anonymous
  • anonymous
Hey guys Solve the double integral by first converting to polar coordinates (for no apparent reason)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1370937938423:dw|
experimentX
  • experimentX
\[ \int_0^2 \int_0^\pi r^2 r d\theta dr\]
dan815
  • dan815
|dw:1370939222012:dw|

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dan815
  • dan815
callisto u know what ur name means?
dan815
  • dan815
|dw:1370939374299:dw|
anonymous
  • anonymous
Starting with the Y integration\[y1=0 \rightarrow r1=0\]\[y2=\sqrt{4-x^2}\]\[y2^2=4-x^2\]\[x^2+y2^2=4\]\[r2^2=4\]\[r2=2\]\[R=\int\limits_{0}^{2}\]
dan815
  • dan815
yes
anonymous
  • anonymous
Then the X integration\[x1=r1cos \theta \]\[-2=r2\cos \theta \]\[r1=-2\sec \theta \]\[\rightarrow r2=2\sec \theta \] \[\theta = \int\limits_{-2\sec \theta}^{2\sec \theta }\]
dan815
  • dan815
wut no forget that stuff
anonymous
  • anonymous
IDK how to do that portion
dan815
  • dan815
just think in polar coords
anonymous
  • anonymous
So if I have for example -2 to 2 should I just automatically think from 0 to pi?
dan815
  • dan815
|dw:1370940131643:dw|
dan815
  • dan815
u only look at the inital bounds to see what the region u are integrating over
anonymous
  • anonymous
So I should always usually be able to visualize the thetas?
dan815
  • dan815
yeah
anonymous
  • anonymous
Aight, you earned it :D
dan815
  • dan815
sometimes u might not be able but then they usually give u a visualizable radius and u can write theta as a function of radius
anonymous
  • anonymous
Grr you should medal me, I'm only one ranking away from my dream number
anonymous
  • anonymous
;D
anonymous
  • anonymous
Dream Number.. AHa I see what you mean :D

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