Convert Cartesian To Cylindrical Limits

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farso
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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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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?<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f609.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":wink:" title="Wink :wink:" data-smilie="2"data-shortname=":wink:" /><br /> <br /> 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]?<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 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]. </div> </div> </blockquote><br /> 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?[/tex][/tex][/tex][/tex]
 
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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?<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f609.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":wink:" title="Wink :wink:" data-smilie="2"data-shortname=":wink:" /><br /> <br /> 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]?[/tex][/tex][/tex][/tex]
[tex][tex][tex][tex] <br /> Sorry, I am unsure what you mean here? <br /> <br /> <blockquote data-attributes="" data-quote="gabbagabbahey" data-source="post: 2695089" cite="https://www.physicsforums.com/goto/post?id=2695089" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> gabbagabbahey said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 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? </div> </div> </blockquote><br /> Ok, I think I see what you mean here... Would it be:<br /> <br /> [tex]\rho \leq 1[/tex]<br /> [tex]x^{2}+y^{2} \leq 1[/tex]<br /> [tex]= \rho cos^{2}(\varphi) + \rho sin^{2}(\varphi)[/tex]<br /> <br /> If this is correct, I am still unsure how to get [tex]\varphi[/tex]?<br /> <br /> Thanks for the speedy response![/tex][/tex][/tex][/tex]
 
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]<br /> <br /> 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<br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Ok, I think I see what you mean here... Would it be:<br /> <br /> [tex]\rho \leq 1[/tex]<br /> [tex]x^{2}+y^{2} \leq 1[/tex]<br /> [tex]= \rho cos^{2}(\varphi) + \rho sin^{2}(\varphi)[/tex]<br /> <br /> If this is correct, I am still unsure how to get [tex]\varphi[/tex]?<br /> <br /> Thanks for the speedy response! </div> </div> </blockquote><br /> 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].[/tex]
 
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]<br /> <br /> 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[/tex]
[tex]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?<br /> <br /> Thanks[/tex]
 
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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
 
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.