Problem 14
Question
In Problems 13-18, find div \(\mathbf{F}\) and curl \(\mathbf{F}\). $$ \mathbf{F}(x, y, z)=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k} $$
Step-by-Step Solution
Verified Answer
div \( \mathbf{F} = 2x + 2y + 2z \); curl \( \mathbf{F} = \mathbf{0} \).
1Step 1: Understand the Problem
We need to calculate both the divergence and the curl of the vector field \( \mathbf{F}(x, y, z) = x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k} \). The divergence gives us a scalar field from a vector field, while the curl gives us another vector field.
2Step 2: Apply Divergence Formula
The divergence of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \) is given by \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \). For our function, \( P = x^2 \), \( Q = y^2 \), and \( R = z^2 \).
3Step 3: Calculate Partial Derivatives for Divergence
Find the partial derivatives: \[ \frac{\partial P}{\partial x} = \frac{\partial}{\partial x}(x^2) = 2x \] \[ \frac{\partial Q}{\partial y} = \frac{\partial}{\partial y}(y^2) = 2y \] \[ \frac{\partial R}{\partial z} = \frac{\partial}{\partial z}(z^2) = 2z \]
4Step 4: Calculate Divergence
Substitute the partial derivatives into the divergence formula: \[ abla \cdot \mathbf{F} = 2x + 2y + 2z \]
5Step 5: Apply Curl Formula
The curl of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \) is given by \( abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k}\) for our function, \( P = x^2 \), \( Q = y^2 \), and \( R = z^2 \).
6Step 6: Calculate Partial Derivatives for Curl
Find the relevant partial derivatives: \[ \frac{\partial R}{\partial y} = \frac{\partial}{\partial y}(z^2) = 0 \] \[ \frac{\partial Q}{\partial z} = \frac{\partial}{\partial z}(y^2) = 0 \] \[ \frac{\partial P}{\partial z} = \frac{\partial}{\partial z}(x^2) = 0 \] \[ \frac{\partial R}{\partial x} = \frac{\partial}{\partial x}(z^2) = 0 \] \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(y^2) = 0 \] \[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x^2) = 0 \]
7Step 7: Calculate Curl
Substitute the partial derivatives into the curl formula: \[ abla \times \mathbf{F} = \left( 0 - 0 \right) \mathbf{i} + \left( 0 - 0 \right) \mathbf{j} + \left( 0 - 0 \right) \mathbf{k} = \mathbf{0} \]
8Step 8: Final Step: Compile the Results
The divergence of the vector field \( \mathbf{F} \) is \( 2x + 2y + 2z \), and the curl is \( \mathbf{0} \).
Key Concepts
DivergenceCurlPartial Derivatives
Divergence
The divergence of a vector field is a scalar that gives us an idea of how much the field spreads out or converges at a point. In mathematical terms, it's an operation that measures the magnitude of a vector field's source or sink at a given point. To find the divergence of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \), we use the formula: \[ abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \] In our exercise, we have \( P = x^2 \), \( Q = y^2 \), and \( R = z^2 \). By calculating the partial derivatives for each component:
- \( \frac{\partial P}{\partial x} = 2x \)
- \( \frac{\partial Q}{\partial y} = 2y \)
- \( \frac{\partial R}{\partial z} = 2z \)
Curl
Curl is another operation in vector calculus. It measures the rotation or twist of a vector field in three-dimensional space. Think of it like the swirl one might see in a whirlpool. Mathematically, the curl of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \) is found using the formula:\[ abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k} \] For our vector field with \( P = x^2 \), \( Q = y^2 \), \( R = z^2 \), we calculate the following partial derivatives:
- \( \frac{\partial R}{\partial y} = 0 \)
- \( \frac{\partial Q}{\partial z} = 0 \)
- \( \frac{\partial P}{\partial z} = 0 \)
- \( \frac{\partial R}{\partial x} = 0 \)
- \( \frac{\partial Q}{\partial x} = 0 \)
- \( \frac{\partial P}{\partial y} = 0 \)
Partial Derivatives
Partial derivatives are crucial in vector calculus as they allow us to understand how functions change with respect to one variable while keeping other variables constant. They are especially useful when dealing with functions of multiple variables.For a function \( f(x, y, z) \), the partial derivative with respect to \( x \) is expressed as \( \frac{\partial f}{\partial x} \). This represents the rate at which the function changes as \( x \) changes, while \( y \) and \( z \) remain constant. Similarly, we have partial derivatives with respect to \( y \) and \( z \).In our vector field \( \mathbf{F}(x, y, z) = x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k} \), each component function (like \( x^2 \), \( y^2 \), and \( z^2 \)) was differentiated with respect to its corresponding variable in the calculation of both divergence and curl.
- \( \frac{\partial}{\partial x}(x^2) = 2x \)
- \( \frac{\partial}{\partial y}(y^2) = 2y \)
- \( \frac{\partial}{\partial z}(z^2) = 2z \)
Other exercises in this chapter
Problem 13
\(\int_{C}(x+y+z) d x+x d y-y z d z ; C\) is the line segment from \((1,2,1)\) to \((2,1,0)\).
View solution Problem 13
Evaluate each line integral. \(\int_{C}(x+y+z) d x+x d y-y z d z ; C\) is the line segment from \((1,2,1)\) to \((2,1,0)\).
View solution Problem 14
\(\int_{C} x z d x+(y+z) d y+x d z ; C\) is the curve \(x=e^{t}\), \(y=e^{-t}, z=e^{2 t}, 0 \leq t \leq 1 .\)
View solution Problem 14
Find the mass of the surface \(z=1-\left(x^{2}+y^{2}\right) / 2\) over \(0 \leq x \leq 1,0 \leq y \leq 1\), if \(\delta(x, y, z)=k x y\).
View solution