Question

Calculate \(\nabla \cdot \mathbf{D}\) at the point specified if \((a) \mathbf{D}=\left(1 / z^{2}\right)\left[10 x y z \mathbf{a}_{x}+\right.\) \(\left.5 x^{2} z \mathbf{a}_{y}+\left(2 z^{3}-5 x^{2} y\right) \mathbf{a}_{z}\right]\) at \(P(-2,3,5) ;(b) \mathbf{D}=5 z^{2} \mathbf{a}_{\rho}+10 \rho z \mathbf{a}_{z}\) at \(P\left(3,-45^{\circ}, 5\right) ;(c) \mathbf{D}=2 r \sin \theta \sin \phi \mathbf{a}_{r}+r \cos \theta \sin \phi \mathbf{a}_{\theta}+r \cos \phi \mathbf{a}_{\phi}\) at \(P\left(3,45^{\circ},-45^{\circ}\right) .\)

Short answer

Question: Calculate the divergence of the vector field \(\mathbf{D}\) at the specified points in Cartesian, cylindrical, and spherical coordinates. Solution: a) In Cartesian coordinates, the divergence at point \((-2, 3, 5)\) is \(\nabla \cdot \mathbf{D}(-2, 3, 5) = 320\). b) In cylindrical coordinates, the divergence at point \((3, -45^\circ, 5)\) is \(\nabla \cdot \mathbf{D}(3, -45^\circ, 5) = \frac{340}{3}\). c) In spherical coordinates, the divergence at point \((3, 45^\circ, -45^\circ)\) is \(\nabla \cdot \mathbf{D}(3, 45^\circ, -45^\circ) = -6 + \frac{-1 + \sqrt{2}}{3}\).

Step-by-step solution

Calculate the divergence of \(\mathbf{D}\) in Cartesian coordinates#a)

To calculate the divergence of \(\mathbf{D}\) in Cartesian coordinates, we need to compute the following partial derivatives: \(\frac{\partial}{\partial x} (10 x y z)\), \(\frac{\partial}{\partial y} (5 x^2 z)\), and \(\frac{\partial}{\partial z} \left(2 z^3 - 5 x^2 y\right)\). We find the following: \(\frac{\partial}{\partial x} (10 x y z) = 10 y z\) \(\frac{\partial}{\partial y} (5 x^2 z) = 5 x^2\) \(\frac{\partial}{\partial z} \left(2 z^3 - 5 x^2 y\right) = 6z^2\) Now we compute the cartesian divergence of \(\mathbf{D}\) as: \(\nabla \cdot \mathbf{D} = 10 y z + 5 x^2 + 6 z^2\)

Evaluate the cartesian divergence of \(\mathbf{D}\) at the given point#a)

We find the cartesian divergence of \(\mathbf{D}\) at the given point \(P(-2, 3, 5)\) by substituting the coordinates of the point into the resulting equation from Step 1: \(\nabla \cdot \mathbf{D}(-2, 3, 5) = 10(3)(5) + 5(-2)^2 + 6(5)^2 = 150 + 20 + 150 = 320\)

Calculate the divergence of \(\mathbf{D}\) in cylindrical coordinates#b)

In cylindrical coordinates, we have \(\mathbf{D} = 5z^2 \mathbf{a}_{\rho} + 10\rho z \mathbf{a}_{z}\). To compute the divergence, we use: \(\nabla \cdot \mathbf{D} = \frac{1}{\rho}\frac{\partial}{\partial \rho} (\rho D_\rho) + \frac{\partial}{\partial z} (D_z)\) Computing the partial derivatives and applying the formula, we get: \(\nabla \cdot \mathbf{D} = \frac{1}{\rho}\frac{\partial}{\partial \rho} (5 \rho z^2) + \frac{\partial}{\partial z} (10 \rho z)\) \(\nabla \cdot \mathbf{D} = \frac{1}{\rho}(10 z^2) + 10 \rho\) \(\nabla \cdot \mathbf{D} = 10\left( \frac{z^2}{\rho} + \rho \right)\)

Evaluate the cylindrical divergence of \(\mathbf{D}\) at the given point#b)

The given point in cylindrical coordinates is \(P(3, -45^\circ, 5)\). We convert the angle to radians: \(-45^\circ = -\frac{\pi}{4}\). Now we substitute the coordinates into the divergence equation from Step 3: \(\nabla \cdot \mathbf{D}(3, -\frac{\pi}{4}, 5) = 10\left( \frac{25}{3} + 3 \right) = 10\left( \frac{34}{3} \right) = \frac{340}{3}\)

Calculate the divergence of \(\mathbf{D}\) in spherical coordinates#c)

In spherical coordinates, we have \(\mathbf{D} = 2r \sin \theta \sin \phi \mathbf{a}_{r} + r \cos \theta \sin \phi \mathbf{a}_{\theta} + r \cos \phi \mathbf{a}_{\phi}\). To compute the divergence, we use: \(\nabla \cdot \mathbf{D} = \frac{1}{r^2}\frac{\partial}{\partial r} (r^2 D_r) + \frac{1}{r \sin \theta}\frac{\partial}{\partial \theta} (\sin \theta D_\theta) + \frac{1}{r \sin \theta}\frac{\partial}{\partial \phi} (D_\phi)\) Computing the partial derivatives and applying the formula, we get: \(\nabla \cdot \mathbf{D} = \frac{1}{r^2}\frac{\partial}{\partial r} \left(r^2 (2r \sin \theta \sin \phi) \right) + \frac{1}{r \sin \theta}\frac{\partial}{\partial \theta} \left( \sin \theta (r \cos \theta \sin \phi) \right) + \frac{1}{r \sin \theta}\frac{\partial}{\partial \phi} (r \cos \phi)\) \(\nabla \cdot \mathbf{D} = \frac{1}{r^2}(6r^2 \sin \theta \sin \phi) + \frac{1}{r \sin \theta}(\sin^2 \theta \sin \phi) + \frac{1}{r \sin \theta}(-r \sin \phi)\) \(\nabla \cdot \mathbf{D} = 6 \sin \theta \sin \phi + \frac{\sin \theta \sin \phi - \sin \phi}{r}\)

