Sarahisme Posted August 21, 2006 Share Posted August 21, 2006 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 Link to comment Share on other sites More sharing options...

Norman Albers Posted August 21, 2006 Share Posted August 21, 2006 Radius does not go through zero unless you've got some truly cool physics. Link to comment Share on other sites More sharing options...

Sarahisme Posted August 21, 2006 Author Share Posted August 21, 2006 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? Link to comment Share on other sites More sharing options...

matt grime Posted August 21, 2006 Share Posted August 21, 2006 It does, but your r's go from minus something to plus something, and negative radii are not used. Link to comment Share on other sites More sharing options...

Sarahisme Posted August 21, 2006 Author Share Posted August 21, 2006 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] Link to comment Share on other sites More sharing options...

Norman Albers Posted August 21, 2006 Share Posted August 21, 2006 Reconsider your limits on the second one. There is less here than meets the eye in the stated domain. Link to comment Share on other sites More sharing options...

Norman Albers Posted August 21, 2006 Share Posted August 21, 2006 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 +/-. Link to comment Share on other sites More sharing options...

Sarahisme Posted August 22, 2006 Author Share Posted August 22, 2006 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] Link to comment Share on other sites More sharing options...

Norman Albers Posted August 22, 2006 Share Posted August 22, 2006 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. Link to comment Share on other sites More sharing options...

Sarahisme Posted August 22, 2006 Author Share Posted August 22, 2006 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... Link to comment Share on other sites More sharing options...

Norman Albers Posted August 22, 2006 Share Posted August 22, 2006 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). Link to comment Share on other sites More sharing options...

Sarahisme Posted August 23, 2006 Author Share Posted August 23, 2006 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 Link to comment Share on other sites More sharing options...

Norman Albers Posted August 23, 2006 Share Posted August 23, 2006 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. Link to comment Share on other sites More sharing options...

Sarahisme Posted August 24, 2006 Author Share Posted August 24, 2006 [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] Link to comment Share on other sites More sharing options...

Norman Albers Posted August 24, 2006 Share Posted August 24, 2006 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. Link to comment Share on other sites More sharing options...

Severian Posted August 24, 2006 Share Posted August 24, 2006 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] Link to comment Share on other sites More sharing options...

Norman Albers Posted August 24, 2006 Share Posted August 24, 2006 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. Link to comment Share on other sites More sharing options...

Norman Albers Posted August 24, 2006 Share Posted August 24, 2006 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. Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now