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Chapter 16: Problem 2

Find (a) the curl and (b) the divergence of the vector field. $$ \mathbf{F}(x, y, z)=x^{3} y z^{2} \mathbf{j}+y^{4} z^{3} \mathbf{k} $$

Short Answer

Expert verified
Curl: \((4y^3 z^3 - 2x^3 y z) \mathbf{i} + 3x^2 y z^2 \mathbf{k}\); Divergence: \(x^3 z^2 + 3y^4 z^2\).

Step by step solution

01

Understand Vector Field Components

The given vector field \( \mathbf{F}(x, y, z) = x^{3} y z^{2} \mathbf{j} + y^{4} z^{3} \mathbf{k} \) does not have an \( i \)-component, simplifying our calculations.Identify components: \( F_x = 0 \), \( F_y = x^3 y z^2 \), and \( F_z = y^4 z^3 \).
02

Calculate Curl of Vector Field

The curl of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \) is given by \( abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k} \).Substitute the components into the formula:- \( \frac{\partial R}{\partial y} = \frac{\partial}{\partial y} (y^4 z^3) = 4y^3 z^3 \)- \( \frac{\partial Q}{\partial z} = \frac{\partial}{\partial z} (x^3 y z^2) = 2x^3 y z \).Thus, the \( i \)-component is \( 4y^3 z^3 - 2x^3 y z \).- \( \frac{\partial P}{\partial z} = 0 \), since \( P = 0 \).- \( \frac{\partial R}{\partial x} = \frac{\partial}{\partial x} (y^4 z^3) = 0 \).Thus, the \( j \)-component is \( 0 \).- \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (x^3 y z^2) = 3x^2 y z^2 \).- \( \frac{\partial P}{\partial y} = 0 \), since \( P = 0 \).Thus, the \( k \)-component is \( 3x^2 y z^2 \).Therefore, \( abla \times \mathbf{F} = (4y^3 z^3 - 2x^3 y z) \mathbf{i} + 0 \mathbf{j} + 3x^2 y z^2 \mathbf{k} \).
03

Calculate Divergence of Vector Field

The divergence of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \) is given by \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \).Substitute the components into the formula:- \( \frac{\partial P}{\partial x} = 0 \), since \( P = 0 \).- \( \frac{\partial Q}{\partial y} = \frac{\partial}{\partial y} (x^3 y z^2) = x^3 z^2 \).- \( \frac{\partial R}{\partial z} = \frac{\partial}{\partial z} (y^4 z^3) = 3y^4 z^2 \).Thus, \( abla \cdot \mathbf{F} = 0 + x^3 z^2 + 3y^4 z^2 = x^3 z^2 + 3y^4 z^2 \).

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Key Concepts

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

Curl of a Vector Field
In vector calculus, the curl of a vector field helps us understand the rotation at a point in 3D space. It's particularly useful in physics for studying fluid dynamics and electromagnetism.
For a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \), the curl is given by the expression \( abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k} \).
Here's a breakdown:
  • The \( i \)-component involves the difference of the partial derivative of \( R \) with respect to \( y \) and \( Q \) with respect to \( z \).
  • The \( j \)-component is the difference between partials of \( P \) with respect to \( z \) and \( R \) with respect to \( x \).
  • The \( k \)-component involves the difference of \( Q \) with respect to \( x \) and \( P \) with respect to \( y \).
In our exercise:
  • Calculate \( \frac{\partial R}{\partial y} \) and \( \frac{\partial Q}{\partial z} \) to find the \( i \)-component.
  • The \( j \)-component simplifies here as the terms are zero due to \( P = 0 \).
  • For the \( k \)-component, find the difference between \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \).
This calculation reveals where and how the field rotates.
Divergence of a Vector Field
Divergence in a vector field illustrates how a vector field diverges or converges at a given point. It's a scalar-valued operation that is essential in fields like fluid dynamics and electromagnetics.
If you have a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \), then the divergence formula is \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \).
Let's break it down:
  • The first term involves the partial derivative of \( P \) with respect to \( x \).
  • The second term is the partial derivative of \( Q \) with respect to \( y \).
  • The final term involves \( R \) with respect to \( z \).
In this vector field:
  • Since \( P = 0 \), its derivative with respect to \( x \) is zero.
  • Calculate \( \frac{\partial Q}{\partial y} = x^3 z^2 \).
  • Find \( \frac{\partial R}{\partial z} = 3y^4 z^2 \).
Adding these values provides the measure of how much the vector field spreads out or contracts in space.
Partial Derivatives
Partial derivatives help us determine how a function changes as one of several variables changes while the others remain constant. They're crucial in multivariable calculus, providing insights for both divergence and curl.
For a function \( f(x, y, z) \), the partial derivative with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \), finds the rate of change in the \( x \)-direction while keeping other variables constant.
Consider a few key points:
  • The notation \( \frac{\partial}{\partial y} R \) expresses how \( R \) changes as \( y \) alone changes, ignoring variations in \( x \) or \( z \).
  • These derivatives can be plugged into the curl and divergence formulas, simplifying the process of obtaining these values.
  • In the exercise, observe how partial derivatives play a role in determining the rotation and divergence values for vector fields.
They are the building blocks for higher-order multi-variable functions, making them indispensable for engineers and scientists working in fields where changes occur across multiple dimensions.

