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

Short Answer

Expert verified
The divergence is \( yz + 2x^{2}yz^{2} + 3y^{2}z^{2} \).

Step by step solution

01

Understanding the Expression

We are given a vector field \( \mathbf{F} \) where \( \mathbf{F}(x, y, z) = x y z \mathbf{i} + x^{2} y^{2} z^{2} \mathbf{j} + y^{2} z^{3} \mathbf{k} \). Our objective is to find the divergence of this vector field, which measures the rate at which the vector field is spreading out from a point.
02

Divergence Formula

The divergence of a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \) is given by \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \). Here, \( P = xy z \), \( Q = x^{2}y^{2}z^{2} \), and \( R = y^{2}z^{3} \).
03

Computing Partial Derivatives

1. **Partial Derivative of P with respect to x**: \( \frac{\partial}{\partial x} (xyz) = yz \).2. **Partial Derivative of Q with respect to y**: \( \frac{\partial}{\partial y} (x^{2}y^{2}z^{2}) = 2x^{2}yz^{2} \).3. **Partial Derivative of R with respect to z**: \( \frac{\partial}{\partial z} (y^{2}z^{3}) = 3y^{2}z^{2} \).
04

Calculating the Divergence

Substitute the partial derivatives into the divergence formula: \[abla \cdot \mathbf{F} = yz + 2x^{2}yz^{2} + 3y^{2}z^{2}.\]
05

Simplifying the Expression

Combine like terms and factors: \[ abla \cdot \mathbf{F} = yz + 2x^{2}yz^{2} + 3y^{2}z^{2} \] amely,- The term \( yz \) is already simplified, - The terms \( 2x^{2}yz^{2} \text{ and } 3y^{2}z^{2} \) are distinct and do not combine.

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Key Concepts

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

Vector Fields
A vector field is a function that assigns a vector to each point in space. Imagine wind speed at various points in the atmosphere; the speed and direction at each point can be described by a vector. In mathematics, we depict vector fields using vector functions, such as \( \mathbf{F}(x, y, z) = x y z \mathbf{i} + x^{2} y^{2} z^{2} \mathbf{j} + y^{2} z^{3} \mathbf{k} \).
The components \( P, Q, R \) represent the magnitude and direction of the field in the x, y, and z directions, respectively. Here's why vector fields are important:
  • Visualization: They help in visualizing how a vector quantity changes in space.
  • Applications: They are used in physics and engineering to describe force fields, fluid flow, and more.
Understanding vector fields aids in grasping more complex topics like divergence, as they form the foundation of scalar and vector calculus.
Partial Derivatives
Partial derivatives deal with functions of multiple variables, allowing us to see how the function changes concerning one variable at a time, while keeping others constant. This concept is crucial in multivariable calculus.
For example, consider the partial derivative of \( P = xy z \) with respect to \( x \):
  • To find \( \frac{\partial}{\partial x} (xyz) \), treat \( y \) and \( z \) as constants, resulting in \( yz \). This shows the sensitivity of \( P \) to changes in \( x \).
Other examples from our exercise include the partial derivatives of \( Q \) and \( R \):
  • For \( Q = x^{2}y^{2}z^{2} \), \( \frac{\partial}{\partial y} (x^{2}y^{2}z^{2}) = 2x^{2}yz^{2} \).
  • And for \( R = y^{2}z^{3} \), \( \frac{\partial}{\partial z} (y^{2}z^{3}) = 3y^{2}z^{2} \).
Derivatives like these allow us to measure how the vector field changes in each respective dimension, which is especially useful when calculating divergence.
Divergence Formula
Divergence is a scalar value that measures how much a vector field spreads out from or converges at a point. If you've seen water flowing and it seems like more water is flowing out from a source point, that is similar to divergence in a vector field.
The formula \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \) gives us a handy way to compute divergence. It sums up the partial derivatives of each component:
  • \( \frac{\partial P}{\partial x} = yz \)
  • \( \frac{\partial Q}{\partial y} = 2x^2yz^2 \)
  • \( \frac{\partial R}{\partial z} = 3y^2z^2 \)
By substituting these derivatives into the divergence formula, we simplify to find the overall divergence: \( abla \cdot \mathbf{F} = yz + 2x^2yz^2 + 3y^2z^2 \). This tells us how much and in which way the vector field fluxes or moves through a tiny volume around a point.

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Most popular questions from this chapter

For the following exercises, Fourier's law of heat transfer states that the heat flow vector \(\mathrm{F}\) at a point is proportional to the negative gradient of the temperature; that is, \(\mathbf{F}=-k \nabla T\), which means that heat energy flows hot regions to cold regions. The constant \(k>0\) is called the conductivity, which has metric units of joules per meter per second-kelvin or watts per meter-kelvin. A temperature function for region \(D\) is given. Use the divergence theorem to find net outward heat flux \(\iint_{S} \mathbf{F} \cdot \mathbf{N} d S=-k \iint_{S} \nabla T \cdot \mathbf{N} d S\) across the boundary \(S\) of \(D\), where \(k=1$$T(x, y, z)=100+x+2 y+z ; D=\\{(x, y, z): 0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1\\}\)

Compute divergence \(\mathbf{F}=(\sinh x) \mathbf{i}+(\cosh y) \mathbf{j}-x y z \mathbf{k}\).

For the following exercises, use geometric reasoning to evaluate the given surface integrals. $$ \iint_{S}(z \mathbf{k}) \cdot d \mathbf{S}, \text { where } S \text { is disc } x^{2}+y^{2} \leq 9 \text { on plane } z=4, \text { oriented with unit normal vectors pointing upward } $$

For the following exercises, write formulas for the vector fields with the given properties. Give a formula \(\mathbf{F}(x, y)=M(x, y) \mathbf{i}+N(x, y) \mathbf{j}\) for the vector field in a plane that has the properties that \(\mathbf{F}=0\) at \((0,0)\) and that at any other point \((a, b), \mathbf{F}\) is tangent to circle \(x^{2}+y^{2}=a^{2}+b^{2}\) and points in the clockwise direction with magnitude \(|\mathbf{F}|=\sqrt{a^{2}+b^{2}}\).

For the following exercises, evaluate \(\iint_{S} \mathbf{F} \cdot \mathbf{N} d s\) for vector field \(F\), where \(\mathbf{N}\) is an outward normal vector to surface \(S\). \(\mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}\), and \(S\) is hemisphere \(z=\sqrt{1-x^{2}-y^{2}} .\)

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