Convert Cartesian To Cylindrical Limits

In summary: And you're welcome...BTW, I am using KA Stroud, Advanced Engineering Maths, but cannot find it in there still.I'm sorry, but I don't have access to that text. I'm glad you found what you were looking for, though.
  • #1
farso
15
0

Homework Statement



Hi

I am converting from Cartesian co-ordinates to cylindrical co-ordinates systems and can do the conversion fine, but am unsure about how to convert the limits.

Q)

Region V is given in Cartesian by: [tex]F(x,y,z) = xi+yj+z(x^{2}+y^{2})k[/tex]

Where [tex]x^{2}+y^{2} \leq 1[/tex]
and [tex]0 \leq z \leq 2[/tex]

Homework Equations




[tex] x = \rho cos(\varphi), y = \rho sin(\varphi), z = z[/tex]
[tex] cos^{2}(x)+sin^{2}(x) = 1[/tex]


The Attempt at a Solution



[tex] F(\rho,\varphi,z) = \rho cos(\varphi)\widehat{\rho} + \rho sin(\varphi)\widehat{\varphi} + z\rho^{2}\widehat{z}[/tex]

I know that z has stayed the same as it has not changed, but I am not sure how to do the [tex]\rho[/tex] and [tex]\varphi[/tex].

If you could help me on how to do this id really appreciate it.

Many thanks in advance
 
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  • #2
farso said:
[tex] F(\rho,\varphi,z) = \rho cos(\varphi)\widehat{\rho} + \rho sin(\varphi)\widehat{\varphi} + z\rho^{2}\widehat{z}[/tex]

It looks as though you've magically replaced [itex]\hat{\mathbf{x}}[/itex] and [itex]\hat{\mathbf{y}}[/itex] with [tex]\hat{\mathbf{\rho}}[/itex] and [tex]\hat{\mathbf{\varphi}}[/itex]...are you really allowed to do that?:wink:

What are [itex]\hat{\mathbf{x}}[/itex] and [itex]\hat{\mathbf{y}}[/itex] in terms of [tex]\hat{\mathbf{\rho}}[/itex] and [tex]\hat{\mathbf{\varphi}}[/itex]?

I know that z has stayed the same as it has not changed, but I am not sure how to do the [tex]\rho[/tex] and [tex]\varphi[/tex].

Well, your other boundary is [itex]x^2+y^2\leq 1[/itex]...so, what is [itex]x^2+y^2[/itex] is cylindrical coordinates?
 
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  • #3
gabbagabbahey said:
It looks as though you've magically replaced [itex]\hat{\mathbf{x}}[/itex] and [itex]\hat{\mathbf{y}}[/itex] with [tex]\hat{\mathbf{\rho}}[/itex] and [tex]\hat{\mathbf{\varphi}}[/itex]...are you really allowed to do that?:wink:

What are [itex]\hat{\mathbf{x}}[/itex] and [itex]\hat{\mathbf{y}}[/itex] in terms of [tex]\hat{\mathbf{\rho}}[/itex] and [tex]\hat{\mathbf{\varphi}}[/itex]?

Sorry, I am unsure what you mean here?

gabbagabbahey said:
Well, your other boundary is [itex]x^2+y^2\leq 1[/itex]...so, what is [itex]x^2+y^2[/itex] is cylindrical coordinates?

Ok, I think I see what you mean here... Would it be:

[tex]\rho \leq 1[/tex]
[tex]x^{2}+y^{2} \leq 1[/tex]
[tex] = \rho cos^{2}(\varphi) + \rho sin^{2}(\varphi)[/tex]

If this is correct, I am still unsure how to get [tex]\varphi[/tex]?

Thanks for the speedy response!
 
  • #4
farso said:
Sorry, I am unsure what you mean here?

When you substitute [itex]x=\rho\cos\varphi[/itex] and [itex]y=\rho\sin\varphi[/itex] into [itex]\textbf{F}=x\hat{\mathbf{x}}+y\hat{\mathbf{y}}+z(x^2+y^2)\hat{\mathbf{z}}[/itex], you get [itex]\textbf{F}=\rho\cos\varphi\hat{\mathbf{x}}+\rho\sin\varphi\hat{\mathbf{y}}+z\rho^2\hat{\mathbf{z}}[/itex], not [itex]\textbf{F}=\rho\cos\varphi\hat{\mathbf{\rho}}+\rho\sin\varphi\hat{\mathbf{\varphi}}+z\rho^2\hat{\mathbf{z}}[/itex]

[tex]\hat{\mathbf{x}}\neq\hat{\mathbf{\rho}} \;\;\;\;\;\;\text{and}\;\;\;\;\;\;\; \hat{\mathbf{y}}\neq\hat{\mathbf{\varphi}}[/itex]

In your text, you should have derived expressions of the form [itex]\hat{\mathbf{x}}=f(\rho,\theta)\hat{\mathbf{\rho}}+g(\rho\theta)\hat{\mathbf{\varphi}}[/itex] and [itex]\hat{\mathbf{y}}=F(\rho,\theta)\hat{\mathbf{\rho}}+G(\rho\theta)\hat{\mathbf{\varphi}}[/itex] ...use those
Ok, I think I see what you mean here... Would it be:

[tex]\rho \leq 1[/tex]
[tex]x^{2}+y^{2} \leq 1[/tex]
[tex] = \rho cos^{2}(\varphi) + \rho sin^{2}(\varphi)[/tex]

If this is correct, I am still unsure how to get [tex]\varphi[/tex]?

