# Area Differential in Cartesian and Polar Coordinates

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In summary, the area differential in Cartesian coordinates is given by dxdy, while in polar coordinates it is r dr dθ. To convert between the two, trigonometric functions are used to obtain the conversions x=r cos θ and y=r sin θ. By differentiating both with respect to the polar angle θ, the differentials become dx=cos θ dr-r sin θ dθ and dy=sin θ dr+r cos θ dθ. Using the total derivatives dx,dy, etc. or exterior derivatives, the area differential can be expressed as dA=dx dy = det J dr dθ, where det J is the determinant of the Jacobian matrix. This can be simplified to
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TL;DR Summary
How do we get rdrdt from dxdy?
The area differential ##dA## in Cartesian coordinates is ##dxdy##.
The area differential ##dA## in polar coordinates is ##r dr d\theta##.
How do we get from one to the other and prove that ##dxdy## is indeed equal to ##r dr d\theta##?

##dxdy=r dr d\theta##
The trigonometric functions are used to obtain the conversions ##x=r \cos \theta ## and ##y=r \sin \theta ## . If we differentiate both with respect to the polar angle ##\theta##,
$$dx d\theta = \cos \theta dr d\theta -r \sin \theta\; , \;dy d\theta = \sin \theta dr d\theta +r \cos \theta$$
The differentials then become
$$dx= \cos \theta dr -r \sin \theta d\theta \; , \;dy = \sin \theta dr + r \cos \theta dr \theta$$
which to me does not make any sense.
It seems as though finding ##dr d\theta ## is what we must do, but how? What am I missing?
How do we prove that indeed ...
$$dx dy = r dr d\theta$$
Is it in Griffiths? On what page?

Last edited by a moderator:
O yea, you have to partially differentiate one coordinate with respect to the other.
That's not in Griffiths!
I totally forgot.
Thanks.
That was driving me crazy.
That method was always a great mystery to me, because the professor just kind of told us to think about how it is correct .
What if we thought too much about it and it no longer makes any sense ?
I was thinking about taking a partial derivative, but it didn't "click."
Griffths has a God awful method for converting coordinate differentials.
I suggest everyone not use that method.
He actually brings out the old ##\Delta x## thing and takes the limit.
Hideous.

Last edited:
Zap said:
That's not in Griffiths!
It's in most calculus textbooks ...

Zap said:
O yea, you have to partially differentiate one coordinate with respect to the other.
That's not in Griffiths!
I totally forgot.
Thanks.
That was driving me crazy.
That method was always a great mystery to me, because the professor just kind of told us to think about how it is correct .
What if we thought too much about it and it no longer makes any sense ?
I was thinking about taking a partial derivative, but it didn't "click."
Griffths has a God awful method for converting coordinate differentials.
I suggest everyone not use that method.
He actually brings out the old ##\Delta x## thing and takes the limit.
Hideous.
You use the total derivatives dx,dy, etc. aka exterior derivative and then sub-in.

Actually, I'm still not getting it. The wiki page shows this:
,
but it leaves a lot to the imagination.
I was thinking of changing coordinates, not differentials. When changing coordinates you can do the $$\hat x = \frac{ \frac{\partial y(x)}{\partial x} }{ |\frac{ \partial y(x)}{ \partial x } |}$$ thingy to find the unit vectors. I guess we are still changing coordinates, but to find the differentials, so maybe the method is similar.

I think someone messed with the OP, because it's not really legible anymore. I was just asking how to get from ##dxdy## to ##rdrd\theta##.

Zap said:
Actually, I'm still not getting it. The wiki page shows this:
,
but it leaves a lot to the imagination.
I was thinking of changing coordinates, not differentials. When changing coordinates you can do the $$\hat x = \frac{ \frac{\partial y(x)}{\partial x} }{ |\frac{ \partial y(x)}{ \partial x } |}$$ thingy to find the unit vectors. I guess we are still changing coordinates, but to find the differentials, so maybe the method is similar.

I think someone messed with the OP, because it's not really legible anymore. I was just asking how to get from ##dxdy## to ##rdrd\theta##.
I guess that someone was you together with a software bug. The MathJax code sometimes gets lost if you mix it with BB-code like "center" in the edit mode. Anyway, the calculation goes:
\begin{align*}
x&=r \cos \varphi \; , \; y= r\sin \varphi \\ &\Longrightarrow \\
\det J &= \left| \dfrac{\partial (x,y)}{\partial (r,\varphi)} \right| = \left| \begin{bmatrix} \dfrac{\partial x}{\partial r} & \dfrac{\partial x}{\partial \varphi} \\ \dfrac{\partial y}{\partial r} & \dfrac{\partial y}{\partial \varphi} \end{bmatrix} \right| = \left| \begin{bmatrix} \cos \varphi & -r \sin \varphi \\ \sin \varphi & r \cos \varphi \end{bmatrix} \right| =r\cos^2 \varphi +r \sin^2 \varphi = r \\
&\Longrightarrow \\
dA&= dx \,dy = \left(\det J \right) \,dr \, d\varphi = r \,dr \,d\varphi
\end{align*}

Weird. One might think it is better just to memorize it.
I guess I'm confused on this part ##dxdy=(detJ)drdφ##.
I'm not really sure where the det J is coming from. This question has been asked a few times on google, so I will continue reading up on it, but I'm still not entirely getting it.

