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May 16th, 2017, 12:03 PM   #1
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A question about cylindrical coordinates

So cylindrical coordinates are basically a 3D representation of polar coordinates. Since with polar coordinates we can use basic trigonometry to figure out the length of the triangle sides x or y and therefore convert from polar to 2D rectangular, the same is true for converting from cylindrical to 3D rectangular.

The formulas for converting cylindrical to 3D rectangular are :
x = r cos theta
y = r sin theta
z = z

Now, my questions is: WHY is y = r sin theta ?

If theta is the angle between x and r , why is it applicable to y? Am I looking at this the wrong way? The way I see it, y isn't even part of the triangle that contains the theta angle. When looking for cos theta , we get it by dividing it's adjacent side (x) with the hypotenuse (r) . But when looking for sin theta, shouldn't the opposite side we need in the equation be a line drawn between x and r ?

When I can't figure something like this out it just keeps bugging me and I can't move on with the materia and just apply the formula. Can anyone help clear this up?
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May 24th, 2017, 11:52 AM   #2
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Cylindrical Coordinate

Question: Why is $\displaystyle y=r \sin \theta$?

Answer:
If you take a look at the graph I've attached, you can see that $\displaystyle x$ and $\displaystyle y$ are related by the triangle in the $\displaystyle xy$ plane.

We can derive a relationship in terms of the angle $\displaystyle \theta$ by solving two trig functions.

Let:
$\displaystyle \cos \theta = \frac{x}{r}$ and $\displaystyle \sin \theta = \frac{y}{r}$.

Solving each of these equations for $\displaystyle x$ and $\displaystyle y$ yields their conversion equations.

$\displaystyle x=r \cos \theta$ and $\displaystyle y = r \sin \theta$

The part you may be stuck on, is seeing that the bottom leg of the triangle is equal to $\displaystyle y$. This is because it is parallel to the $\displaystyle y$ axis. It lets us use 1 triangle, instead of switching to a second triangle along the $\displaystyle y$ axis.

Hope that helps!
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June 9th, 2017, 01:54 PM   #3
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There are several ways to answer this. The simplest is to imagine the are bounded by the arcs $\displaystyle r= r_0$ and$\displaystyle r=r_0+ \delta r$ and by the radial straight lines $\displaystyle \theta= \theta_0$ and $\displaystyle \theta= \theta_0+ \delta \theta$. The radial straight lines have, of course, length $\displaystyle \delta r$ while the two curved sides have length $\displaystyle r\delta \theta$ and $\displaystyle (r+ \delta r)(\theta+ \delta \theta)$. That is almost a rectangle (it would be if it weren't for those curved sides!) so its area is almost $\displaystyle (r\delta \theta)(\delta r)$ which, in the limit as both $\displaystyle \delta r$ and $\displaystyle \delta\theta$ go to 0, goes to the "differential of area" $\displaystyle r drd\theta$.

Another way is to write $\displaystyle x= r \cos(\theta)$ so that $\displaystyle dx= \cos(\theta)dr- r \sin(\theta) d\theta$ and $\displaystyle y= r \sin(\theta)$ so that $\displaystyle dy= \sin(\theta)dr+ r \cos(\theta) d\theta$. Now we want to find $\displaystyle dxdy$ in terms of $\displaystyle r$, $\displaystyle \theta$, $\displaystyle dr$, and $\displaystyle d\theta$. But multiplication of "differentials" is anti-commutative. In particular, $\displaystyle drdr= d\theta d\theta= 0$ and $\displaystyle drd\theta= -d\theta dr$: $\displaystyle dxdy= (\cos(\theta) dr- r\sin(\theta) d\theta)(\sin(\theta)dr+ r \cos(\theta)d\theta)= r \cos^2(\theta)drd\theta- (-r \sin^2(\theta)d\theta dr= (r^2 \cos^2(\theta)+ r^2 \sin^2(\theta))drd\theta= r drd\theta.$

That can be seen perhaps more clearly by thinking of that product as the determinant of the "differential matrix":
$\displaystyle \left|\begin{array} \cos(\theta) & \sin(\theta) \\ -r \sin(\theta) & r \cos(\theta)\end{array}\right|drd\theta= r dr d\theta$.

Last edited by skipjack; June 9th, 2017 at 09:23 PM.
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