Divergence And Curl Calculation With Proof Of Spherical Coordinate Orthogonality

This article delves into vector calculus, exploring concepts such as the divergence and curl of a vector field derived from a scalar potential function. We will specifically focus on calculating these quantities for the vector field F = ∇(xy³z²) at the point (1, -1, 1). Furthermore, we will provide a rigorous proof demonstrating the orthogonality of the spherical coordinate system. Understanding these concepts is crucial in various fields, including physics, engineering, and computer graphics, where vector fields and coordinate systems play a fundamental role. This exploration not only strengthens one's grasp of vector calculus but also highlights its practical applications in diverse scientific and technological domains. Mastering these techniques allows for a deeper understanding of phenomena governed by vector fields and coordinate transformations.

Part b: Finding Divergence and Curl of ${\vec{F}}$

Let's consider the vector field F defined as the gradient of the scalar function ${f(x, y, z) = xy^3z^2}$. That is, ${\vec{F} = \nabla(xy^3z^2)}$. Our goal is to determine the divergence (div F) and the curl (curl F) of this vector field at the specific point (1, -1, 1). These operations provide valuable insights into the behavior of the vector field; the divergence measures the field's tendency to expand or contract at a point, while the curl measures its rotational tendency. These concepts are foundational in understanding fluid dynamics, electromagnetism, and other areas of physics and engineering.

Calculating ${\vec{F}}$

First, we need to compute the gradient of the scalar function ${f(x, y, z) = xy^3z^2}$. The gradient, denoted by ${\nabla f}$, is a vector field whose components are the partial derivatives of ${f}$ with respect to each coordinate direction. Specifically, ${\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right)}$. Calculating the partial derivatives:

• ${\frac{\partial f}{\partial x} = y^3z^2}$
• ${\frac{\partial f}{\partial y} = 3xy^2z^2}$
• ${\frac{\partial f}{\partial z} = 2xy^3z}$

Therefore, the vector field F is given by:

\begin{equation} \vec{F} = \nabla(xy^3z2) = \langle y^3z2, 3xy^2z2, 2xy^3z \rangle \end{equation}

This expression for F is crucial for the subsequent calculations of divergence and curl. Each component of F represents the field's strength in the corresponding coordinate direction, and understanding these components is essential for visualizing the vector field's behavior.

Calculating div ${\vec{F}}$

The divergence of a vector field F = ${\<P, Q, R\>}$ is a scalar function that measures the flux density of the vector field at a given point. It is defined as:

\begin{equation} \mathrm{div} \ \vec{F} = \nabla \cdot \vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \end{equation}

where ${P = y^3z^2}$, ${Q = 3xy^2z^2}$, and ${R = 2xy^3z}$ are the components of F. Now, we compute the necessary partial derivatives:

• ${\frac{\partial P}{\partial x} = \frac{\partial}{\partial x}(y^3z^2) = 0}$
• ${\frac{\partial Q}{\partial y} = \frac{\partial}{\partial y}(3xy^2z^2) = 6xyz^2}$
• ${\frac{\partial R}{\partial z} = \frac{\partial}{\partial z}(2xy^3z) = 2xy^3}$

Thus, the divergence of F is:

\begin{equation} \mathrm{div} \ \vec{F} = 0 + 6xyz^2 + 2xy^3 = 6xyz^2 + 2xy^3 \end{equation}

Now, we evaluate the divergence at the point (1, -1, 1):

\begin{equation} \mathrm{div} \ \vec{F}(1, -1, 1) = 6(1)(-1)(1)^2 + 2(1)(-1)^3 = -6 - 2 = -8 \end{equation}

The divergence at (1, -1, 1) is -8, indicating that the vector field is converging or contracting at this point. This negative value signifies that, at this location, the inflow of the vector field is greater than the outflow. Understanding the divergence at specific points is essential for analyzing the overall behavior of the vector field and its impact on the surrounding space.

