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Multivariable integral with unclear bounds

  1. Aug 22, 2016 #1
    1. The problem statement, all variables and given/known data

    Hi everybody! I'm trying to solve the following problem and I am unsure about what I did:

    Calculate ##\int_M f(x,y) dx dy## with ##M = \{ (x,y) \in \mathbb{R}^2: x^2 + y^2 \leq 1, x \geq 0, y \geq 0 \}## and ##f(x,y) = xy^2##.

    2. Relevant equations

    One equation I'd like to discuss comes from the script of my teacher and is:

    ##\int_{x^2 + y^2 \leq R^2} f(x,y) dx dy = \int_0^R r \bigg( \int_0^{2π} f(r \cos \phi, r \sin \phi) d \phi \bigg) dr##

    He claims that works when ##f: \mathbb{R}^2 \to \mathbb{R}## is continuous and ##R > 0##. In general the conversion to polar coordinates is no problem for me, but here I'm wondering where does the first ##r## on the right-hand side come from?

    3. The attempt at a solution

    So at first I tried to use the equation given above without too much thinking, and I repeatedly came up with ##0## as a result. Then I tried to derive the bounds from the conditions present in ##M##, and here's my go:

    ##x^2 + y^2 \leq 1 \implies r^2 (\cos^2 \phi + \sin^2 \phi) \leq 1 \implies 0 \leq r \leq 1##

    ##r \cos \phi \geq 0 \implies \cos \phi \geq 0 \implies -\frac{\pi}{2} \leq \phi \leq \frac{\pi}{2}##
    ##r \sin \phi \geq 0 \implies \sin \phi \geq 0 \implies 0 \leq \phi \leq \pi##
    ##\implies 0 \leq \phi \leq \frac{\pi}{2}##

    Is that correct? Then I solved the integral with the bounds that I found:

    ##\int_M f(x,y) dx dy = \int_0^{\pi/2} \bigg( \int_0^1 r^4 \cos \phi \sin^2 \phi\ dr \bigg) d\phi##
    ##= \int_0^{\pi/2} \bigg[ \frac{r^5}{5} \cos \phi \sin^2 \phi \bigg]_0^1 d\phi##
    ##= \int_0^{\pi/2} \frac{1}{5} \cos \phi \sin^2 \phi\ d\phi##
    ##= \frac{1}{5} \bigg[ \frac{\sin^3 \phi}{3} \bigg]_0^{\pi/2}## (I skipped the substitution part)
    ##= \frac{1}{15}##

    As you can see, I put the ##r## of the equation above inside the integral, though I am not sure why it is there. Is that the correct way to tackle the problem in general?


    Thanks a lot in advance for your answers, always glad to read your comments.


    Julien.
     
  2. jcsd
  3. Aug 22, 2016 #2

    Mark44

    Staff: Mentor

    If I understand your question correctly, the "extra" r comes from conversion from an iterated integral in Cartesian (or rectangular) form, to polar form. In general, an integral of the form ##\int \int f(x, y) dx dy## becomes ##\int \int g(r, \theta) r~drd\theta##.
     
  4. Aug 22, 2016 #3
    Hi @Mark44 and thanks for your answer. Yes that was my first question, I can see now where this ##r## comes from. :)


    Julien.
     
  5. Aug 22, 2016 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Have you really never seen the change-of-variable formulas for double or triple integrals? If not, you should study that material; it is entirely standard, and of great importance in future studies; it also can save you hours of work.

    See, eg.,
    http://tutorial.math.lamar.edu/Classes/CalcIII/ChangeOfVariables.aspx
    or
    http://math.oregonstate.edu/home/pr...ulusQuestStudyGuides/vcalc/change/change.html
     
  6. Aug 23, 2016 #5
    Hi @Ray Vickson! No I had seen them, something in my head didn't connect when seeing that formula in the middle of the script of my teacher. But what about the integral? Did I do it correctly, especially the bounds bit?

    Thanks for the links. :)


    Julien.
     
  7. Aug 23, 2016 #6

    SammyS

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    Staff Emeritus
    Science Advisor
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    Gold Member

    Yes. It looks correct to me .
     
  8. Aug 23, 2016 #7
    @SammyS That's great, thanks a lot for reviewing my work!


    Julien.
     
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