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Chapter 6: Problem 225

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

Short Answer

Expert verified
The divergence of \( \mathbf{F} \) is \( 2x + 2y + 2z \).

Step by step solution

01

Understanding the Divergence

The divergence of a vector field \(\mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}\) is calculated using the formula: \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \). For the given vector field \(\mathbf{F}(x, y, z) = x^2 \mathbf{i} + y^2 \mathbf{j} + z^2 \mathbf{k}\), identify \(P = x^2\), \(Q = y^2\), and \(R = z^2\).
02

Partial Derivative with respect to x

Find the partial derivative of \(P = x^2\) with respect to \(x\). This is \(\frac{\partial P}{\partial x} = \frac{\partial}{\partial x}(x^2) = 2x\).
03

Partial Derivative with respect to y

Find the partial derivative of \(Q = y^2\) with respect to \(y\). This is \(\frac{\partial Q}{\partial y} = \frac{\partial}{\partial y}(y^2) = 2y\).
04

Partial Derivative with respect to z

Find the partial derivative of \(R = z^2\) with respect to \(z\). This is \(\frac{\partial R}{\partial z} = \frac{\partial}{\partial z}(z^2) = 2z\).
05

Sum of Partial Derivatives

Add the partial derivatives obtained: \(\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} = 2x + 2y + 2z\).

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Key Concepts

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

Vector Calculus
Vector calculus is a branch of mathematics that studies vector fields and how they can be differentiated and integrated. It's used extensively in physics and engineering to understand fields like electromagnetic and gravitational fields.
When working with vector fields, we often want to determine how a field behaves over space. One key operation in vector calculus is the 鈥榙ivergence鈥� of a vector field, which helps us understand how the field spreads outward from a certain point.
  • Vector Fields: A vector field assigns a vector to every point in space. For example, in the exercise, the vector field \( \mathbf{F}(x, y, z) = x^2 \mathbf{i} + y^2 \mathbf{j} + z^2 \mathbf{k} \) assigns a different vector to each point \((x, y, z)\).
  • Gradient and Divergence: Whereas a gradient measures how a scalar field changes, divergence measures how a vector field's flow spreads out. The formula given for divergence in vector calculus is \( abla \cdot \mathbf{F} \).
Partial Derivatives
Partial derivatives are a fundamental concept in calculus. They represent the rate of change of a function with respect to one variable while keeping others constant.
This concept is crucial when dealing with functions of multiple variables, like our vector field \( \mathbf{F}(x, y, z) \). To find the divergence, you use partial derivatives for each component of the vector field:
  • Partial Derivative with respect to \(x\): For \(P = x^2\), the rate of change in the x-direction is \(2x\).
  • Partial Derivative with respect to \(y\): For \(Q = y^2\), the rate of change in the y-direction is \(2y\).
  • Partial Derivative with respect to \(z\): For \(R = z^2\), the rate of change in the z-direction is \(2z\).
Understanding partial derivatives helps calculate how each part of a multivariable function behaves.
Vector Field Analysis
Vector field analysis allows us to examine how vector fields behave in space. By calculating the divergence, we gain insight into the nature of the vector field.
The divergence operation in vector fields measures the "spread" of vectors from a point. If the divergence is positive, vectors are spreading out from the point, resembling a source. If negative, they converge.
  • Divergence Calculation: To find the divergence of the vector field, sum the partial derivatives: \( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} = 2x + 2y + 2z \).
  • Interpretation of Divergence: A positive divergence here means that the vector field spreads out uniformly with increasing \(x\), \(y\), or \(z\). This reflects a field emanating uniformly from each point.
Vector field analysis helps in understanding electromagnetic, fluid flow, and other physical phenomena.

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Most popular questions from this chapter

Use the divergence theorem to evaluate \(\rrbracket_{S} \mathbf{F} \cdot d \mathbf{S}\), where \(\mathbf{F}(x, y, z)=x y \mathbf{i}-\frac{1}{2} y^{2} \mathbf{j}+z \mathbf{k}\) and \(S\) is the surface consisting of three pieces: \(z=4-3 x^{2}-3 y^{2}, 1 \leq z \leq 4\) on the top; \(x^{2}+y^{2}=1,0 \leq z \leq 1\) on the sides; and \(z=0\) on the bottom.

David and Sandra plan to evaluate line integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) along a path in the \(x y\) -plane from (0,0) to \((1,\) 1). The force field is \(\mathbf{F}(x, y)=(x+2 y) \mathbf{i}+\left(-x+y^{2}\right) \mathbf{j}\). David chooses the path that runs along the \(x\) -axis from (0, 0) to (1,0) and then runs along the vertical line \(x=1\) from (1, 0 ) to the final point (1,1) . Sandra chooses the direct path along the diagonal line \(y=x\) from (0,0) to (1,1) . Whose line integral is larger and by how much?

For the following exercises, without using Stokes' theorem, calculate directly both the flux of \(\operatorname{curl} \mathbf{F} \cdot \mathbf{N}\) over the given surface and the circulation integral around its boundary, assuming all boundaries are oriented clockwise as viewed from above. \(\mathbf{F}(x, y, z)=z \mathbf{i}+x \mathbf{j}+y \mathbf{k} ; \quad S \quad\) is \(\quad\) hemisphere \(z=\left(a^{2}-x^{2}-y^{2}\right)^{1 / 2}\)

Use the surface integral in Stokes' theorem to \(\begin{array}{llll}\text { calculate } & \text { the } & \text { circulation } & \text { of } & \text { field }\end{array}\) \(\mathbf{F}(x, y, z)=x^{2} y^{3} \mathbf{i}+\mathbf{j}+z \mathbf{k}\) around \(C,\) which is the intersection of cylinder \(x^{2}+y^{2}=4\) and hemisphere \(x^{2}+y^{2}+z^{2}=16, z \geq 0, \quad\) oriented counterclockwise when viewed from above.

Find the line integral of \(\mathbf{F}(x, y, z)=12 x^{2} \mathbf{i}-5 x y \mathbf{j}+x z \mathbf{k}\) over path \(C\) defined by \(y=x^{2}, \quad z=x^{3}\) from point (0,0,0) to point \((2,4,8) .\)
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