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Homework Help: Change of variables for multiple integrals (2)

  1. Jan 18, 2008 #1
    Q1: Suppose B=[0,1]x[0,2]x[0,3]x[0,4] in R4, and that C=[0,1]x[0,1]x[0,1]x[0,1]. Given that
    ∫ ∫ ∫ ∫f(x)=d4x=(2pi)4
    What is the value of
    ∫ ∫ ∫ ∫ f(x1,2x2,3x3,4x4) d4x?

    Define x=G(u)=(u1,u2/2,u3/3,u4/4)

    ∫ ∫ ∫ ∫ f(x1,2x2,3x3,4x4) d4x

    by change of variables theorem,
    =(1/24) ∫ ∫ ∫ ∫ f(u1,u2,u3,u4) d4u

    =(1/24)∫ ∫ ∫ ∫ f(u)d4u


    I don't understand the last step...
    We are only given that
    ∫ ∫ ∫ ∫f(x)=d4x=(2pi)4
    NOT that
    ∫ ∫ ∫ ∫f(u)=d4u=(2pi)4
    I will tell you what I am thinking about. Here, u=(u1,u2,u3,u4) and x=(x1,x2,x3,x4) CANNOT be treated as dummy variables since they have a relationship x=G(u) used to define the transformation G, but the last step of the solution seems to treat that x=u, which makes me feel very uncomfortable...

    Can someone explain? I would really appreciate it!:smile:
    Last edited: Jan 18, 2008
  2. jcsd
  3. Jan 18, 2008 #2


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    Do you know what a "dummy variable" is?
    [tex]\int_0^1 f(x) dx= \int_0^1 f(y)dy= \int_0^1 f(u)du[/tex]
    In other words, it doesn't matter what the variable is called.

    They have simply switched the first integral from "x" to "u" so as not to confuse it with the "x" in the second integral.
  4. Jan 18, 2008 #3
    Dummy variable means we can replace by another letter not already in use while not changing the answer, right?

    But x=G(u)=(u1,u2/2,u3/3,u4/4), there are explicit dependencies between u and x, does that matter?
  5. Jan 18, 2008 #4
    What if, instead of x = G(u), you had used t = G(u)?
  6. Jan 18, 2008 #5
    Well, I don't think this can happen at all...

    The question is
    "What is the value of
    ∫ ∫ ∫ ∫ f(x1,2x2,3x3,4x4) d4x?"

    So we need a transformation that involves x, namely, x=G(u), right?
  7. Jan 18, 2008 #6
    There is no difference between the two. It doesn't matter that the variable you end up with is different than the variable you started with. You have an integral that is of the exact same form. Suppose it were originally given as this:

    ∫ ∫ ∫ ∫f(z)d4z=(2pi)4

    Would that help you out?
  8. Jan 19, 2008 #7
    Yes, this will sort of help.

    The thing that really bothers me was that there is a relationship between x and u, so replacing x with u makes me feel uncomfortable...but I think I am OK now...
  9. Jan 19, 2008 #8


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    If you replace "x" by "t" (or any other symbol) and then make that replacement all the way through it should be obvious that it doesn't make any difference at all.
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