nice
  • nice
Double integrals in polar coordinates?? I Know nothing about it :S
Mathematics
schrodinger
  • schrodinger
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Owlfred
  • Owlfred
Hoot! 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
  • anonymous
Sometimes 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
nice
  • nice
aha,, and how I can apply it ? can you give me an example ?

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anonymous
  • anonymous
I am not that good with it, just looking at my lecture notes, but I can try
anonymous
  • anonymous
The height of the hemisphere about the x-y-plane is given by z =sqrt(1 − x^2-y^2) Find the volume under the unit hemisphere with positive z-coordinate and positive y-coordinate.
anonymous
  • anonymous
region of integration is 1> x^2+y^2 (as it has to be + under sqrt) and y>=0
anonymous
  • anonymous
now first challenge is to change the region of integration
nice
  • nice
aha...
anonymous
  • anonymous
x^2+y^2 in polar is =r^2 (as sin^2 +cos^2=1)
anonymous
  • anonymous
so r has to be between 0 and 1
anonymous
  • anonymous
y>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?
nice
  • nice
aha ..
anonymous
  • anonymous
\[\int\limits_{0}^{\pi}\int\limits_{0}^{1}\sqrt{1-r ^{2}} drd \theta\]
anonymous
  • anonymous
so this is what we have to do now
anonymous
  • anonymous
it is not, because of the change of integral dxdy becomes r drdθ
anonymous
  • anonymous
\[rsqrt{(1-r ^{2})} \] this is what you have to integrate
anonymous
  • anonymous
do you know how?
nice
  • nice
one minute, hoe it comes r from 0 to 1 ,, since r^2<1
nice
  • nice
how**
anonymous
  • anonymous
1-x^2-y^2 has to be + as it is under sqrt x^2+y^2= (rcosθ)^2+(rsinθ)^2=r^2 so 1-r^2 has to be +
nice
  • nice
1-r^2>0 1>r^2 -1
anonymous
  • anonymous
u are right but the example says: positive z-coordinate and positive y-coordinate.
nice
  • nice
ahaaa I get it
anonymous
  • anonymous
the answer is 1/3 pi
nice
  • nice
Thanks alot I understood the idea
anonymous
  • anonymous
cool!
nice
  • nice
I have another ques. ?
anonymous
  • anonymous
do you or do you not? :)
anonymous
  • anonymous
go on
nice
  • nice
in another topic
anonymous
  • anonymous
Im just a 1st year maths student, so my knowledge has limits :)
anonymous
  • anonymous
but I can try
nice
  • nice
how to find the projection of a=(1,3) on b=(4,0,2) ??
anonymous
  • anonymous
no clue
anonymous
  • anonymous
post it as a different question
nice
  • nice
ok, thanks for trying :)

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