Thanks for the speedy response!

Well, you know that [itex]0\leq\rho\leq 1[/itex], so these are your integration limits for [itex]\rho[/itex]. As for [itex]\varphi[/itex], there are no further restrictions on its value, so you just go with the conventional [itex]0\leq\varphi\leq 2\pi[/itex].
 
  • #5
gabbagabbahey said:
When you substitute [itex]x=\rho\cos\varphi[/itex] and [itex]y=\rho\sin\varphi[/itex] into [itex]\textbf{F}=x\hat{\mathbf{x}}+y\hat{\mathbf{y}}+z(x^2+y^2)\hat{\mathbf{z}}[/itex], you get [itex]\textbf{F}=\rho\cos\varphi\hat{\mathbf{x}}+\rho\sin\varphi\hat{\mathbf{y}}+z\rho^2\hat{\mathbf{z}}[/itex], not [itex]\textbf{F}=\rho\cos\varphi\hat{\mathbf{\rho}}+\rho\sin\varphi\hat{\mathbf{\varphi}}+z\rho^2\hat{\mathbf{z}}[/itex]

[tex]\hat{\mathbf{x}}\neq\hat{\mathbf{\rho}} \;\;\;\;\;\;\text{and}\;\;\;\;\;\;\; \hat{\mathbf{y}}\neq\hat{\mathbf{\varphi}}[/itex]

In your text, you should have derived expressions of the form [itex]\hat{\mathbf{x}}=f(\rho,\theta)\hat{\mathbf{\rho}}+g(\rho\theta)\hat{\mathbf{\varphi}}[/itex] and [itex]\hat{\mathbf{y}}=F(\rho,\theta)\hat{\mathbf{\rho}}+G(\rho\theta)\hat{\mathbf{\varphi}}[/itex] ...use those
I have searched my notes, and my textbooks, but do not have anything which seems to cover this. Is there a particular name for this operation that I may be able to search up and check out some examples?

Thanks
 
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  • #6
I have a hard time believing that your course textbook doesn't cover unit vector conversions. What text are you using for this course?

Anyways, see here
 
  • #7
gabbagabbahey said:
I have a hard time believing that your course textbook doesn't cover unit vector conversions. What text are you using for this course?

Anyways, see here

Thats amazing, thanks. I hope I have now got it correct.

So...
[tex]F(\rho,\varphi,z) = \rho cos(\varphi)\widehat{i} + \rho sin(\varphi)\widehat{j} + z\rho^{2}\widehat{z}[/tex]

and

[tex]\widehat{\rho} = \widehat{i} cos(\varphi) + \widehat{j} sin(\varphi)[/tex]

Subbing that in I am just left with:
[tex]F(\rho,\varphi,z) = \rho \widehat{\rho} + z\rho^{2}\widehat{z}[/tex]

Is that looking to be correct?

Thanks for all your help, you have been very good to me.

BTW, I am using KA Stroud, Advanced Engineering Maths, but cannot find it in there still.
 
  • #8
Looks good to me!:approve:
 

1. What is the purpose of converting Cartesian to cylindrical limits?

Converting Cartesian to cylindrical limits allows us to express points in a 3-dimensional space using a different coordinate system. This can be useful in various scientific fields, such as physics and engineering.

2. How do you convert Cartesian coordinates to cylindrical coordinates?

To convert Cartesian coordinates (x,y,z) to cylindrical coordinates (r,θ,z), we can use the following equations:
r = √(x² + y²)
θ = arctan(y/x)
z = z

3. What are the limitations of converting Cartesian to cylindrical coordinates?

One limitation of converting Cartesian to cylindrical coordinates is that it can only be used for points in a 3-dimensional space. It also assumes that the origin of the coordinate system is at the center of the cylinder.

4. Can you convert cylindrical coordinates back to Cartesian coordinates?

Yes, it is possible to convert cylindrical coordinates back to Cartesian coordinates using the following equations:
x = r cos(θ)
y = r sin(θ)
z = z

5. How can converting Cartesian to cylindrical coordinates be applied in real-life situations?

Converting Cartesian to cylindrical coordinates can be useful in various real-life situations, such as in the design and construction of cylindrical structures, calculating fluid flow in cylindrical pipes, and analyzing celestial objects that have cylindrical symmetry.

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