Calculating curl ${\vec{F}}$

The curl of a vector field F = ${\<P, Q, R\>}$ is a vector field that measures the rotation or circulation of F at a given point. It is defined as:

\begin{equation} \mathrm{curl} \ \vec{F} = \nabla \times \vec{F} = \left\langle \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right\rangle \end{equation}

We have ${P = y^3z^2}$, ${Q = 3xy^2z^2}$, and ${R = 2xy^3z}$. We compute the required partial derivatives:

• ${\frac{\partial R}{\partial y} = \frac{\partial}{\partial y}(2xy^3z) = 6xy^2z}$
• ${\frac{\partial Q}{\partial z} = \frac{\partial}{\partial z}(3xy^2z^2) = 6xy^2z}$
• ${\frac{\partial P}{\partial z} = \frac{\partial}{\partial z}(y^3z^2) = 2y^3z}$
• ${\frac{\partial R}{\partial x} = \frac{\partial}{\partial x}(2xy^3z) = 2y^3z}$
• ${\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(3xy^2z^2) = 3y^2z^2}$
• ${\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(y^3z^2) = 3y^2z^2}$

Now, we can construct the curl of F:

\begin{equation} \mathrm{curl} \ \vec{F} = \langle 6xy^2z - 6xy^2z, 2y^3z - 2y^3z, 3y^2z2 - 3y^2z2 \rangle = \langle 0, 0, 0 \rangle \end{equation}

Thus, the curl of F is the zero vector field. This result indicates that the vector field F is irrotational, meaning it has no rotational tendency. Such vector fields are often associated with conservative forces in physics, where the work done by the force is independent of the path taken.

Now, we evaluate the curl at the point (1, -1, 1), but since the curl is the zero vector field, it will be zero at any point:

\begin{equation} \mathrm{curl} \ \vec{F}(1, -1, 1) = \langle 0, 0, 0 \rangle \end{equation}

This confirms that at the point (1, -1, 1), the vector field F exhibits no rotational behavior.

Part c: Proving the Orthogonality of the Spherical Coordinate System

The spherical coordinate system is a three-dimensional coordinate system that defines the position of a point using three coordinates: ${\rho}$ (the radial distance from the origin), ${\theta}$ (the azimuthal angle), and ${\phi}$ (the polar angle). To prove that the spherical coordinate system is orthogonal, we need to show that the unit vectors associated with each coordinate direction are mutually orthogonal, meaning their dot products are zero. This orthogonality is a crucial property that simplifies many calculations in physics and engineering, particularly when dealing with problems involving spherical symmetry.

Spherical Coordinates and Unit Vectors

In spherical coordinates, a point in space is represented by ${(\rho, \theta, \phi)}$, where:

• ${\rho \ge 0}$ is the distance from the origin to the point.
• ${0 \le \theta < 2\pi}$ is the angle in the xy-plane measured from the positive x-axis.
• ${0 \le \phi \le \pi}$ is the angle from the positive z-axis to the point.

The transformation from spherical to Cartesian coordinates is given by:

\begin{equation} x = \rho \sin\phi \cos\theta \end{equation}

\begin{equation} y = \rho \sin\phi \sin\theta \end{equation}

\begin{equation} z = \rho \cos\phi \end{equation}

The unit vectors in the spherical coordinate system are denoted by ${\hat{\rho}}$, ${\hat{\theta}}$, and ${\hat{\phi}}$. They point in the direction of increasing ${\rho}$, ${\theta}$, and ${\phi}$, respectively. To find these unit vectors, we first compute the partial derivatives of the position vector r = ${\&lt;x, y, z\&gt;}$ with respect to ${\rho}$, ${\theta}$, and ${\phi}$:

\begin{equation} \frac{\partial \vec{r}}{\partial \rho} = \langle \sin\phi \cos\theta, \sin\phi \sin\theta, \cos\phi \rangle \end{equation}

\begin{equation} \frac{\partial \vec{r}}{\partial \theta} = \langle -\rho \sin\phi \sin\theta, \rho \sin\phi \cos\theta, 0 \rangle \end{equation}

\begin{equation} \frac{\partial \vec{r}}{\partial \phi} = \langle \rho \cos\phi \cos\theta, \rho \cos\phi \sin\theta, -\rho \sin\phi \rangle \end{equation}