Evaluate the spherical divergence of \(\mathbf{D}\) at the given point#c)

The given point in spherical coordinates is \(P(3, 45^\circ, -45^\circ)\). We convert the angles to radians: \(45^\circ = \frac{\pi}{4}\) and \(-45^\circ = -\frac{\pi}{4}\). Now we substitute the coordinates into the divergence equation from Step 5: \(\nabla \cdot \mathbf{D}(3, \frac{\pi}{4}, -\frac{\pi}{4}) = 6 \sin \left(\frac{\pi}{4}\right) \sin \left(-\frac{\pi}{4}\right) + \frac{\sin \left(\frac{\pi}{4}\right) \sin \left(-\frac{\pi}{4}\right) - \sin \left(-\frac{\pi}{4}\right)}{3}\) \(\nabla \cdot \mathbf{D}(3, \frac{\pi}{4}, -\frac{\pi}{4}) = 6 \left(\frac{1}{\sqrt{2}}\right) \left(-\frac{1}{\sqrt{2}}\right) + \frac{\left(\frac{1}{\sqrt{2}}\right) \left(-\frac{1}{\sqrt{2}}\right) -\left(-\frac{1}{\sqrt{2}}\right)}{3}\) \(\nabla \cdot \mathbf{D}(3, \frac{\pi}{4}, -\frac{\pi}{4}) = -6 + \frac{-1 + \sqrt{2}}{3}\) So the divergences at the specified points are: a) \(\nabla \cdot \mathbf{D}(-2, 3, 5) = 320\) b) \(\nabla \cdot \mathbf{D}(3, -45^\circ, 5) = \frac{340}{3}\) c) \(\nabla \cdot \mathbf{D}(3, 45^\circ, -45^\circ) = -6 + \frac{-1 + \sqrt{2}}{3}\)

Key concepts

Cartesian Coordinate System

In the study of vector calculus, the Cartesian coordinate system is fundamental. It involves three orthogonal axes: the x-axis, y-axis, and z-axis, which allow for representation of points in a three-dimensional space using a set of three coordinates \( (x, y, z) \).

In this exercise, we calculate the divergence of a vector field in Cartesian coordinates, which requires the computation of partial derivatives with respect to each of these three axes. For example, finding the divergence at a point involves substituting the specific x, y, and z values into the divergence formula. This system is especially useful because it simplifies to straightforward partial derivatives and simple multiplication when applying the divergence operator.

Cylindrical Coordinate System

When dealing with symmetries like those found in cylinders, the cylindrical coordinate system is more beneficial. It expresses a point in space with three coordinates \( (\rho, \phi, z) \), where \( \rho \) is the radial distance from the origin, \( \phi \) is the azimuthal angle about the z-axis, and \( z \) is the same as in the Cartesian system.

In this system, to calculate divergence, we need to know how to compute the derivatives with respect to \( \rho \) and \( z \) and understand how to adjust for the coordinate system's radial nature. This is evident in the exercise where specific formulas for divergence in cylindrical coordinates are employed, which take into account these nuances to evaluate the vector field.

Spherical Coordinate System

For problems involving spherical symmetry, the spherical coordinate system is extremely effective. Here, a point in space is described by three coordinates \( (r, \theta, \phi) \), where \( r \) is the radial distance from the origin, \( \theta \) is the polar angle from the z-axis, and \( \phi \) is the azimuthal angle in the x-y plane.

The divergence in spherical coordinates requires an understanding of this system's complexity, as seen in the provided exercise. The divergence operator in spherical coordinates includes derivatives with respect to each of these coordinates, and it requires specific multiplicative factors, such as \( r^2 \) and \( r \sin \theta \) to account for the changing volume element sizes in spherical coordinates.

Scalar Field Divergence

The divergence of a scalar field is not conventionally defined because divergence, by definition, applies to vector fields. However, the concept of divergence can be thought of as a measure of the rate at which the density of a physical quantity, represented by the vector field, changes in space.

In our exercise, divergence is calculated for a vector field across different coordinate systems. This divergence helps us understand how much the vector field is 'spreading out' from a point, which would be analogous to the spreading of a scalar quantity like temperature or pressure if they were modeled as vector fields.

Vector Calculus

Vector calculus is a mathematical discipline that deals with the differentiation and integration of vector fields, often in two or three-dimensional Euclidean space. Key operations in vector calculus include gradient, divergence, and curl.

In the context of this exercise, the focus is on the divergence operator, denoted as \( abla \cdot \), which measures the magnitude of a source or sink at a given point in a vector field. This is a crucial concept for understanding physical phenomena such as the flow of fluids or the electromagnetic field generated by charges.

Partial Derivatives

Partial derivatives represent the rate at which a function changes as one variable changes, whilst keeping the other variables constant. In the context of the Cartesian coordinate system, we compute partial derivatives with respect to \( x \) and \( y \) for a vector field \( \mathbf{D} \) as shown in the exercise steps.

Determining the partial derivatives is a necessary step in calculating divergence in any coordinate system because it allows us to measure how the vector field varies in each direction around a point. It’s also essential in both cylindrical and spherical coordinates, although the variables with respect to which we differentiate change according to the system used.