polar integrals

[Sarahisme]

Hi all,

 

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

 

what do you guys think?

 

cheers

 

Sarah

 

 

for (a) i get:

 

[math] I = \int_0^{2\pi}\int_{-1}^1f®rdrd\theta [/math]

 

 

for (b) i get:

 

[math] I = \int_{-\pi/4}^{\pi/4}\int_{\frac{-1}{cos\theta}}^{\frac{1}{cos\theta}}f®rdrd\theta [/math]

 

for © i get:

 

[math] I = \int_{0}^{2\pi}\int_{0}^{cos\theta}f(tan\theta)rdrd\theta [/math]

 

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

 

-sarah

[Norman Albers]

Radius does not go through zero unless you've got some truly cool physics.

[Sarahisme]

Radius does not go through zero unless you've got some truly cool physics.

 

i thought i had it going from 0 to another value...?

 

which part of the question are you referring too btw?

[matt grime]

It does, but your r's go from minus something to plus something, and negative radii are not used.

[Sarahisme]

It does, but your r's go from minus something to plus something, and negative radii are not used.

 

right yep, i should have seen that.

 

ok so how does this look....??

 

(a) [math] I = \int_0^{2\pi}\int_{0}^1f®rdrd\theta [/math]

 

 

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

 

© [math] I = \int_{0}^{2\pi}\int_{0}^{cos\theta}f(tan\theta)rdrd\theta [/math]

[Norman Albers]

Reconsider your limits on the second one. There is less here than meets the eye in the stated domain.

[Norman Albers]

My apologies. I was outfoxed by the second statement. I guess the domain describes a square and you see this. Now, a few hours later it looks to me like you have one-fourth the answer, not one-half like you thought. You write for one quadrant, if both <x,y> go +/-.

[Sarahisme]

ok things are getting confusing.... this is what i have done for part (b) now:

 

when put into polar coordinates:

[math] \int \int f(\sqrt{x^2+y^2})dxdy = \int \int f®rdrd\theta [/math]

 

And then using the condtions that [math] |y| \leq |x| [/math]

 

So then [math] |rsin\theta| \leq |rcos\theta| [/math]

 

[math] \Rightarrow |tan\theta| \leq 1 [/math]

 

So [math] -\pi/4 \leq \theta \leq \pi/4 [/math]

 

So there are the limits for theta.

 

Then using the other condition that [math] |x| \leq 1 [/math]

 

[math] |rcos\theta| \leq 1 [/math]

 

[math] \frac{-1}{cos\theta} \leq r \leq \frac{1}{cos\theta} [/math]

 

So then the integral becomes…

 

[math] I = \int^{\pi/4}_{-pi/4} \int^{1/cos\theta}_{-1/cos\theta} f®rdrd\theta [/math]

[Norman Albers]

You have psyched out what makes me slow. I guess we have to break down the logic to independence and dependence, in the new set of variables. So it's not just all <x,y> in a square, but rather a chosen half of this! Congratulations, thanks for showing me something. The challenge is to know how to proceed in this process.

[Sarahisme]

You have psyched out what makes me slow. I guess we have to break down the logic to independence and dependence, in the new set of variables. So it's not just all <x,y> in a square, but rather a chosen half of this! Congratulations, thanks for showing me something. The challenge is to know how to proceed in this process.

 

 

lol, you've lost me...

[Norman Albers]

I'm trying to say that we try to separate integration variables but when they cannot separate you have to express the integration limits as you have. The domain being described in (b) is |y|'s less than |x|'s, and that is right triangles on their sides, yah? Thus one-half of a total square. Laughing is good - I am continually surprised at how I cannnot ass'u'me people see things similarly. I used the word 'dependent' because the range of |y| depends on |x|. This is a bit strange because they are both "independent variables" of f(x,y).

[Sarahisme]

I'm trying to say that we try to separate integration variables but when they cannot separate you have to express the integration limits as you have. The domain being described in (b) is |y|'s less than |x|'s, and that is right triangles on their sides, yah? Thus one-half of a total square. Laughing is good - I am continually surprised at how I cannnot ass'u'me people see things similarly. I used the word 'dependent' because the range of |y| depends on |x|. This is a bit strange because they are both "independent variables" of [i']f(x,y).[/i]

 

lol, ok i see. i like your method of thinking... its quite unique

[Norman Albers]

Why is this unique? I am 57 years old and an accomplished piano rebuilder with an MS(Stanford) and BS(Princeton) . Vocabulary changes but challenges of expression don't. How do you say it? Did you vizualize the domain or did you proceed as you showed me? I was outfoxed. Remember in high school it was the word problems to be expressed algebraically that separated the masses.

[Sarahisme]

[math] I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \int_0^{\frac{1}{cos(\theta)}} f®drd\theta + \int_{\frac{\pi}{4}}^{\frac{-\pi}{4}} \int_{\frac{-1}{cos(\theta)}}^{0} fdrd\theta [/math]

[Norman Albers]

Thanks for further lessons in laTex. Do you see the two triangles? You know what you're doing but it is a bad habit to write negative radii.

[Severian]

What is wrong with negative r? I would have written it as:

 

[math] I = \int^{\pi/4}_{-\pi/4} \int^{1/\cos\theta}_{-1/\cos\theta} f(|r|)|r|drd\theta [/math]

 

which is almost exactly what Sarah has (apart from the modulus).

 

Edit: After thinking about this some more, I agree with Norman. Formally you should be writing [math]r=x \cos \theta[/math] and [math]y=r \sin \theta[/math]. You could in principle allow r to become negative and restrict [math]\theta[/math] but it is nicer to restrict r, because then your Jacobian is [math]J=|r|=r[/math] and your integral becomes:

 

[math] I = \int^{\pi/4}_{-\pi/4} \int^{1/\cos\theta}_{0} f®rdrd\theta + \int^{5\pi/4}_{3\pi/4} \int^{1/\cos\theta}_{0} f®rdrd\theta [/math]

[Norman Albers]

It is consistent to cover the plane with radius positive and angle covering 2pi. It's just style, here, but since when do we speak physics of negative radii? String people do a neato thing doing 1/r beyond the small Planck limit, but I don't see physical sense in negative values. Like I said it's only a matter of self-consistent notation here.

[Norman Albers]

Thank you, Severian. I certainly had not recalled "Jacobian" here, but I dig it. The task at hand is to construct a one-to-one mapping of coordinate transform, in an efficient easily notated manner. We all have wonderfully convoluted minds so knock yourselves out conceptualizing things, but communicate them simply to me. [Now about the possibilities of negative radii... you probably don't want to open this box with me here, considering my latest ruminations on black holes, down in the kitchen of Speculations. I am on day 21 of editorial review at JMP] Sarahisme: All this exchange is good, and both of you have shown me how to notate things I never have. We must know formally how to follow our nose when we need to. We must trail good strings behind us when we wander into cubist landscapes.