Esercizi sul Rotore

Esercizi sul Rotore

+Esercizi sul Rotore

Versione italiana

Esercizi sul Rotore

Definizione

Il rotore è un operatore che misura la tendenza di un campo vettoriale a ruotare attorno a un punto. È definito per campi vettoriali tridimensionali e viene denotato con il simbolo \nabla \times \mathbf{F}$\nabla \times \mathbf{F}$, dove \mathbf{F}$\mathbf{F}$ è un campo vettoriale.

Formula

Se \mathbf{F} = (F_x, F_y, F_z)$\mathbf{F} = (F_x, F_y, F_z)$ è un campo vettoriale definito in uno spazio tridimensionale, il rotore di \mathbf{F}$\mathbf{F}$ è dato da:

\nabla \times \mathbf{F} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
F_x & F_y & F_z
\end{vmatrix}

$$\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z \end{vmatrix}$$

Dove:

• \mathbf{i}, \mathbf{j}, \mathbf{k}$\mathbf{i}, \mathbf{j}, \mathbf{k}$ sono i versori delle coordinate cartesiane,
• F_x, F_y, F_z$F_x, F_y, F_z$ sono le componenti del campo vettoriale \mathbf{F}$\mathbf{F}$.

Espandendo il determinante, otteniamo:

\nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right)

$$\nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right)$$

Proprietà del Rotore

1. Linearità: Il rotore è un operatore lineare. Se \mathbf{F}$\mathbf{F}$ e \mathbf{G}$\mathbf{G}$ sono campi vettoriali e a$a$ è una costante, allora:

   \nabla \times (a\mathbf{F} + \mathbf{G}) = a(\nabla \times \mathbf{F}) + (\nabla \times \mathbf{G})

   $$\nabla \times (a\mathbf{F} + \mathbf{G}) = a(\nabla \times \mathbf{F}) + (\nabla \times \mathbf{G})$$

2. Rotore di un Gradiente: Il rotore del gradiente di una funzione scalare è sempre zero:

   \nabla \times (\nabla f) = 0

   $$\nabla \times (\nabla f) = 0$$

   Questo implica che i campi conservativi (campi derivanti da un potenziale scalare) hanno un rotore nullo.

3. Rotore di un Campo Vettoriale Costante: Se \mathbf{F}$\mathbf{F}$ è un campo vettoriale costante, allora:

   \nabla \times \mathbf{F} = 0

   $$\nabla \times \mathbf{F} = 0$$

Applicazioni del Rotore

1. Fluidodinamica: In fluidodinamica, il rotore di un campo di velocità rappresenta la vorticità del fluido. Un campo di velocità con rotore diverso da zero indica la presenza di vortici e turbolenze.

2. Elettromagnetismo: Nelle equazioni di Maxwell, il rotore del campo elettrico e del campo magnetico è utilizzato per descrivere come i campi elettrici e magnetici interagiscono e si propagano. Ad esempio, la legge di Faraday dell'induzione elettromagnetica è espressa come:

   \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}

   $$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$$

3. Ingegneria Meccanica: Il rotore è utilizzato per analizzare il comportamento di sistemi meccanici e per studiare il movimento rotatorio di corpi rigidi.

Esercizi

Esercizio 1: Calcolo del Rotore

Calcola il rotore del campo vettoriale \mathbf{F} = (y^2, x^2, z)$\mathbf{F} = (y^2, x^2, z)$.

Soluzione:

Utilizziamo la formula del rotore:

\nabla \times \mathbf{F} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
y^2 & x^2 & z
\end{vmatrix}

$$\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y^2 & x^2 & z \end{vmatrix}$$

Espandendo il determinante, otteniamo:

\nabla \times \mathbf{F} = \left( \frac{\partial z}{\partial y} - \frac{\partial x^2}{\partial z}, \frac{\partial y^2}{\partial z} - \frac{\partial z}{\partial x}, \frac{\partial x^2}{\partial x} - \frac{\partial y^2}{\partial y} \right)

$$\nabla \times \mathbf{F} = \left( \frac{\partial z}{\partial y} - \frac{\partial x^2}{\partial z}, \frac{\partial y^2}{\partial z} - \frac{\partial z}{\partial x}, \frac{\partial x^2}{\partial x} - \frac{\partial y^2}{\partial y} \right)$$

Calcolando le derivate parziali:

\nabla \times \mathbf{F} = \left( 0 - 0, 0 - 0, 2 - 2 \right) = (0, 0, 0)

$$\nabla \times \mathbf{F} = \left( 0 - 0, 0 - 0, 2 - 2 \right) = (0, 0, 0)$$

Quindi, il rotore del campo vettoriale è zero.

