anonymous
  • anonymous
Help with integration with polar coordinates? The main problem I'm having is finding the bounds in polar coordinates. The problem is: Integration of 3sqrt(x^2+y^2)dydx from 0 to sqrt(4-x^2) and from -2 to 2. If someone would tell me how to convert the bounds to polar coordinates, that would be great because I know the integrand converted is 3sqrt(r^2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
do you know the answer? final answer?
anonymous
  • anonymous
answer choices: a) 18 b) 15pi/2 c) 8pi d) 4pi e) none of the above but I don't know the final answer.
anonymous
  • anonymous
hmmmmm let me check my working first. are you 2nd year?

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anonymous
  • anonymous
3rd year
anonymous
  • anonymous
that was weird!
anonymous
  • anonymous
my sister have this in her 2nd year, so idk if the problem in 3rd year is different.
anonymous
  • anonymous
I got 16pi...
anonymous
  • anonymous
no no, it's 8pi, answer is c
anonymous
  • anonymous
thank you, but can you tell me how to convert the bounds or how you got the answer?
anonymous
  • anonymous
okay, theta bound: y bounds from y=0 to y=sqrt(4-x^2) <-> x^2+y^2=4=2^2 which is the equation of circle center (0,0) radius 2 so if you draw this on paper, you'll se y bounds from y=0 to upper hafl of the circle. then you can see theta run from 0 to pi 0<=theta<=pi
anonymous
  • anonymous
r bound: r runs from center (0,0) to circle x^2+y^2=4 so 0<=r<=2
anonymous
  • anonymous
so: omega={ (r,theta): 0 <= r <=2, 0 <= theta <= pi } then do the integration.

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