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anonymous
 5 years ago
Double integrals in polar coordinates??
I Know nothing about it :S
anonymous
 5 years ago
Double integrals in polar coordinates?? I Know nothing about it :S

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Owlfred
 5 years ago
Best ResponseYou've already chosen the best response.0Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sometimes integrals are easier to evaluate using polar coordinates (r, θ), where x = r cos θ and y = r sin θ. that is the basic idea behind it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0aha,, and how I can apply it ? can you give me an example ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am not that good with it, just looking at my lecture notes, but I can try

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The height of the hemisphere about the xyplane is given by z =sqrt(1 − x^2y^2) Find the volume under the unit hemisphere with positive zcoordinate and positive ycoordinate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0region of integration is 1> x^2+y^2 (as it has to be + under sqrt) and y>=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now first challenge is to change the region of integration

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^2+y^2 in polar is =r^2 (as sin^2 +cos^2=1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so r has to be between 0 and 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y>0 implies that rsinθ>0 r is always >0 here (0 to 1) so sinθ has to be >0 implies θ is from 0 to pi clear so far?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{\pi}\int\limits_{0}^{1}\sqrt{1r ^{2}} drd \theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so this is what we have to do now

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is not, because of the change of integral dxdy becomes r drdθ

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[rsqrt{(1r ^{2})} \] this is what you have to integrate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one minute, hoe it comes r from 0 to 1 ,, since r^2<1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01x^2y^2 has to be + as it is under sqrt x^2+y^2= (rcosθ)^2+(rsinθ)^2=r^2 so 1r^2 has to be +

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01r^2>0 1>r^2 1<r<1 ??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u are right but the example says: positive zcoordinate and positive ycoordinate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks alot I understood the idea

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have another ques. ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you or do you not? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Im just a 1st year maths student, so my knowledge has limits :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how to find the projection of a=(1,3) on b=(4,0,2) ??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0post it as a different question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, thanks for trying :)
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