Esercizio 2: Rotore in un Campo Vorticoso

Calcola il rotore del campo vettoriale \mathbf{F} = (-y, x, 0)$\mathbf{F} = (-y, x, 0)$.

Soluzione:

Utilizziamo la formula del rotore:

\nabla \times \mathbf{F} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
-y & x & 0
\end{vmatrix}

$$\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -y & x & 0 \end{vmatrix}$$

Espandendo il determinante, otteniamo:

\nabla \times \mathbf{F} = \left( \frac{\partial (0)}{\partial y} - \frac{\partial (x)}{\partial z}, \frac{\partial (-y)}{\partial z} - \frac{\partial (0)}{\partial x}, \frac{\partial (x)}{\partial x} - \frac{\partial (-y)}{\partial y} \right)

$$\nabla \times \mathbf{F} = \left( \frac{\partial (0)}{\partial y} - \frac{\partial (x)}{\partial z}, \frac{\partial (-y)}{\partial z} - \frac{\partial (0)}{\partial x}, \frac{\partial (x)}{\partial x} - \frac{\partial (-y)}{\partial y} \right)$$

Calcolando le derivate parziali:

1. \frac{\partial (0)}{\partial y} = 0$\frac{\partial (0)}{\partial y} = 0$
2. \frac{\partial (x)}{\partial z} = 0$\frac{\partial (x)}{\partial z} = 0$
3. \frac{\partial (-y)}{\partial z} = 0$\frac{\partial (-y)}{\partial z} = 0$
4. \frac{\partial (0)}{\partial x} = 0$\frac{\partial (0)}{\partial x} = 0$
5. \frac{\partial (x)}{\partial x} = 1$\frac{\partial (x)}{\partial x} = 1$
6. \frac{\partial (-y)}{\partial y} = -1$\frac{\partial (-y)}{\partial y} = -1$

Sostituendo i risultati:

\nabla \times \mathbf{F} = \left( 0 - 0, 0 - 0, 1 - (-1) \right) = (0, 0, 2)

$$\nabla \times \mathbf{F} = \left( 0 - 0, 0 - 0, 1 - (-1) \right) = (0, 0, 2)$$

Quindi, il rotore del campo vettoriale \mathbf{F} = (-y, x, 0)$\mathbf{F} = (-y, x, 0)$ è:

\nabla \times \mathbf{F} = (0, 0, 2)

$$\nabla \times \mathbf{F} = (0, 0, 2)$$

Esercizio 3: Calcolo del Rotore di un Campo Vettoriale

Calcola il rotore del campo vettoriale \mathbf{F} = (2xy, x^2, z)$\mathbf{F} = (2xy, x^2, z)$.

Soluzione:

Utilizziamo la formula del rotore in coordinate cartesiane:

\nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) $\nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right)$

Calcoliamo ciascun componente:

1. Primo componente:
   \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = \frac{\partial z}{\partial y} - \frac{\partial (x^2)}{\partial z} = 0 - 0 = 0 $\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = \frac{\partial z}{\partial y} - \frac{\partial (x^2)}{\partial z} = 0 - 0 = 0$

2. Secondo componente:
   \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} = \frac{\partial (2xy)}{\partial z} - \frac{\partial z}{\partial x} = 0 - 0 = 0 $\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} = \frac{\partial (2xy)}{\partial z} - \frac{\partial z}{\partial x} = 0 - 0 = 0$

3. Terzo componente:
   \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = \frac{\partial (x^2)}{\partial x} - \frac{\partial (2xy)}{\partial y} = 2x - 2x = 0 $\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = \frac{\partial (x^2)}{\partial x} - \frac{\partial (2xy)}{\partial y} = 2x - 2x = 0$

Quindi, il rotore è:

\nabla \times \mathbf{F} = (0, 0, 0) $\nabla \times \mathbf{F} = (0, 0, 0)$

Esercizio 4: Rotore di un Campo Vettoriale in Coordinate Polari

Calcola il rotore del campo vettoriale \mathbf{F} = (0, r^2, 0)$\mathbf{F} = (0, r^2, 0)$ in coordinate polari.

