# Convert Cartesian To Cylindrical Limits

1. Apr 29, 2010

### farso

1. The problem statement, all variables and given/known data

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 $$0 \leq z \leq 2$$

2. Relevant equations

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

3. 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 im 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

2. Apr 29, 2010

It looks as though you've magically replaced $\hat{\mathbf{x}}$ and $\hat{\mathbf{y}}$ with $$\hat{\mathbf{\rho}}[/itex] and [tex]\hat{\mathbf{\varphi}}[/itex]....are you really allowed to do that? What are $\hat{\mathbf{x}}$ and $\hat{\mathbf{y}}$ in terms of [tex]\hat{\mathbf{\rho}}[/itex] and [tex]\hat{\mathbf{\varphi}}[/itex]? Well, your other boundary is $x^2+y^2\leq 1$....so, what is $x^2+y^2$ is cylindrical coordinates? Last edited: Apr 29, 2010 3. Apr 29, 2010 ### farso Sorry, Im unsure what you mean here? Ok, I think I see what you mean here... Would it be: [tex]\rho \leq 1$$
$$x^{2}+y^{2} \leq 1$$
$$= \rho cos^{2}(\varphi) + \rho sin^{2}(\varphi)$$

If this is correct, Im still unsure how to get $$\varphi$$?

Thanks for the speedy response!

4. Apr 29, 2010

### gabbagabbahey

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 $\textbf{F}=\rho\cos\varphi\hat{\mathbf{\rho}}+\rho\sin\varphi\hat{\mathbf{\varphi}}+z\rho^2\hat{\mathbf{z}}$

$$\hat{\mathbf{x}}\neq\hat{\mathbf{\rho}} \;\;\;\;\;\;\text{and}\;\;\;\;\;\;\; \hat{\mathbf{y}}\neq\hat{\mathbf{\varphi}}[/itex] In your text, you should have derived expressions of the form $\hat{\mathbf{x}}=f(\rho,\theta)\hat{\mathbf{\rho}}+g(\rho\theta)\hat{\mathbf{\varphi}}$ and $\hat{\mathbf{y}}=F(\rho,\theta)\hat{\mathbf{\rho}}+G(\rho\theta)\hat{\mathbf{\varphi}}$ ...use those Well, you know that $0\leq\rho\leq 1$, so these are your integration limits for $\rho$. As for $\varphi$, there are no further restrictions on its value, so you just go with the conventional $0\leq\varphi\leq 2\pi$. 5. Apr 30, 2010 ### farso I have searched my notes, and my text books, but do not have anything which seems to cover this. Is there a particular name for this operation that I may be able to search up and check out some examples? Thanks Last edited: Apr 30, 2010 6. Apr 30, 2010 ### gabbagabbahey I have a hard time believing that your course textbook doesn't cover unit vector conversions. What text are you using for this course? Anyways, see here 7. Apr 30, 2010 ### farso Thats amazing, thanks. I hope I have now got it correct. So... [tex]F(\rho,\varphi,z) = \rho cos(\varphi)\widehat{i} + \rho sin(\varphi)\widehat{j} + z\rho^{2}\widehat{z}$$

and

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

Subbing that in Im 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.

8. Apr 30, 2010

### gabbagabbahey

Looks good to me!