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Problems on double integrals

  1. Jan 18, 2016 #1
    I started learning double integrals in this semester but I got some problems on it.
    Is it matter if the order of integration interchange?
    For example, V=∫f(x,y)dydx, can we rewrite the equation as V=∫f(x,y)dxdy?
    If not, could you explain on it little bit:sorry:
    Thanks!
     
  2. jcsd
  3. Jan 18, 2016 #2

    blue_leaf77

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    In terms of appearance, ##dx\hspace{1mm}dy## and ##dy\hspace{1mm}dx## bear no difference in mathematical meaning. However, the order with which the integral calculation is carried out, in general, cannot be interchanged without changing the given form of the integration limits. For example, a question asks you to integrate a function ##f(x,y) = y## over an area bounded by ##x^2+y^2 \leq 9##, ##x>0##, and ##y>0##, which is the first quadrant of a circle with radius 3. The limits of the integral is not fixed for any value of ##x## and ##y##. In this case there are two ways to perform the integral, the first one is to integrate over a vertical stripe located at a fixed ##x## (which must lies within the said limiting area) and then integrate over ##x## from 0 to 3. The first integral over the stripe is obviously performed along ##y## direction from ##y=0## to ##y=\sqrt{9-x^2}##, this looks like
    $$
    \int_0^3 \int_0^\sqrt{9-x^2} ydy\hspace{1mm}dx
    $$
    In that integral, you will calculate the ##y## integration first followed by ##x## integration. However, you can also flip the order by integrating over ##x## first. But if you do this, you will have to change the limit as well. This means, you will now be calculating integral over a horizontal stripe of fixed ##y## first from ##x=0## to ##x=\sqrt{9-y^2} ##, and then integral over ##y## from 0 to 3.
    $$
    \int_0^3 y\int_0^\sqrt{9-y^2} dx\hspace{1mm}dy
    $$
    You should get identical answers from both paths. Note that, after you learn change of coordinate, you will see that this integral can be done in polar coordinates with fixed integral limits, which means the order of integration will not matter anymore.
     
  4. Jan 19, 2016 #3

    Svein

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    Yes, usually you can switch the order of integration, but not always. The restrictions are described in Fubini's theorem (https://en.wikipedia.org/wiki/Fubini's_theorem).
     
  5. Jan 19, 2016 #4

    Ssnow

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  6. Jan 20, 2016 #5
    The iterated integrals need not be the same — even on a domain of integration that is "both x-simple and y-simple" — it if it permitted to be infinite in at least one direction.

    Here is a standard counterexample: consider this function:

    f(x,y) = e−xy − 2e−2xy

    on the domain 1 ≤ x < ∞, 0 ≤ y ≤ 1.

    Note that f(x,y) > 0 everywhere on this domain.

    The two iterated integrals cannot be evaluated explicitly in terms of elementary functions, but using numerical approximation I get

    ∫ (∫ f(x,y) dy) dx ≈ 0.17048+

    whereas

    ∫ (∫ f(x,y) dy) dx ≈ 0.52266+
    , showing that Fubini's theorem can fail unless certain conditions are met — see Svein's link above to the Wikipedia article.
     
  7. Jan 21, 2016 #6

    Svein

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    Sorry: f(x,0) = -1...
     
  8. Jan 21, 2016 #7
    Oops, did I say "Note that f(x,y) > 0 everywhere on this domain" ?

    Wrong! Sorry about that.

    Setting

    f(x,y) = e−xy − 2e−2xy = 0 ​

    to see where f(x,y) changes sign, first multiply by e2xy, which definitely is positive everywhere, to get:

    exy = 2​

    and so the expression changes sign along the locus of

    xy = ln(2)​

    where that passes through the domain 1 ≤ x < ∞, 0 ≤ y ≤ 1. In other words, the curve

    S = {(x, ln(2)/x) | x ≥ 1}.​

    (To prove this, it suffices to check that the critical points of the graph of z = f(x,y) do not intersect the curve S. This is left to the reader.

    Numerical experiment suggests the minimum on the domain 1 ≤ x < ∞, 0 ≤ y ≤ 1 is -1 (the value taken at all points (x, 0) for x ≥ 1).

    Tricky!
     
  9. Jan 22, 2016 #8

    Ssnow

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    In my answer I did not regard the improper integrals ...
     
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