# Vector Fields in Cartesian and Cylindrical Coordinates, The Curl

• sriracha
In summary, the necessary information for the conversion from cylindrical to Cartesian coordinates has been provided, and the answer in Cartesian coordinates is -ix-jy+2kz. However, WolframAlpha has given a different result, so it would be helpful to confirm if the way the conversion was done is correct.
sriracha
All necessary information is attached except the answer in Cartesian coordinates, which is -ix-jy+2kz and my work converting back from cylindrical to Cartesian, which I used WolframAlpha for, as the trig is a mess (that is, if the way I am doing this is correct).

http://www.wolframalpha.com/input/?i=%28icos%28tan^-1%28y%2Fx%29%29%2Bjsin%28tan^-1%28y%2Fx%29%29%29%28-sqrt%28x^2%2By^2%29%28cos^2%28tan^-1%28y%2Fx%29%29-sin^2%28tan^-1%28y%2Fx%29%29%29%29%2B%28-isin%28tan^-1%28y%2Fx%29%29%2Bjcos%28tan^-1%28y%2Fx%29%29%29%282z%28cos^2%28tan^-1%28y%2Fx%29%29-sin^2%28tan^-1%28y%2Fx%29%29%29%29&a=i_Variable

WolframAlpha, however, just gave me this mess back. Would appreciate if someone would let me know if the way I'm going about this is correct. I have a strong feeling the matrix I'm using for reconversion is off.

Edit: I already posted this in the physics section, but I got no replies, because it is probably more appropriate to place here. See the attachment on the original thread at https://www.physicsforums.com/showthread.php?t=588902

for the full problem.The answer in Cartesian coordinates is -ix-jy+2kz.To convert from cylindrical to Cartesian coordinates, you can use the following transformation matrix: \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} \cos(\theta) & -\sin(\theta) & 0 \\ \sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} r \\ \theta \\ z \end{bmatrix}.Using this transformation matrix, you can obtain the Cartesian coordinates of the point in question by plugging in the given cylindrical coordinates:\begin{bmatrix} x \\ y \\ z \end{bmatrix}= \begin{bmatrix} \cos(\tan^{-1}(y/x)) & -\sin(\tan^{-1}(y/x)) & 0 \\ \sin(\tan^{-1}(y/x)) & \cos(\tan^{-1}(y/x)) & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \sqrt{x^2 + y^2} \\ \tan^{-1}(y/x) \\ 2z \end{bmatrix}= \begin{bmatrix} -x \\ -y \\ 2z \end{bmatrix}.Therefore, the answer in Cartesian coordinates is -ix-jy+2kz.

## 1. What is a vector field?

A vector field is a mathematical concept that describes a vector quantity, such as force or velocity, at every point in a given space. It is represented by arrows, where the length and direction of the arrow indicate the magnitude and direction of the vector at that point.

## 2. What are Cartesian coordinates?

Cartesian coordinates are a system of representing points in a two-dimensional or three-dimensional space using x, y, and z coordinates. The x-axis is horizontal, the y-axis is vertical, and the z-axis is perpendicular to the x and y axes. This system is often used to describe vector fields in a flat space.

## 3. What are cylindrical coordinates?

Cylindrical coordinates are a system of representing points in a three-dimensional space using r, θ, and z coordinates. The r coordinate represents the distance from the origin, the θ coordinate represents the angle in a polar coordinate system, and the z coordinate represents the height or depth. This system is often used to describe vector fields in a cylindrical or circular space.

## 4. What is the curl of a vector field?

The curl of a vector field is a mathematical operation that describes the rotation or circulation of the vector field at a given point. It is represented by a vector and can be thought of as the tendency of the vector field to swirl around a point.

## 5. How do you calculate the curl of a vector field in Cartesian and cylindrical coordinates?

The curl of a vector field can be calculated by taking the partial derivatives of the vector components with respect to each coordinate and then rearranging them in a specific formula. In Cartesian coordinates, the formula is (∂Fz/∂y - ∂Fy/∂z)i + (∂Fx/∂z - ∂Fz/∂x)j + (∂Fy/∂x - ∂Fx/∂y)k. In cylindrical coordinates, the formula is (1/r)(∂(rFz)/∂θ - ∂Fθ/∂z)i + (1/r)(∂Fz/∂r - ∂Fr/∂z)j + (1/r)(∂(rFθ)/∂r - ∂Fθ/∂r)k.

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