These partial derivatives represent tangent vectors to the coordinate curves. To obtain the unit vectors, we need to normalize these vectors by dividing each by its magnitude:

\begin{equation} \hat{\rho} = \frac{\frac{\partial \vec{r}}{\partial \rho}}{\left|\frac{\partial \vec{r}}{\partial \rho}\right|} = \frac{\langle \sin\phi \cos\theta, \sin\phi \sin\theta, \cos\phi \rangle}{1} = \langle \sin\phi \cos\theta, \sin\phi \sin\theta, \cos\phi \rangle \end{equation}

\begin{equation} \hat{\theta} = \frac{\frac{\partial \vec{r}}{\partial \theta}}{\left|\frac{\partial \vec{r}}{\partial \theta}\right|} = \frac{\langle -\rho \sin\phi \sin\theta, \rho \sin\phi \cos\theta, 0 \rangle}{\rho \sin\phi} = \langle -\sin\theta, \cos\theta, 0 \rangle \end{equation}

\begin{equation} \hat{\phi} = \frac{\frac{\partial \vec{r}}{\partial \phi}}{\left|\frac{\partial \vec{r}}{\partial \phi}\right|} = \frac{\langle \rho \cos\phi \cos\theta, \rho \cos\phi \sin\theta, -\rho \sin\phi \rangle}{\rho} = \langle \cos\phi \cos\theta, \cos\phi \sin\theta, -\sin\phi \rangle \end{equation}

Proving Orthogonality

To prove the orthogonality of the spherical coordinate system, we need to show that the dot products of the unit vectors are zero. That is, we need to demonstrate that ${\hat{\rho} \cdot \hat{\theta} = 0}$, ${\hat{\rho} \cdot \hat{\phi} = 0}$, and ${\hat{\theta} \cdot \hat{\phi} = 0}$.

1. ${\hat{\rho} \cdot \hat{\theta}}$

\begin{equation} \hat{\rho} \cdot \hat{\theta} = (\sin\phi \cos\theta)(-\sin\theta) + (\sin\phi \sin\theta)(\cos\theta) + (\cos\phi)(0) \end{equation}

\begin{equation} = -\sin\phi \cos\theta \sin\theta + \sin\phi \sin\theta \cos\theta = 0 \end{equation}

2. ${\hat{\rho} \cdot \hat{\phi}}$

\begin{equation} \hat{\rho} \cdot \hat{\phi} = (\sin\phi \cos\theta)(\cos\phi \cos\theta) + (\sin\phi \sin\theta)(\cos\phi \sin\theta) + (\cos\phi)(-\sin\phi) \end{equation}

\begin{equation} = \sin\phi \cos\phi \cos^2\theta + \sin\phi \cos\phi \sin^2\theta - \cos\phi \sin\phi \end{equation}

\begin{equation} = \sin\phi \cos\phi(\cos^2\theta + \sin^2\theta) - \cos\phi \sin\phi \end{equation}

Since ${\cos^2\theta + \sin^2\theta = 1}$:

\begin{equation} = \sin\phi \cos\phi - \cos\phi \sin\phi = 0 \end{equation}

3. ${\hat{\theta} \cdot \hat{\phi}}$

\begin{equation} \hat{\theta} \cdot \hat{\phi} = (-\sin\theta)(\cos\phi \cos\theta) + (\cos\theta)(\cos\phi \sin\theta) + (0)(-\sin\phi) \end{equation}

\begin{equation} = -\cos\phi \sin\theta \cos\theta + \cos\phi \cos\theta \sin\theta = 0 \end{equation}

Since all dot products are zero, we have proven that the unit vectors ${\hat{\rho}}$, ${\hat{\theta}}$, and ${\hat{\phi}}$ are mutually orthogonal. Therefore, the spherical coordinate system is indeed orthogonal.

In summary, we have calculated the divergence and curl of the vector field F = ∇(xy³z²) at the point (1, -1, 1), finding that div F(1, -1, 1) = -8 and curl F(1, -1, 1) = ${\&lt;0, 0, 0\&gt;}$. Additionally, we provided a detailed proof demonstrating the orthogonality of the spherical coordinate system by showing that the dot products of the unit vectors ${\hat{\rho}}$, ${\hat{\theta}}$, and ${\hat{\phi}}$ are zero. These results underscore the importance of vector calculus and coordinate systems in mathematical and scientific applications. Understanding divergence and curl helps analyze vector field behavior, while the orthogonality of coordinate systems simplifies problem-solving in various contexts. These concepts are essential tools for anyone working in fields that rely on spatial analysis and mathematical modeling.