Surface Integral, Polar Coordinates

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  • #1
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Homework Statement


Express [itex]f(x,y) = \frac{1}{\sqrt{x^{2} + y^{2}}}\frac{y}{\sqrt{x^{2} + y^{2}}}e^{-2\sqrt{x^2 + y^2}}[/itex] in terms of the polar coordinates [itex]\rho[/itex] and [itex]\phi[/itex] and then evaluate the integral of [itex]f(x,y)[/itex] over a circle of radius 1 centered at the origin.

Homework Equations


[itex]y = \rho \sin(\phi)[/itex]
[itex]\rho = \sqrt{x^{2} + y^{2}}[/itex]

The Attempt at a Solution



Using the above relevant equations;

[itex]f(\rho,\phi) = \frac{\sin(\phi)}{\rho}e^{-2\rho}[/itex].

I am confused about what the second half of this question is asking me to do. The integral of a function gives the area under the curve, so surely evaluating an integral over a circle, merely gives the area of the circle?

It also looks like it would make more sense to evaluate the integral of [itex]f(\rho,\phi)[/itex] than [itex]f(x,y)[/itex].

EDIT - I just had a rather obvious thought. This is akin to asking me to find the mass of a circle, with density given by the function... If that's correct, then I think i'm ok.

EDIT2 - Also, with regards to performing the integral of [itex]f(x,y)[/itex] the fact that it is a circle radius 1 means that [itex]\sqrt{x^2 + y^2} = 1[/itex], right? So that makes my job much easier.

I'd appreciate some guidance about what this question is asking of me.

Thanks.
 
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Answers and Replies

  • #2
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You can calculate an integral in any coordinate system you wish. But often problems have nice symmetries that massively simplify the integration itself.

This problem is a nice example. What you have to realize is that ##f(x,y)## and ##f(\rho, \phi)## are in fact absolutely the same, just written in different coordinate system. So calculating the integral of ##f(x,y)## in cartesian coordinates brings you to the same result as calculating the integral of ##f(\rho ,\phi)## in polar coordinates. And that is exactly what the second part of your problem wants to demonstrate.
 
  • #3
555
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You can calculate an integral in any coordinate system you wish. But often problems have nice symmetries that massively simplify the integration itself.

This problem is a nice example. What you have to realize is that ##f(x,y)## and ##f(\rho, \Phi)## are in fact absolutely the same, just written in different coordinate system. So calculating the integral of ##f(x,y)## in cartesian coordinates brings you to the same result as calculating the integral of ##f(\rho ,\Phi)## in polar coordinates. And that is exactly what the second part of your problem wants to demonstrate.

I can see that, but i'm struggling to wrap my head around the limits in the cartesian case. The polar coordinates make much more sense when building the mental picture...
 
  • #4
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Sure they so, that's the point. :)

Anway, in cartesian coordinate system, you just have to be a bit more careful. Since the problem says, that the radius of the circle is ##1##, this in cartesian coordinates means (as you already mentioned) ##x^2+y^2=1##.
And you are trying to calculate ##\int \int f(x,y)dydx##. What are the boundaries of both integrals? (Hint: One integral is trivial, the other isn't.)
 
  • #5
Ray Vickson
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Homework Statement


Express [itex]f(x,y) = \frac{1}{\sqrt{x^{2} + y^{2}}}\frac{y}{\sqrt{x^{2} + y^{2}}}e^{-2\sqrt{x^2 + y^2}}[/itex] in terms of the polar coordinates [itex]\rho[/itex] and [itex]\phi[/itex] and then evaluate the integral of [itex]f(x,y)[/itex] over a circle of radius 1 centered at the origin.

Homework Equations


[itex]y = \rho \sin(\phi)[/itex]
[itex]\rho = \sqrt{x^{2} + y^{2}}[/itex]

The Attempt at a Solution



Using the above relevant equations;

[itex]f(\rho,\phi) = \frac{\sin(\phi)}{\rho}e^{-2\rho}[/itex].

