Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How do we integrate with respect to (dx)^2 (not dx)?

  1. Nov 4, 2013 #1
    [tex] ∫x(dx)^2=?[/tex]

    and what is the difference between dx^2 and (dx)^2?
     
  2. jcsd
  3. Nov 4, 2013 #2

    pwsnafu

    User Avatar
    Science Advisor

    Where's this from? I've seen ##dx^2## but not ##(dx)^2##. The only way I can parse the latter is with wedge products
     
    Last edited: Nov 4, 2013
  4. Nov 4, 2013 #3

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The integral is zero.
    i.e. it has no contribution to the overall sum.

    But lets be careful: where have you seen this come up?
     
  5. Nov 5, 2013 #4
    Isn't that just short-hand notation for a double integral, i.e.,

    [tex]\iint x dx dx[/tex]
     
  6. Nov 5, 2013 #5

    pwsnafu

    User Avatar
    Science Advisor

    I don't think you are allowed to have the same variable for multiple integrals. A double integral is integrating with respect to the area element.
     
  7. Nov 5, 2013 #6
    Ok, but the reason I suggested this is because I recall in one of my books on integral equations, the author, when converting from IVP to Volterra Equations, use the syntax such as:

    [tex]\int_0^x \int_0^x f(t) dt dt=\int_0^x (x-t) f(t)dt[/tex]

    however, to be fair, he never used the syntax [itex](dt)^2[/itex] so if I am in error, my apologies.
     
  8. Nov 5, 2013 #7
    I was trying to find the moment of inertia of a disk rotating at the axis through its center and perpendicular to it.

    i got an integral for some function with (dx)^2 in it.
     
  9. Nov 5, 2013 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I strongly suspect that you have misread
    [tex]\int_0^x\int_0^x f(t)d\tau dt[/tex]

    Neither "[itex]dtdt[/itex]" nor "[itex](dx)^2[/itex]" has any meaning.
     
  10. Nov 5, 2013 #9
    With all due respect Hall, I have not. It's from "A First Course in Integral Equations" by Abdul-Majid Wazwaz. It's on p. 20 and includes also, the expression:

    [tex]\int_0^x \int_0^x \int_0^x f(t)dt dt dt=1/2 \int_0^x (x-t)^2 f(t)dt[/tex]
     
    Last edited: Nov 5, 2013
  11. Nov 5, 2013 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Then it is just bad notation.
     
  12. Nov 5, 2013 #11
    Dear lord, that's awful. I'd read the lhs as
    [tex]\int_0^x \int_0^x \int_0^x f(t)dt dt dt
    = \int_0^x \int_0^x \left( \int_0^x f(t)dt \right) dt_1 \; dt_2
    = \int_0^x \int_0^x H(x) dt_1 \; dt_2
    = x^2 H(x)
    = x^2 \int_0^x f(t) dt [/tex]

    while, what was probably meant is
    [tex]
    \int_0^x \int_0^{t_2} \int_0^{t_1} f(t) dt \; dt_1 \; dt_2 = \frac{1}{2} \int_0^x (x-t)^2 f(t) dt.
    [/tex]
     
  13. Nov 5, 2013 #12
    Yikes, that's awful notation.
     
  14. Nov 5, 2013 #13

    pwsnafu

    User Avatar
    Science Advisor

    That's the Cauchy formula for repeated integration. And you are not allowed to write it as dtdtdt. You must keep the variables separate.
    ##\int_{a}^x \int_a^{t_1}\ldots \int_a^{t_{n-1}}f(t_n) \, dt_n \, dt_{n-1} \ldots dt_1 = \frac{1}{(n-1)!} \int_a^x (x-t)^{n-1} f(t) \, dt.##
     
    Last edited: Nov 5, 2013
  15. Nov 5, 2013 #14
    Ok. Thanks! That makes sense to me now. Sorry for causing trouble.
     
  16. Nov 5, 2013 #15

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Please show how you derived that.
     
  17. Nov 5, 2013 #16
    [tex]I=\int_0^R dmr^2 dm=σπ[(r+dr)^2-r^2][/tex]
    and then I put [tex]dm[/tex] in first.
     
    Last edited: Nov 6, 2013
  18. Nov 5, 2013 #17
    [tex]I=\int_0^R dmr^2 ;dm=σπ[(r+dr)^2-r^2][/tex]
    and then I put [tex]dm[/tex] in first.
     
    Last edited: Nov 6, 2013
  19. Nov 6, 2013 #18

    pwsnafu

    User Avatar
    Science Advisor

    What does ##(r+dr)^2## mean?

    Edit: I think it should read ##dm = 2\sigma \pi r \, dr##, which is compatible with what you wrote if we take ##(dr)^2 = 0##.

    Was there a textbook or a teacher who said you could square differentials? What made you think ##(r+dr)^2## is well defined?
     
    Last edited: Nov 6, 2013
  20. Nov 6, 2013 #19
    i tried to find the area of a ring by subtracting the area of smaller circle from larger circle.
    The difference in the radii is dr.

    area of ring [tex] =π[(r+dr)^2 - r^2][/tex]
     
  21. Nov 6, 2013 #20

    pwsnafu

    User Avatar
    Science Advisor

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How do we integrate with respect to (dx)^2 (not dx)?
  1. Integrating dx/dv^2 ? (Replies: 8)

  2. Integration of dx^2 (Replies: 8)

Loading...