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

Having trouble understanding Domains and Multi Variable Functions

  1. Sep 28, 2011 #1
    Hey guys, I'm doing some multivariable calculus atm, and I need some help with the Domains of some multivariable functions...
    1) f(x,y) = 3x^2 + 2y
    The problem I'm having here is I basically forget the definition of domain... would it be for all x and y even though there are two whole quadrants this function doesn't hit? (looking at the x-y plane here..)

    2) f x(x,y) = 6x
    I'm looking at the partial derivative in regards to x here. Would I just assume this is a single variable function and only write the Domain in terms of x? If not, what would I write for the Domain?

    3)f y(x,y) = 2
    This is the partial derivative in terms of y, I know on the x-y plane it would be defined for all x... is it the same here?
  2. jcsd
  3. Sep 28, 2011 #2


    User Avatar
    Homework Helper

    The domain of a function is the set of all values for which the function is defined. Are there any real values of x and y for which you cannot compute 3x2 + 2y ?

    It is not a single variable function, or it wouldn't have been written as f(x,y) . If there were a term for f(x,y) which only depended on y alone, would it appear in fx ?

    Yes, same issue, different variable...

    ("Multivariate calculus ATM"? I have to evaluate a triple integral to get money out of the machine now!?)
  4. Sep 28, 2011 #3
    Hahaha atm/at the moment, same thing! (god wouldn't that be terrifying!!)

    Ok! I think I understand. I'm just trying to see where you CAN plug in numbers for...
    For 1) you can plug in any x,y and get an answer
    For 2) any x, y works, even though there's no y, it just means it works for any y
    For 3) any x, y works because it's just constant, like in single variable if we have y=2 it works for all x
    So for each function the Domain would be any x, y.

    Now if we were looking at the range for 3), would it be z is only defined at 2?
  5. Sep 28, 2011 #4


    User Avatar
    Homework Helper

    What fx = 6x implies is that after integrating with respect to x , we would have f( x, y ) = 3x2 + g(y) ; instead of an "arbitrary constant" as in single-variable integration, we get an "arbitrary function" which has no dependence on the variable of integration.

    So as far as the domain goes, f( x , y ) would be defined for all real x , provided that y is in the domain of g(y) . In other words, there could still be whole lines or bands which all not in the domain of f( x, y ) because g(y) is undefined for those values of y (regardless of whether terms involving x are defined).

    By the same reasoning, since fy = 2 , integration in y tells us that
    f( x , y ) = 2y + g(x) . We can always calculate 2y , but if g(x) = 1/x , for instance, then the domain of f( x , y ) would not include the line x = 0 .
  6. Sep 28, 2011 #5
    Ok! I get it now, thanks a lot for all the help! I'm starting to dread integration of multivariable functions now though :P
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook