Question

Determine whether the following vector fields are conservative on the specified region. If so, determine a potential function. Let \(R^{*}\) and \(D^{*}\) be open regions of \(\mathbb{R}^{2}\) and \(\mathbb{R}^{3}\), respectively, that do not include the origin. $$\mathbf{F}=\langle y z, x z, x y\rangle \text { on } \mathbb{R}^{3}$$

Short answer

The given vector field is conservative on the specified region of \(\mathbb{R}^{3}\), and the potential function is \(\phi(x, y, z) = xyz\).

Step-by-step solution

Calculate the curl of the given vector field

The curl of a vector field \(\mathbf{F} = \langle F_1, F_2, F_3 \rangle\) can be calculated using the following determinant involving partial derivatives: $$ \nabla \times \mathbf{F} = \left\langle \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right\rangle $$ For our given vector field, \(\mathbf{F} = \langle y z, x z, x y \rangle\). Let's compute the curl for our vector field: $$ \nabla \times \mathbf{F} = \left\langle \frac{\partial (x y)}{\partial y} - \frac{\partial (x z)}{\partial z}, \frac{\partial (y z)}{\partial z} - \frac{\partial (x y)}{\partial x}, \frac{\partial (x z)}{\partial x} - \frac{\partial (y z)}{\partial y} \right\rangle = \left\langle x - x, y - y, z - z \right\rangle $$ So we have \(\nabla \times \mathbf{F} = \left\langle 0, 0, 0 \right\rangle\).

Determine if the vector field is conservative

Since the curl of the vector field is zero, i.e., \(\nabla \times \mathbf{F} = \left\langle 0, 0, 0 \right\rangle\), the vector field is conservative on the specified region of \(\mathbb{R}^{3}\).

Find a potential function

To find the potential function, we need to solve the following equations for a scalar function \(\phi(x, y, z)\): 1. \(\frac{\partial \phi}{\partial x} = yz\) 2. \(\frac{\partial \phi}{\partial y} = xz\) 3. \(\frac{\partial \phi}{\partial z} = xy\) Now, integrate each equation with respect to their respective variables. 1. Integrating equation (1) with respect to \(x\): \(\phi(x, y, z) = \int yz \, dx = xyz + g(y, z)\) 2. Integrating equation (2) with respect to \(y\): \(\phi(x, y, z) = \int xz \, dy = xyz + h(x, z)\) 3. Integrating equation (3) with respect to \(z\): \(\phi(x, y, z) = \int xy \, dz = xyz + k(x, y)\) Notice that we get the same term \(xyz\) when we integrate each equation. Let's combine these results: $$ \phi(x, y, z) = xyz + g(y, z) + h(x, z) + k(x, y) $$ From this equation, we see that \(g(y,z) = 0\), \(h(x,z)= 0\), and \(k(x,y) = 0\). Thus, the potential function is given by: $$ \phi(x, y, z) = xyz $$ We have determined that the vector field is conservative and the potential function is \(\phi(x, y, z) = xyz\).

Key concepts

Curl of a Vector Field

The concept of the **curl** of a vector field is fundamental in vector calculus. It provides a measure of the rotational motion in a field, capturing how the field "curls" around a point. Whenever you have a vector field \( \mathbf{F} = \langle F_1, F_2, F_3 \rangle \), the curl \( abla \times \mathbf{F} \) is determined by calculating partial derivatives and setting them into a cross product-like structure:\[abla \times \mathbf{F} = \left\langle \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right\rangle\]If the curl evaluates to zero, as it did in our vector field \( \mathbf{F} = \langle yz, xz, xy \rangle \), then the field has no rotational movement about any point in the region and is termed **conservative**. This indicates that the vector field can be described by a scalar potential function.

Potential Function

In vector calculus, a **potential function** is a scalar function whose gradient yields a given vector field. So, when a vector field is conservative, it can be expressed as the gradient of a potential function.For a given vector field \( \mathbf{F} = \langle F_1, F_2, F_3 \rangle \), the potential function \( \phi(x, y, z) \) satisfies:

• \( \frac{\partial \phi}{\partial x} = F_1 \)
• \( \frac{\partial \phi}{\partial y} = F_2 \)
• \( \frac{\partial \phi}{\partial z} = F_3 \)

To find this, integrate each component with respect to its variable, up to a constant function of the other variables. For example, integrating \( \frac{\partial \phi}{\partial x} = yz \) with respect to \( x \) gives \( \phi(x, y, z) = xyz + g(y, z) \). Continue similarly for each of the other variables.In our solved exercise, these integrations all reveal a similar structure, leading to the **potential function** \( \phi(x, y, z) = xyz \). This function encapsulates the conservative nature of our vector field, allowing us to explore further calculations.

Partial Derivatives

**Partial derivatives** are crucial in multivariable calculus. They provide an understanding of how a function changes as one specific variable changes, holding others constant.When dealing with vector fields and potential functions, partial derivatives are used to connect the field components to the potential function. For the field \( \mathbf{F} = \langle yz, xz, xy \rangle \):

• The partial derivative \( \frac{\partial F_1}{\partial x} \) gives information about how \( F_1 = yz \) changes with \( x \).
• Similarly, \( \frac{\partial F_2}{\partial y} \) shows how \( F_2 = xz \) changes with \( y \), and so on.

In finding a potential function, partial derivatives allow you to write equations like \( \frac{\partial \phi}{\partial x} = yz \). Solving these involves integrating with respect to single variables, providing insight into the structure and behavior of potential functions. By mastering partial derivatives, you'll have a powerful toolset to tackle problems in multidimensional calculus and physics.