Double integrals in polar coordinates??
I Know nothing about it :S

- nice

Double integrals in polar coordinates??
I Know nothing about it :S

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- 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

aha,, and how I can apply it ?
can you give me an example ?

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## More answers

- anonymous

I am not that good with it, just looking at my lecture notes, but I can try

- 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

region of integration is 1> x^2+y^2 (as it has to be + under sqrt)
and y>=0

- anonymous

now first challenge is to change the region of integration

- nice

aha...

- anonymous

x^2+y^2 in polar is =r^2 (as sin^2 +cos^2=1)

- anonymous

so r has to be between 0 and 1

- 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

aha ..

- anonymous

\[\int\limits_{0}^{\pi}\int\limits_{0}^{1}\sqrt{1-r ^{2}} drd \theta\]

- anonymous

so this is what we have to do now

- anonymous

it is not, because of the change of integral dxdy becomes r drdθ

- anonymous

\[rsqrt{(1-r ^{2})} \]
this is what you have to integrate

- anonymous

do you know how?

- nice

one minute, hoe it comes r from 0 to 1 ,, since r^2<1

- nice

how**

- 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

1-r^2>0
1>r^2
-1

- anonymous

u are right but the example says:
positive z-coordinate and
positive y-coordinate.

- nice

ahaaa
I get it

- anonymous

the answer is 1/3 pi

- nice

Thanks alot
I understood the idea

- anonymous

cool!

- nice

I have another ques. ?

- anonymous

do you or do you not? :)

- anonymous

go on

- nice

in another topic

- anonymous

Im just a 1st year maths student, so my knowledge has limits :)

- anonymous

but I can try

- nice

how to find the projection of a=(1,3) on b=(4,0,2) ??

- anonymous

no clue

- anonymous

post it as a different question

- nice

ok, thanks for trying :)

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