I am confused about what the second half of this question is asking me to do. The integral of a function gives the area under the curve, so surely evaluating an integral over a circle, merely gives the area of the circle?

It also looks like it would make more sense to evaluate the integral of [itex]f(\rho,\phi)[/itex] than [itex]f(x,y)[/itex].

EDIT - I just had a rather obvious thought. This is akin to asking me to find the mass of a circle, with density given by the function... If that's correct, then I think i'm ok.

EDIT2 - Also, with regards to performing the integral of [itex]f(x,y)[/itex] the fact that it is a circle radius 1 means that [itex]\sqrt{x^2 + y^2} = 1[/itex], right? So that makes my job much easier.

I'd appreciate some guidance about what this question is asking of me.

Thanks.

Is your wording of the question exactly as it was given to you? The reason I ask is that there is an ambiguity: "circle" really means a one-dimensional curve, while "disc" means the two-dimensional area whose boundary is the circle. So, are you supposed to do a one-dimensional integral over a circle, or a two-dimensional integral over a disc? I know you entitled your thread "surface integral...", but is that really what you are supposed to do?
 
  • #6
555
19
Sure they so, that's the point. :)

Anway, in cartesian coordinate system, you just have to be a bit more careful. Since the problem says, that the radius of the circle is ##1##, this in cartesian coordinates means (as you already mentioned) ##x^2+y^2=1##.
And you are trying to calculate ##\int \int f(x,y)dydx##. What are the boundaries of both integrals? (Hint: One integral is trivial, the other isn't.)

I think I need to evaluate the integral for x between -1 and 1, and y between [itex]\sqrt{1-x^{2}}[/itex] and [itex]-\sqrt{1-x^{2}}[/itex]
 
  • #7
555
19
Is your wording of the question exactly as it was given to you? The reason I ask is that there is an ambiguity: "circle" really means a one-dimensional curve, while "disc" means the two-dimensional area whose boundary is the circle. So, are you supposed to do a one-dimensional integral over a circle, or a two-dimensional integral over a disc? I know you entitled your thread "surface integral...", but is that really what you are supposed to do?

The wording of the question is exactly as written.

I think it means to perform the double integral over a disc, because that is something we have focused on a few weeks ago, and I see no mention of the distinction you are making in the course document I have.
 
  • #8
Ray Vickson
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The wording of the question is exactly as written.

I think it means to perform the double integral over a disc, because that is something we have focused on a few weeks ago, and I see no mention of the distinction you are making in the course document I have.

Google 'circle' and 'disc'.
 
  • #9
HallsofIvy
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I would feel a lot better about this if it had said "over a disk of radius 1 centered at the origin". A "disk" is the set of points inside the circle. But you cannot integrate a scalar function over a curve- you have to have a vector valued function or a gradient of a scalar function.
 
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  • #10
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Ray pointed out a very important question. There is of course a huge difference (in shape and therefore also) in integrating. When this is clear, you can continue.

If it is a disk:
I think I need to evaluate the integral for x between -1 and 1, and y between 1−x2−−−−−√\sqrt{1-x^{2}} and −1−x2−−−−−√
Correct, but the result will be zero. (Question: Why?)
Use x from -1 to 1 and y from 0 to ##\sqrt {1-x^2}## and than multiply the whole integral by two. Ig got ##2\frac{e^2-1}{e^2}## and you hope you get the same.
 
  • #11
555
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Google 'circle' and 'disc'.

I appreciate the difference - I have sent an email to my lecturer asking for clarification, but am going to proceed with the surface integral interpretation of the question, because it makes more sense in the context of the course.
 
  • #12
555
19
Ray pointed out a very important question. There is of course a huge difference (in shape and therefore also) in integrating. When this is clear, you can continue.

If it is a disk:

Correct, but the result will be zero. (Question: Why?)
Use x from -1 to 1 and y from 0 to ##\sqrt {1-x^2}## and than multiply the whole integral by two. Ig got ##2\frac{e^2-1}{e^2}## and you hope you get the same.

the area below the x axis will cancel out with the area above, giving zero.

Thanks for your help.
 

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