Soluzione:

In coordinate polari, il rotore è dato da:

\nabla \times \mathbf{F} = \left( \frac{1}{r} \frac{\partial (r F_z)}{\partial r} - \frac{\partial F_r}{\partial z}, \frac{\partial F_r}{\partial z} - \frac{\partial F_z}{\partial r}, \frac{1}{r} \frac{\partial (r F_\theta)}{\partial r} - \frac{1}{r} \frac{\partial F_r}{\partial \theta} \right) $\nabla \times \mathbf{F} = \left( \frac{1}{r} \frac{\partial (r F_z)}{\partial r} - \frac{\partial F_r}{\partial z}, \frac{\partial F_r}{\partial z} - \frac{\partial F_z}{\partial r}, \frac{1}{r} \frac{\partial (r F_\theta)}{\partial r} - \frac{1}{r} \frac{\partial F_r}{\partial \theta} \right)$

Dove F_r = 0$F_r = 0$, F_\theta = r^2$F_\theta = r^2$, e F_z = 0$F_z = 0$.

Calcoliamo ciascun componente:

1. Primo componente:
   \frac{1}{r} \frac{\partial (r \cdot 0)}{\partial r} - \frac{\partial (0)}{\partial z} = 0 - 0 = 0 $\frac{1}{r} \frac{\partial (r \cdot 0)}{\partial r} - \frac{\partial (0)}{\partial z} = 0 - 0 = 0$

2. Secondo componente:
   \frac{\partial (0)}{\partial z} - \frac{\partial (0)}{\partial r} = 0 $\frac{\partial (0)}{\partial z} - \frac{\partial (0)}{\partial r} = 0$

3. Terzo componente:
   \frac{1}{r} \frac{\partial (r \cdot r^2)}{\partial r} - \frac{1}{r} \frac{\partial (0)}{\partial \theta} = \frac{1}{r} \frac{\partial (r^3)}{\partial r} - 0 = \frac{1}{r} \cdot 3r^2 = 3r $\frac{1}{r} \frac{\partial (r \cdot r^2)}{\partial r} - \frac{1}{r} \frac{\partial (0)}{\partial \theta} = \frac{1}{r} \frac{\partial (r^3)}{\partial r} - 0 = \frac{1}{r} \cdot 3r^2 = 3r$

Quindi, il rotore del campo vettoriale in coordinate polari è:

\nabla \times \mathbf{F} = (0, 0, 3r) $\nabla \times \mathbf{F} = (0, 0, 3r)$

Esercizio 5: Rotore di un Campo Vettoriale in Coordinate Sferiche

Calcola il rotore del campo vettoriale \mathbf{F} = (r^2 \sin \theta, r^2 \cos \theta, z)$\mathbf{F} = (r^2 \sin \theta, r^2 \cos \theta, z)$ in coordinate sferiche.

Soluzione:

In coordinate sferiche, il rotore è dato da:

\nabla \times \mathbf{F} = \left( \frac{1}{r \sin \theta} \left( \frac{\partial (F_\phi \sin \theta)}{\partial \theta} - \frac{\partial F_\theta}{\partial \phi} \right), \frac{1}{r} \left( \frac{\partial F_r}{\partial \phi} - \frac{\partial (F_\phi r)}{\partial r} \right), \frac{1}{r} \left( \frac{\partial (F_\theta r)}{\partial r} - \frac{\partial F_r}{\partial \theta} \right) \right) $\nabla \times \mathbf{F} = \left( \frac{1}{r \sin \theta} \left( \frac{\partial (F_\phi \sin \theta)}{\partial \theta} - \frac{\partial F_\theta}{\partial \phi} \right), \frac{1}{r} \left( \frac{\partial F_r}{\partial \phi} - \frac{\partial (F_\phi r)}{\partial r} \right), \frac{1}{r} \left( \frac{\partial (F_\theta r)}{\partial r} - \frac{\partial F_r}{\partial \theta} \right) \right)$

Dove F_r = r^2 \sin \theta$F_r = r^2 \sin \theta$, F_\theta = r^2 \cos \theta$F_\theta = r^2 \cos \theta$, e F_\phi = z$F_\phi = z$.