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Most popular questions from this chapter

Prove the identity, assuming that the appropriate partial derivatives exist and are continuous. If \(f\) is a scalar field and \(\mathbf{F}, \mathbf{G}\) are vector fields, then \(f \mathbf{F}, \mathbf{F} \cdot \mathbf{G},\) and \(\mathbf{F} \times \mathbf{G}\) are defined by $$ \begin{aligned}(f \mathbf{F})(x, y, z) &=f(x, y, z) \mathbf{F}(x, y, z) \\\\(\mathbf{F} \cdot \mathbf{G})(x, y, z) &=\mathbf{F}(x, y, z) \cdot \mathbf{G}(x, y, z) \\\\(\mathbf{F} \times \mathbf{G})(x, y, z) &=\mathbf{F}(x, y, z) \times \mathbf{G}(x, y, z) \end{aligned} $$ $$ \operatorname{div}(\mathbf{F} \times \mathbf{G})=\mathbf{G} \cdot \operatorname{curl} \mathbf{F}-\mathbf{F} \cdot \operatorname{curl} \mathbf{G} $$

Find the area of the surface with vector equation \(\mathbf{r}(u, v)=\left\langle\cos ^{3} u \cos ^{3} v, \sin ^{3} u \cos ^{3} v, \sin ^{3} v\right\rangle, 0 \leqslant u \leqslant \pi\) \(0 \leqslant v \leqslant 2 \pi .\) State your answer correct to four decimal places.

Determine whether the points \(P\) and \(Q\) lie on the given surface. $$ \begin{array}{l}{\mathbf{r}(u, v)=\left\langle 1+u-v, u+v^{2}, u^{2}-v^{2}\right\rangle} \\ { P(1,2,1), Q(2,3,3)}\end{array} $$

(a) Evaluate the line integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r},\) where \(\mathbf{F}(x, y, z)=x \mathbf{i}-z \mathbf{j}+y \mathbf{k}\) and \(C\) is given by \(\mathbf{r}(t)=2 t \mathbf{i}+3 t \mathbf{j}-t^{2} \mathbf{k},-1 \leqslant t \leqslant 1 \leqslant 1\) (b) Illustrate part (a) by using a computer to graph \(C\) and the vectors from the vector field corresponding to \(t=\pm 1\) and \(\pm \frac{1}{2}\) (as in Figure 13 ).

We have seen that all vector fields of the form \(\mathbf{F}=\nabla g\) satisfy the equation curl \(\mathbf{F}=\mathbf{0}\) and that all vector fields of the form \(\mathbf{F}=\) curl \(\mathbf{G}\) satisfy the equation div \(\mathbf{F}=0\) (assuming continuity of the appropriate partial derivatives). This suggests the question: are there any equations that all functions of the form \(f=\) div \(\mathbf{G}\) must satisfy? Show that the answer to this question is "No" by proving that every continuous function \(f\) on \(\mathbb{R}^{3}\) is the divergence of some vector field. [Hint: Let \(\mathbf{G}(x, y, z)=\langle g(x, y, z), 0,0\rangle,\) where \(\left.g(x, y, z)=\int_{0}^{x} f(t, y, z) d t .\right]\)
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