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Find the divergence of the field. $$\mathbf{F}=(x-y+z) \mathbf{i}+(2 x+y-z) \mathbf{j}+(3 x+2 y-2 z) \mathbf{k}$$

Short Answer

Expert verified
The divergence of the field is 0.

Step by step solution

01

Understand the Divergence Operator

The divergence of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \) is given by the scalar combination of partial derivatives: \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \). This operator measures the 'spread' or 'outflow' of a field at a given point.
02

Identify the Component Functions

From the given vector field \( \mathbf{F} = (x-y+z) \mathbf{i} + (2x+y-z) \mathbf{j} + (3x+2y-2z) \mathbf{k} \), identify the component functions: \( P = x - y + z \), \( Q = 2x + y - z \), \( R = 3x + 2y - 2z \).
03

Compute Partial Derivative of P with respect to x

Calculate \( \frac{\partial P}{\partial x} \). Since \( P = x - y + z \) and \( x \) is the variable, differentiate: \( \frac{\partial}{\partial x}(x - y + z) = 1 \).
04

Compute Partial Derivative of Q with respect to y

Calculate \( \frac{\partial Q}{\partial y} \). With \( Q = 2x + y - z \), differentiate with respect to \( y \): \( \frac{\partial}{\partial y}(2x + y - z) = 1 \).
05

Compute Partial Derivative of R with respect to z

Calculate \( \frac{\partial R}{\partial z} \). For \( R = 3x + 2y - 2z \), differentiate with respect to \( z \): \( \frac{\partial}{\partial z}(3x + 2y - 2z) = -2 \).
06

Sum the Partial Derivatives

Sum the results of the partial derivatives to find the divergence: \( abla \cdot \mathbf{F} = 1 + 1 - 2 \).
07

Final Calculation

Perform the final arithmetic operation: \( abla \cdot \mathbf{F} = 1 + 1 - 2 = 0 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Fields
In mathematics, a vector field is an assignment of a vector to each point in a subset of space. Think of it as a region in space where each location is associated with a vector that has both a direction and a magnitude. Vector fields are commonly used to model various types of flows and forces.
  • The wind velocity at each point in the atmosphere creates a vector field.
  • The magnetic field surrounding a magnet is another example of a vector field.
In our example, the vector field is given by \( \mathbf{F}=(x-y+z) \mathbf{i}+(2 x+y-z) \mathbf{j}+(3 x+2 y-2 z) \mathbf{k} \). Each of these components, \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), corresponds to the field's effect in the x, y, and z directions respectively. Understanding the nature of vector fields really helps us grasp phenomena like fluid flow or electromagnetic fields in physics.
Partial Derivatives
Partial derivatives play a crucial role in understanding how functions change when involving multiple variables. In essence, they help us examine how a complicated function varies when we vary one of its input variables while keeping others constant.
  • If you have a function \( f(x, y) \), then \( \frac{\partial f}{\partial x} \) shows how \( f \) changes as \( x \) changes, holding \( y \) constant.
  • Similarly, \( \frac{\partial f}{\partial y} \) gives you the change in \( f \) when varying \( y \) alone.
In our problem, we calculate the partial derivatives of the vector field components \( P, Q, \) and \( R \) with respect to their respective variables \( x, y, \) and \( z \). This yields important insights into how each component contributes to the field's divergence.
Divergence Operator
The divergence operator is a key tool in vector calculus. It helps us measure the magnitude of a source or sink at a given point in a vector field, effectively indicating how much a field is spreading out or converging at that location.
  • For a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \), the divergence is \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \).
  • A positive divergence indicates a source or originating point, whereas a negative divergence suggests a sink or disappearing point.
In our example, the computed divergence is zero \((abla \cdot \mathbf{F} = 0)\), illustrating a situation where the vector field has a balanced distribution. It neither sources nor sinks at any point in the space, signifying a perfectly conserved field. This balance is useful to understand in contexts like fluid mechanics or electromagnetic theory.

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Most popular questions from this chapter

Green's first formula \(\quad\) Suppose that \(f\) and \(g\) are scalar functions with continuous first- and second-order partial derivatives throughout a region \(D\) that is bounded by a closed piecewise smooth surface \(S\). Show that $$\iint_{S} f \nabla g \cdot \mathbf{n} d \sigma=\iiint_{D}\left(f \nabla^{2} g+\nabla f \cdot \nabla g\right) d V.$$ Equation (10) is Green's first formula. (Hint: Apply the Divergence Theorem to the field \(\mathbf{F}=f \nabla g .\) )

Let \(S\) be the portion of the cylinder \(y=\ln x\) in the first octant whose projection parallel to the \(y\) -axis onto the \(x z\) -plane is the rectangle \(R_{k}: 1 \leq x \leq e, 0 \leq z \leq 1 .\) Let \(n\) be the unit vector normal to \(S\) that points away from the \(x z\) -plane. Find the flux of \(\mathbf{F}=2 y \mathbf{j}+z \mathbf{k}\) through \(S\) in the direction of \(\mathbf{n}\)

A revealing experiment By experiment, you find that a force field \(\mathbf{F}\) performs only half as much work in moving an object along path \(C_{1}\) from \(A\) to \(B\) as it does in moving the object along path \(C_{2}\) from \(A\) to \(B .\) What can you conclude about F? Give reasons for your answer.

Find the outward flux of the ficld \(\quad \mathbf{F}=x z \mathbf{i}+y z \mathbf{j}+\mathbf{k}\) across the surface of the upper cap cut from the solid sphere \(x^{2}+y^{2}+z^{2} \leq 25\) by the plane \(z=3\).

Use the surface integral in Stokes' Theorem to calculate the flux of the curl of the field \(\mathbf{F}\) across the surface \(S .\) $$\mathbf{F}=(x-y) \mathbf{i}+(y-z) \mathbf{j}+(z-x) \mathbf{k}$$ \(S: \quad \mathbf{r}(r, \theta)=(r \cos \theta) \mathbf{i}+(r \sin \theta) \mathbf{j}+(5-r) \mathbf{k}, 0 \leq r \leq 5, \quad 0 \leq \theta \leq 2 \pi,\) in the direction away from the \(z\)-axis.

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