Calcoliamo ciascun componente:

1. Primo componente:
   \frac{1}{r \sin \theta} \left( \frac{\partial (z \sin \theta)}{\partial \theta} - \frac{\partial (r^2 \cos \theta)}{\partial \phi} \right) = \frac{1}{r \sin \theta} \left( z \cos \theta - 0 \right) = \frac{z \cos \theta}{r \sin \theta} $\frac{1}{r \sin \theta} \left( \frac{\partial (z \sin \theta)}{\partial \theta} - \frac{\partial (r^2 \cos \theta)}{\partial \phi} \right) = \frac{1}{r \sin \theta} \left( z \cos \theta - 0 \right) = \frac{z \cos \theta}{r \sin \theta}$

2. Secondo componente:
   \frac{1}{r} \left( \frac{\partial (r^2 \sin \theta)}{\partial \phi} - \frac{\partial (z r)}{\partial r} \right) = \frac{1}{r} \left( 0 - z \right) = -\frac{z}{r} $\frac{1}{r} \left( \frac{\partial (r^2 \sin \theta)}{\partial \phi} - \frac{\partial (z r)}{\partial r} \right) = \frac{1}{r} \left( 0 - z \right) = -\frac{z}{r}$

3. Terzo componente:
   \frac{1}{r} \left( \frac{\partial (r^2 \cos \theta) r}{\partial r} - \frac{\partial (r^2 \sin \theta)}{\partial \theta} \right) = \frac{1}{r} \left( 2r \cos \theta - r^2 \cos \theta \right) = \frac{2 \cos \theta - r \cos \theta}{r} = \frac{\cos \theta (2 - r)}{r} $\frac{1}{r} \left( \frac{\partial (r^2 \cos \theta) r}{\partial r} - \frac{\partial (r^2 \sin \theta)}{\partial \theta} \right) = \frac{1}{r} \left( 2r \cos \theta - r^2 \cos \theta \right) = \frac{2 \cos \theta - r \cos \theta}{r} = \frac{\cos \theta (2 - r)}{r}$

Quindi, il rotore del campo vettoriale in coordinate sferiche è:

\nabla \times \mathbf{F} = \left( \frac{z \cos \theta}{r \sin \theta}, -\frac{z}{r}, \frac{\cos \theta (2 - r)}{r} \right) $\nabla \times \mathbf{F} = \left( \frac{z \cos \theta}{r \sin \theta}, -\frac{z}{r}, \frac{\cos \theta (2 - r)}{r} \right)$

English version

Rotor Exercises

Definition

The rotor is an operator that measures the tendency of a vector field to rotate around a point. It is defined for three-dimensional vector fields and is denoted by the symbol \nabla \times \mathbf{F}$\nabla \times \mathbf{F}$, where \mathbf{F}$\mathbf{F}$ is a vector field.

Formula

If \mathbf{F} = (F_x, F_y, F_z)$\mathbf{F} = (F_x, F_y, F_z)$ is a vector field defined in a three-dimensional space, the rotor of \mathbf{F}$\mathbf{F}$ is given by:

\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z \end{vmatrix} 

$$\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z \end{vmatrix}$$

Where:

• \mathbf{i}, \mathbf{j}, \mathbf{k}$\mathbf{i}, \mathbf{j}, \mathbf{k}$ are the vectors of the Cartesian coordinates,
• F_x, F_y, F_z$F_x, F_y, F_z$ are the components of the vector field \mathbf{F}$\mathbf{F}$.

Expanding the determinant, we get:

\nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial_x} {\partial y} \right) 

$$\nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial_x} {\partial y} \right)$$

Rotor Properties

1. Linearity: The rotor is a linear operator. If \mathbf{F}$\mathbf{F}$ and \mathbf{G}$\mathbf{G}$ are vector fields and a$a$ is a constant, then:

\nabla \times (a\mathbf{F} + \mathbf{G}) = a(\nabla \times \mathbf{F}) + (\nabla \times \mathbf{G})

$$\nabla \times (a\mathbf{F} + \mathbf{G}) = a(\nabla \times \mathbf{F}) + (\nabla \times \mathbf{G})$$

2. Roost of a Gradient: The roost of the gradient of a scalar function is always zero:

\nabla \times (\nabla f) = 0

$$\nabla \times (\nabla f) = 0$$

This implies that conservative fields (fields arising from a scalar potential) have a zero roost.

