Convert Cartesian To Cylindrical Limits

[farso]

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: $$F(x,y,z) = xi+yj+z(x^{2}+y^{2})k$$

Where $$x^{2}+y^{2} \leq 1$$
and $$0 \leq z \leq 2$$

Homework Equations

$$x = \rho cos(\varphi), y = \rho sin(\varphi), z = z$$
$$cos^{2}(x)+sin^{2}(x) = 1$$

The Attempt at a Solution

$$F(\rho,\varphi,z) = \rho cos(\varphi)\widehat{\rho} + \rho sin(\varphi)\widehat{\varphi} + z\rho^{2}\widehat{z}$$

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

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

Many thanks in advance

 

[gabbagabbahey]

$$F(\rho,\varphi,z) = \rho cos(\varphi)\widehat{\rho} + \rho sin(\varphi)\widehat{\varphi} + z\rho^{2}\widehat{z}$$

It looks as though you've magically replaced $\hat{\mathbf{x}}$ and $\hat{\mathbf{y}}$ 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 $\hat{\mathbf{x}}$ and $\hat{\mathbf{y}}$ 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 $$\rho$$ and $$\varphi$$. </div> </div> </blockquote><br /> Well, your other boundary is $x^2+y^2\leq 1$...so, what is $x^2+y^2$ is cylindrical coordinates?[/tex][/tex][/tex][/tex]

 

[farso]

It looks as though you've magically replaced $\hat{\mathbf{x}}$ and $\hat{\mathbf{y}}$ 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 $\hat{\mathbf{x}}$ and $\hat{\mathbf{y}}$ 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" 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 $x^2+y^2\leq 1$...so, what is $x^2+y^2$ is cylindrical coordinates? </div> </div> </blockquote><br /> Ok, I think I see what you mean here... Would it be:<br /> <br /> $$\rho \leq 1$$<br /> $$x^{2}+y^{2} \leq 1$$<br /> $$= \rho cos^{2}(\varphi) + \rho sin^{2}(\varphi)$$<br /> <br /> If this is correct, I am still unsure how to get $$\varphi$$?<br /> <br /> Thanks for the speedy response![/tex][/tex][/tex][/tex]

 

[gabbagabbahey]

Sorry, I am unsure what you mean here?

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

[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 $\hat{\mathbf{x}}=f(\rho,\theta)\hat{\mathbf{\rho}}+g(\rho\theta)\hat{\mathbf{\varphi}}$ and $\hat{\mathbf{y}}=F(\rho,\theta)\hat{\mathbf{\rho}}+G(\rho\theta)\hat{\mathbf{\varphi}}$ ...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 /> $$\rho \leq 1$$<br /> $$x^{2}+y^{2} \leq 1$$<br /> $$= \rho cos^{2}(\varphi) + \rho sin^{2}(\varphi)$$<br /> <br /> If this is correct, I am still unsure how to get $$\varphi$$?<br /> <br /> Thanks for the speedy response! </div> </div> </blockquote><br /> Well, you know that $0\leq\rho\leq 1$, so these are your integration limits for $\rho$. As for $\varphi$, there are no further restrictions on its value, so you just go with the conventional $0\leq\varphi\leq 2\pi$.[/tex]

 

[farso]

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

[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 $\hat{\mathbf{x}}=f(\rho,\theta)\hat{\mathbf{\rho}}+g(\rho\theta)\hat{\mathbf{\varphi}}$ and $\hat{\mathbf{y}}=F(\rho,\theta)\hat{\mathbf{\rho}}+G(\rho\theta)\hat{\mathbf{\varphi}}$ ...use those[/tex]

$$I have searched my notes, and my textbooks, but do not have anything which seems to cover this. Is there a particular name for this operation that I may be able to search up and check out some examples?<br /> <br /> Thanks$$

 

[gabbagabbahey]

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

 

[farso]

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...
$$F(\rho,\varphi,z) = \rho cos(\varphi)\widehat{i} + \rho sin(\varphi)\widehat{j} + z\rho^{2}\widehat{z}$$

and

$$\widehat{\rho} = \widehat{i} cos(\varphi) + \widehat{j} sin(\varphi)$$

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

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.

 

[gabbagabbahey]

Looks good to me!