itex]
    So then the integrals become:
    [itex] = \frac{b}{2b^{2}a}\int_{0}^{a} 2b^{2}xe^{b^{2}x^{2}}\,dx + \frac{a}{2a^{2}b}\int_{0}^{a} 2a^{2}ye^{a^{2}y^{2}}\,dy [/itex]
    [itex] = \frac{b}{2b^{2}a}[e^{b^{2}a^{2}}-1] + \frac{a}{2a^{2}b}[e^{b^{2}a^{2}}-1] [/itex]
    And finally, I arrived at my final answer:
    [itex] = \frac{(e^{b^{2}a^{2}})-1}{2ab}[/itex]
     
    Last edited: Oct 22, 2014
  19. Oct 22, 2014 #18

    Orodruin

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    Try ##x = at## and ##y = bs##. The integral in the end is the same, but you only need to do it once and have less constants to worry about.
     
  20. Oct 22, 2014 #19
    thanks a lot, I'll try it out
     
  21. Oct 22, 2014 #20
    Also I just realized I made a mistake, the final answer should be:
    [itex] \frac{(e^{a^{2}b^{2}}-1)}{ab} [/itex]
     
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