polar integrals Watch

sarahisme
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#1
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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:

 I = \int_0^{2\pi}\int_{-1}^1f(r)rdrd\theta


for (b) i get:

 I = \int_{-\pi/4}^{\pi/4}\int_{\frac{-1}{cos\theta}}^{\frac{1}{cos\the  ta}}f(r)rdrd\theta

for (c) i get:

 I = \int_{0}^{2\pi}\int_{0}^{cos\the  ta}f(tan\theta)rdrd\theta

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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FWoodhouse
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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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sarahisme
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(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)....

 I = \int_0^{2\pi}\int_{0}^1f(r)rdrd\  theta

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

 I = \int_0^{2\pi}\int_{0}^1f(r)rdrd\  theta

is that is as 'simplified' as much as possible though?
Looks good to me.
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sarahisme
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(Original post by FWoodhouse)
Looks good to me.
sweet. ok, how about this for (b) and (c)....?

(b)   I = 2\int_{-\pi/4}^{\pi/4}\int_0^{\frac{1}{cos\theta}} f(r)rdrd \theta  ]

(c) ] I = \int_{0}^{2\pi}\int_{0}^{cos\the  ta}f(tan\theta)rdrd\theta ]
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jpowell
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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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FWoodhouse
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(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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sarahisme
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ok things are getting confusing.... this is what i have done for part (b) now:

when put into polar coordinates:
 \int \int f(\sqrt{x^2+y^2})dxdy = \int \int f(r)rdrd\theta

And then using the condtions that  |y| \leq |x|

So then  |rsin\theta| \leq |rcos\theta|

 \Rightarrow |tan\theta| \leq 1

So  -\pi/4 \leq \theta \leq \pi/4

So there are the limits for theta.

Then using the other condition that  |x| \leq 1

 |rcos\theta| \leq 1

 \frac{-1}{cos\theta} \leq r \leq \frac{1}{cos\theta}

So then the integral becomes…

 I = \int^{\pi/4}_{-pi/4} \int^{1/cos\theta}_{-1/cos\theta} f(r)rdrd\theta
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FWoodhouse
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But the integral is of f(\sqrt{x^2 + y^2}), i.e. the argument of the function is always positive, so you must have r > 0 at all times.

Also, |tan \theta| < 1 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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sarahisme
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(Original post by FWoodhouse)
But the integral is of f(\sqrt{x^2 + y^2}), i.e. the argument of the function is always positive, so you must have r > 0 at all times.

Also, |tan \theta| < 1 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?

 I = \int^{\pi/4}_{-3\pi/4} \int^{1/cos\theta}_{0} f(r)rdrd\theta

?????????
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