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Double Integral - polar coordinates

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]
    \displaystyle\int\int\sqrt{4-x^2-y^2} dA
    [/tex]

    [tex]
    R{(x,y)|x^2+y^2\leq4 .. 0\leq x}
    [/tex]

    3. The attempt at a solution

    So far i have:


    [tex]
    \displaystyle\int^{\pi}_{0}\int^{r}_{0}\sqrt{4-r^2} rdrd\theta
    [/tex]

    Solving i get:



    [tex]
    \displaystyle\int^{\pi}_{0}\frac{-1}{3}(4-r^2)^{\frac{3}{2}}+\frac{1}{3}(4)^{\frac{3}{2}}d\theta
    [/tex]

    Am i on the right track, and if so how do i integrate the third root term?


    edit//

    ok i forgot the radius goes from 0 to 2... if i sub 2 into r, i can integrate it. But just to check, am i still doing this correctly?
     
    Last edited: Apr 25, 2010
  2. jcsd
  3. Apr 25, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Yes, you should integrate from zero to 2.
     
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