Double Integral - polar coordinates

  • Thread starter exidez
  • Start date
  • #1
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Homework Statement



[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]

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:

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
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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?

Yes, you should integrate from zero to 2.
 

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