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A doubt about double integration

  1. Aug 20, 2004 #1
    Salutations!

    This is my first post. I´m writing here because I have a doubt regarding a solved problem of double integration present in a book of mine. I don´t speak english yet ( :frown: ), but I will try to translate the problem to english. Well, here I go:

    "Calculate int[ (x^2 + y^2) dx dy ] over the region D on the first quadrant of the xy-plane limitated by the hyperboles x^2 - y^2 = 1, x^2 - y^2 = 9, x*y = 2 and x*y = 4".

    Well, using the following transformation: u = x^2 - y^2, v = x*y; we obtain the following Jacobian: J(u,v) = 1 / ( 2 * (u^2 + 4*v^2)^(1/2) ).

    The region of integration Q on the uv-plane is the rectangle: Q = {(u,v) E R² | 1 <= u <= 9 , 2 <= v <= 4}.

    So,

    int[ (x^2 + y^2) dx dy ] = int[ (u^2 + 4*v^2) * ( 1 / ( 2 * (u^2 + 4*v^2)^(1/2) ) ) du dv ] = (1/2) * int[ du dv ] = 8.

    Well, I don´t understand this solution. If you plot the region Q, you will obtain a rectangle lying on the uv-plane, but this rectangle includes (through the previous change of coordinates) the two regions lying on the xy-plane that are limitated by the four hyperboles: the first one lying on the first quadrant, and the second one lying on the third quadrant. But I want only the region on the first quadrant. I can´t see how the definition of the rectangle Q could exclude the region on the third quadrant. And this region must to be excluded, because the change of coordinates must to be injective (otherwise I can´t use the previous method to calculate double integrals).

    I don´t know if you could understand what I wrote, or even if my doubt was clearly exposed (in case of my horrible english have been understood). But I thank in advance for any help! :smile:
     
  2. jcsd
  3. Aug 21, 2004 #2
    Theatre of Fate, welcome to physics forums!

    I don't speak english neither, so I wish you would understand what I say.

    I think what I get from your post is that you are not at all sure that the transformation you use is one-one (because each hyperbola has two disconnected components and each point in (u,v) seems to "correspond" to two points in (x,y) ) and so you are not sure that the "change of variable" formula for integration works.

    I think the point is that we may "restrict" the map/transformation to the first quadrant of the (x,y) plane. Then the map will be one-one and there will be no problem. Specifically, when calculating the Jacobian of the transformation, you need to express (x,y) in terms of (u,v). The transformation is, u = x^2 - y^2, v = x*y. When expressing (x,y) in (u,v), you will find that there is more than one solution (differring by +/- sign). We choose specifically the solution which maps (u,v) to the first quadrant of (x,y) plane and we will get the Jacobian as you mentioned.

    Hope that helps.
     
  4. Aug 21, 2004 #3
    Here's another question regarding double integration:

    I was reading through a proof showing that the area under the standard normal curve = 1, and there's one part of the proof I couldn't follow:

    [tex]I_{x} = \int_ {-\infty}^\infty \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx [/tex]

    [tex]I_{y} = \int_ {-\infty}^\infty \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(y-\mu)^2}{2\sigma^2}} dy[/tex]

    Then you multiply [tex]I_{x}[/tex] with [tex]I_{y}[/tex]:
    [tex]I_{x} \cdot I_{y} = \int_ {-\infty}^\infty \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} \int_ {-\infty}^\infty \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(y-\mu)^2}{2\sigma^2}} dxdy[/tex]

    Then the book goes on to say (or at least I think it says this since I don't have it with me now):

    Since I(x) and I(y) are the same except for the choice of variables:

    [tex]I^2 = \int_ {-\infty}^\infty \int_ {-\infty}^\infty \frac{1}{\sigma^2 2\pi}e^{\frac{-(x-\mu)^2-(y-\mu)^2}{2\sigma^2}[/tex]

    Why is the last step valid? I couldn't find a proof for it.
     
  5. Aug 21, 2004 #4

    Let me just say that you both speak English well enough to be easily understood. :biggrin:

    -Ray.
     
  6. Aug 21, 2004 #5

    arildno

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    Let's first look at a somewhat more general case:
    The assertion is that (for constants a,b,c,d):
    [tex]\int_{a}^{b}\int_{c}^{d}f(x)g(y)dxdy=\int_{a}^{b}f(x)(\int_{c}^{d}g(y)dy)dx=[/tex]
    [tex]\int_{c}^{d}g(y)(\int_{a}^{b}f(x)dx)dy=\int_{a}^{b}f(x)dx\int_{c}^{d}g(y)dy[/tex]

    The equality between the 1. and the 2.&3. terms are consequences of Fubini's theorem (essentially, that double integrals can be computed stepwise; that is we may first integrate with respect to one variable, keeping the other constant, and after that, integrating with respect to the last variable)
    . The equality between the 2&3. terms and the 4.term is a consequence that the integral placed in parentheses within the other integral is a constant,
    which therefore may be extracted from the outer integral.
     
  7. Aug 25, 2004 #6
    Thanks for the reply, but could you explain this in more detail? I don't get what you are trying to say. Thanks.
     
  8. Aug 25, 2004 #7

    HallsofIvy

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    Your English is excellent. Its my (put the language of your choice here) that is terrible!
     
  9. Jun 11, 2005 #8
    Wong, I thank you for the explanation!
     
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