3. Rotor of a Constant Vector Field: If \mathbf{F}$\mathbf{F}$ is a constant vector field, then:

\nabla \times \mathbf{F} = 0

$$\nabla \times \mathbf{F} = 0$$

Applications of the Rotor

1. Fluid Dynamics: In fluid dynamics, the rotor of a velocity field represents the vorticity of the fluid. A velocity field with a non-zero rotor indicates the presence of vortices and turbulence.

2. Electromagnetism: In Maxwell's equations, the rotor of the electric and magnetic fields is used to describe how electric and magnetic fields interact and propagate. For example, Faraday's law of electromagnetic induction is expressed as:

\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}

$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$$

3. Mechanical Engineering: The rotor is used to analyze the behavior of mechanical systems and to study the rotational motion of rigid bodies.

Exercises

Exercise 1: Calculating the Rotor

Calculate the rotor of the vector field \mathbf{F} = (y^2, x^2, z)$\mathbf{F} = (y^2, x^2, z)$.

Solution: We use the rotor formula:

\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 
y^2 & x^2 & z \end{vmatrix}

$$\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y^2 & x^2 & z \end{vmatrix}$$

Expanding the determinant, we get:

 \nabla \times \mathbf{F} = \left( \frac{\partial z}{\partial y} - \frac{\partial x^2}{\partial z}, \frac{\partial y^2}{\partial z} - \frac{\partial z}{\partial x}, \frac{\partial x^2}{\partial x} - \frac{\partial y^2}{\partial y} \right)

$$\nabla \times \mathbf{F} = \left( \frac{\partial z}{\partial y} - \frac{\partial x^2}{\partial z}, \frac{\partial y^2}{\partial z} - \frac{\partial z}{\partial x}, \frac{\partial x^2}{\partial x} - \frac{\partial y^2}{\partial y} \right)$$

Calculating the partial derivatives:

\nabla \times \mathbf{F} = \left( 0 - 0, 0 - 0, 2 - 2 \right) = (0, 0, 0)

$$\nabla \times \mathbf{F} = \left( 0 - 0, 0 - 0, 2 - 2 \right) = (0, 0, 0)$$

So, the curl of the vector field is zero.

Exercise 2: Curl in a Vortex Field

Calculate the curl of the vector field \mathbf{F} = (-y, x, 0)$\mathbf{F} = (-y, x, 0)$.

Solution:
We use the rotor formula:

\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -y & x & \end{vmatrix} 

$$\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -y & x & \end{vmatrix}$$

Expanding the determinant, we get:

\nabla \times \mathbf{F} = \left( \frac{\partial (0)}{\partial y} - \frac{\partial (x)}{\partial z}, \frac{\partial (-y)}{\partial z} - \frac{\partial (0)}{\partial x}, \frac{\partial (x)}{\partial x} - \frac{\partial (-y)}{\partial y} \right) 

$$\nabla \times \mathbf{F} = \left( \frac{\partial (0)}{\partial y} - \frac{\partial (x)}{\partial z}, \frac{\partial (-y)}{\partial z} - \frac{\partial (0)}{\partial x}, \frac{\partial (x)}{\partial x} - \frac{\partial (-y)}{\partial y} \right)$$

Calculating the partial derivatives:

1. \frac{\partial (0)}{\partial y} = 0$\frac{\partial (0)}{\partial y} = 0$
2. \frac{\partial (x)}{\partial z} = 0$\frac{\partial (x)}{\partial z} = 0$
3. \frac{\partial (-y)}{\ partial z} = 0$\frac{\partial (-y)}{\ partial z} = 0$
4. \frac{\partial (0)}{\partial x} = 0$\frac{\partial (0)}{\partial x} = 0$
5. \frac{\partial (x)}{\partial x} = 1$\frac{\partial (x)}{\partial x} = 1$
6. \frac{\partial (-y)}{\partial y} = -1$\frac{\partial (-y)}{\partial y} = -1$
   Substituting the results:

