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anonymous
 5 years ago
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(4x^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)
anonymous
 5 years ago
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(4x^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)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you know the answer? final answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0answer choices: a) 18 b) 15pi/2 c) 8pi d) 4pi e) none of the above but I don't know the final answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmmmm let me check my working first. are you 2nd year?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my sister have this in her 2nd year, so idk if the problem in 3rd year is different.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no no, it's 8pi, answer is c

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you, but can you tell me how to convert the bounds or how you got the answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, theta bound: y bounds from y=0 to y=sqrt(4x^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
 5 years ago
Best ResponseYou've already chosen the best response.0r bound: r runs from center (0,0) to circle x^2+y^2=4 so 0<=r<=2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so: omega={ (r,theta): 0 <= r <=2, 0 <= theta <= pi } then do the integration.
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