# polar integralsWatch

#1
Hi all,

i am little puzzled by these questions, not sure if i am doing it correctly or just blundering along! :P

what do you guys think?

cheers

Sarah

for (a) i get:

for (b) i get:

for (c) i get:

if people want the working just shout, i just don't have quite enough time to put it up at the moment.

-sarah
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12 years ago
#2
For (a), be careful - as it stands, the limits you've got mean that you'll get twice what you want. You can change either the outer or inner limits.
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#3
(Original post by FWoodhouse)
For (a), be careful - as it stands, the limits you've got mean that you'll get twice what you want. You can change either the outer or inner limits.
hmm ok, how does this look now for (a)....

is that is as 'simplified' as much as possible though?
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12 years ago
#4
(Original post by sarahisme)
hmm ok, how does this look now for (a)....

is that is as 'simplified' as much as possible though?
Looks good to me.
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#5
(Original post by FWoodhouse)
Looks good to me.

(b) ]

(c) ]
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12 years ago
#6
b) Did you mean to put a factor of 2 in that? I think your first version without is correct.

c) Looks fine.
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12 years ago
#7
(Original post by jpowell)
b) Did you mean to put a factor of 2 in that? I think your first version without is correct.
That misses out half of the area to be integrated over. (Sketch the area in the x-y plane.) I'm not sure whether in this case it works just to double it, or whether you also need to integrate over r = 3pi/4 to r=5pi/4.
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#8
ok things are getting confusing.... this is what i have done for part (b) now:

when put into polar coordinates:

And then using the condtions that

So then

So

So there are the limits for theta.

Then using the other condition that

So then the integral becomes…

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12 years ago
#9
But the integral is of , i.e. the argument of the function is always positive, so you must have r > 0 at all times.

Also, has more solutions than the ones you gave. I recommend sketching the area to be integrated over in the x-y plane.
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#10
(Original post by FWoodhouse)
But the integral is of , i.e. the argument of the function is always positive, so you must have r > 0 at all times.

Also, has more solutions than the ones you gave. I recommend sketching the area to be integrated over in the x-y plane.
ok i think i sort of see what you mean....

how does this look for part (b) now?

?????????
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