 \nabla \times \mathbf{F} = \left( 0 - 0, 0 - 0, 1 - (-1) \right) = (0, 0, 2)

$$\nabla \times \mathbf{F} = \left( 0 - 0, 0 - 0, 1 - (-1) \right) = (0, 0, 2)$$

So, the curl of the vector field \mathbf{F} = (-y, x, 0)$\mathbf{F} = (-y, x, 0)$ is:

\nabla \times \mathbf{F} = (0, 0, 2)

$$\nabla \times \mathbf{F} = (0, 0, 2)$$

Exercise 3: Computing the Curl of a Vector Field

Compute the curl of the vector field \mathbf{F} = (2xy, x^2, z)$\mathbf{F} = (2xy, x^2, z)$.

Solution:
We use the rotor formula in Cartesian coordinates:
\nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right)$\nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right)$
We calculate each component:

1. First component:
   \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = \frac{\partial z}{\partial y} - \frac{\partial (x^2)}{\partial z} = 0 - 0 = 0$\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = \frac{\partial z}{\partial y} - \frac{\partial (x^2)}{\partial z} = 0 - 0 = 0$
2. Second component:
   \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} = \frac{\partial (2xy)}{\partial z} - \frac{\partial z}{\partial x} = 0 - 0 = 0$\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} = \frac{\partial (2xy)}{\partial z} - \frac{\partial z}{\partial x} = 0 - 0 = 0$
3. Third component :
   \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = \frac{\partial (x^2)}{\partial x} - \frac{\partial (2xy)}{\partial y} = 2x - 2x = 0$\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = \frac{\partial (x^2)}{\partial x} - \frac{\partial (2xy)}{\partial y} = 2x - 2x = 0$

Therefore, the rotor is:

\nabla \times \mathbf{F} = (0, 0, 0) $\nabla \times \mathbf{F} = (0, 0, 0)$

Exercise 4: Curl of a Vector Field in Polar Coordinates

Calculate the curl of the vector field \mathbf{F} = (0, r^2, 0)$\mathbf{F} = (0, r^2, 0)$ in polar coordinates.

Solution:
In polar coordinates, the rotor is given by:
\nabla \times \mathbf{F} = \left( \frac{1}{r} \frac{\partial (r F_z)}{\partial r} - \frac{\partial F_r}{\partial z}, \frac{\partial F_r}{\partial z} - \frac{\partial F_z}{\partial r}, \frac{1}{r } \frac{\partial (r F_\theta)}{\partial r} - \frac{1}{r} \frac{\partial F_r}{\partial \theta} \right)$\nabla \times \mathbf{F} = \left( \frac{1}{r} \frac{\partial (r F_z)}{\partial r} - \frac{\partial F_r}{\partial z}, \frac{\partial F_r}{\partial z} - \frac{\partial F_z}{\partial r}, \frac{1}{r } \frac{\partial (r F_\theta)}{\partial r} - \frac{1}{r} \frac{\partial F_r}{\partial \theta} \right)$
Where F_r = 0$F_r = 0$, F_\theta = r^2$F_\theta = r^2$, and F_z = 0$F_z = 0$.

Let's calculate each component:

1. First component:
   \frac{1}{r} \frac{\partial (r \cdot 0)}{\partial r} - \frac{\partial (0)}{\partial z} = 0 - 0 = 0$\frac{1}{r} \frac{\partial (r \cdot 0)}{\partial r} - \frac{\partial (0)}{\partial z} = 0 - 0 = 0$
2. Second component:
   \frac{\partial (0)}{\partial z} - \frac{\partial (0)}{\partial r} = 0$\frac{\partial (0)}{\partial z} - \frac{\partial (0)}{\partial r} = 0$
3. Third component:
   \frac{1}{r} \frac{\partial (r \cdot r^2)}{\partial r} - \frac{1}{r} \frac{\partial (0)}{\partial \theta} = \frac{1}{r} \frac{\partial (r^3)}{\partial r} - 0 = \frac{1}{r} \cdot 3r^2 = 3r $\frac{1}{r} \frac{\partial (r \cdot r^2)}{\partial r} - \frac{1}{r} \frac{\partial (0)}{\partial \theta} = \frac{1}{r} \frac{\partial (r^3)}{\partial r} - 0 = \frac{1}{r} \cdot 3r^2 = 3r$

So, the curl of the vector field in polar coordinates is:

\nabla \times \mathbf{F} = (0, 0, 3r) $\nabla \times \mathbf{F} = (0, 0, 3r)$

Exercise 5: Curl of a Vector Field in Spherical Coordinates

Calculate the curl of the vector field \mathbf{F} = (r^2 \sin \theta, r^2 \cos \theta, z)$\mathbf{F} = (r^2 \sin \theta, r^2 \cos \theta, z)$ in spherical coordinates.