Zap said:
Weird. One might think it is better just to memorize it.
I guess I'm confused on this part ##dxdy=(detJ)drdφ##.
I'm not really sure where the det J is coming from. This question has been asked a few times on google, so I will continue reading up on it, but I'm still not entirely getting it.
Sort of properties of multilinear maps come down to, or simplify into, the DetJ statement.

Zap
Why is ## dxdy = (cos\theta dr - rsin\theta d\theta)(rcos \theta d\theta + sin\theta dr) ## not allowed? Is it because ##dx## and ##dy## are actually vectors and we have to do some kind of vector product? Or is it allowed, but not very useful, since it produces weird stuff like ##(dr)^2##.

Zap said:
I guess I'm confused on this part ##dxdy=(\det J)drd\varphi##.
Try to convert a basis ##\{\,(1,0),(1,4)\,\}## into ##\{\,(2,5),(-1,7)\,\}##. The functional determinant is the factor which adjusts the volumes. If you measure in inch instead of centimeters, you apply a scaling factor, too, don't you? Now what about square inch and square centimeters?

Zap said:
Why is dxdy=(cosθdr−rsinθdθ)(rcosθdθ+sinθdr)dxdy=(cosθdr−rsinθdθ)(rcosθdθ+sinθdr) not allowed? Is it because dxdx and dydy are actually vectors and we have to do some kind of vector product? Or is it allowed, but not very useful, since it produces weird stuff like (dr)2(dr)2.
It is allowed. You just used the wrong multiplication. If you insist on differential forms, then we have
\begin{align*}
dx \wedge dy &= (\cos \varphi \,dr -r\sin \varphi \, d\varphi) \wedge (\sin \varphi \, dr + r \cos \varphi \, d\varphi)\\
&= (\cos \varphi )\,dr \wedge (r\cos \varphi \, d\varphi) - (r \sin \varphi \, d \varphi) \wedge (\sin \varphi \, dr)\\
&= r\cos^2 \varphi \,dr \wedge d \varphi - r\sin^2 \varphi \,d\varphi \wedge dr \\
&= (r\cos^2 \varphi + r\sin^2\varphi )\, dr \wedge d\varphi \\
&=r \,dr \wedge d\varphi
\end{align*}

Zap said:
Why is ## dxdy = (cos\theta dr - rsin\theta d\theta)(rcos \theta d\theta + sin\theta dr) ## not allowed? Is it because ##dx## and ##dy## are actually vectors and we have to do some kind of vector product? Or is it allowed, but not very useful, since it produces weird stuff like ##(dr)^2##.
You can do it that way. There is an antisymmetric product so
drdr=0 dθdθ=0 drdθ=-dθdr

dxdy=rdrdθ

The determinant give the ratio of the differentials

Wow. I always wished that physics could be taught by mathematicians.
There's a lot for me to digest here, but I am learning.

weirdoguy

## 1. What is the difference between Cartesian and Polar coordinates?

Cartesian coordinates, also known as rectangular coordinates, use two perpendicular axes (x and y) to locate a point in a two-dimensional space. Polar coordinates, on the other hand, use a distance (r) and an angle (θ) to locate a point in a two-dimensional space.

## 2. How do you convert between Cartesian and Polar coordinates?

To convert from Cartesian coordinates to Polar coordinates, you can use the following formulas:
r = √(x² + y²)
θ = tan⁻¹(y/x)
To convert from Polar coordinates to Cartesian coordinates, you can use the following formulas:
x = r cos(θ)
y = r sin(θ)

## 3. What is the purpose of using Polar coordinates?

Polar coordinates are useful for representing points in a circular or curved system, such as in polar graphs or in navigation systems. They can also make certain calculations, such as finding the distance between two points, easier to solve.

## 4. Can you use both Cartesian and Polar coordinates in the same system?

Yes, it is possible to use both Cartesian and Polar coordinates in the same system. This is known as a mixed coordinate system and is often used in engineering and physics applications.

## 5. How does the concept of area differ in Cartesian and Polar coordinates?

In Cartesian coordinates, the area is calculated by multiplying the length and width of a rectangle. In Polar coordinates, the area is calculated by integrating the function r(θ) over a given interval. This means that the area element in Polar coordinates is not a rectangle, but a sector of a circle.

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