Solution:
In spherical coordinates, the rotor is given by:
\nabla \times \mathbf{F} = \left( \frac{1}{r \sin \theta} \left( \frac{\partial (F_\phi \sin \theta)}{\partial \theta} - \frac{\partial F_\theta}{\partial \phi} \right), \frac{1}{r} \left( \frac{\partial F_r}{\partial \phi} - \frac{\partial (F_\phi r)}{\partial r} \right), \frac{1}{r} \left( \frac{\partial (F_\theta r)}{\partial r} - \frac{\partial F_r}{\partial \theta} \right) \right)$\nabla \times \mathbf{F} = \left( \frac{1}{r \sin \theta} \left( \frac{\partial (F_\phi \sin \theta)}{\partial \theta} - \frac{\partial F_\theta}{\partial \phi} \right), \frac{1}{r} \left( \frac{\partial F_r}{\partial \phi} - \frac{\partial (F_\phi r)}{\partial r} \right), \frac{1}{r} \left( \frac{\partial (F_\theta r)}{\partial r} - \frac{\partial F_r}{\partial \theta} \right) \right)$
Where F_r = r^2 \sin \theta$F_r = r^2 \sin \theta$, F_\theta = r^2 \cos \theta$F_\theta = r^2 \cos \theta$, and F_\phi = z$F_\phi = z$.

Let's calculate each component:

1. First component:
   \frac{1}{r \sin \theta} \left( \frac{\partial (z \sin \theta)}{\partial \theta} - \frac{\partial (r^2 \cos \theta)}{\partial \phi} \right) = \frac{1}{r \sin \theta} \left( z \cos \theta - 0 \right) = \frac{z cos \theta}{r \sin \theta}$\frac{1}{r \sin \theta} \left( \frac{\partial (z \sin \theta)}{\partial \theta} - \frac{\partial (r^2 \cos \theta)}{\partial \phi} \right) = \frac{1}{r \sin \theta} \left( z \cos \theta - 0 \right) = \frac{z cos \theta}{r \sin \theta}$
2. Second component:
   \frac{1}{r} \left( \frac{\partial (r^2 \sin \theta)}{\partial \phi} - \frac{\partial (z r)}{\partial r} \right) = \frac{1}{r} \left( 0 - z \right) = -\frac{z}{r}$\frac{1}{r} \left( \frac{\partial (r^2 \sin \theta)}{\partial \phi} - \frac{\partial (z r)}{\partial r} \right) = \frac{1}{r} \left( 0 - z \right) = -\frac{z}{r}$
3. Third component:
   \frac{1}{r} \left( \frac{\partial (r^2 \cos \theta) r}{\partial r} - \frac{\partial (r^2 \sin \theta)}{\partial \theta} \right) = \frac{1}{r} \left( 2r \cos \theta - r^2 \cos \theta \right) = \frac{2 \cos \theta - r \cos \theta}{r} = \frac{\cos \theta (2 - r)}{r}$\frac{1}{r} \left( \frac{\partial (r^2 \cos \theta) r}{\partial r} - \frac{\partial (r^2 \sin \theta)}{\partial \theta} \right) = \frac{1}{r} \left( 2r \cos \theta - r^2 \cos \theta \right) = \frac{2 \cos \theta - r \cos \theta}{r} = \frac{\cos \theta (2 - r)}{r}$
   Thus, the rotor of the vector field in spherical coordinates is: \nabla \times \mathbf{F} = \left( \frac{z \cos \theta}{r \sin \theta}, -\frac{z}{r}, \frac{\cos \theta (2 - r)}{r} \right)$\nabla \times \mathbf{F} = \left( \frac{z \cos \theta}{r \sin \theta}, -\frac{z}{r}, \frac{\cos \theta (2 - r)}{r} \right)$