Chapter 19: Problem 43
Calculate the flux of the vector field through the surface. \(\vec{F}=-5 \vec{r}\) through the sphere of radius 2 centered at the origin.
Short Answer
Expert verified
The flux is \(-160 \pi\).
Step by step solution
01
Identify the vector field
The vector field is given by \( \vec{F} = -5 \vec{r} \), where \( \vec{r} = (x, y, z) \). This represents the vector field as \( \vec{F} = (-5x, -5y, -5z) \).
02
Define the surface
The surface is a sphere centered at the origin with a radius of 2. The equation for this sphere is \( x^2 + y^2 + z^2 = 4 \).
03
Use the Divergence Theorem
The Divergence Theorem states that the flux through a closed surface can be converted to a volume integral of the divergence of \( \vec{F} \) over the volume enclosed by the surface. Thus, \( \iint_S \vec{F} \cdot d\vec{S} = \iiint_V (abla \cdot \vec{F}) \, dV \).
04
Calculate the divergence of F
The divergence of \( \vec{F} = (-5x, -5y, -5z) \) is calculated as follows: \( abla \cdot \vec{F} = \frac{\partial}{\partial x}(-5x) + \frac{\partial}{\partial y}(-5y) + \frac{\partial}{\partial z}(-5z) = -5 - 5 - 5 = -15 \).
05
Set up the volume integral
The volume integral thus becomes \( \iiint_V (-15) \, dV \), where \( V \) is the volume of the sphere. The volume of a sphere is \( \frac{4}{3} \pi (2)^3 = \frac{32}{3} \pi \).
06
Calculate the volume integral
Multiply the divergence by the volume of the sphere: \( (-15) \times \frac{32}{3} \pi = -160 \pi \).
07
Conclude the solution
The flux of the vector field through the surface of the sphere is therefore \( -160 \pi \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Divergence Theorem
The Divergence Theorem is a powerful tool for simplifying the calculation of flux in certain problems. It allows us to convert a surface integral into a volume integral, which is often easier to compute. The theorem states that if you have a vector field \( \vec{F} \) and a closed surface \( S \) enveloping a volume \( V \), the flux through \( S \) is equal to the integral of the divergence of \( \vec{F} \) over \( V \). This is represented mathematically as:
\[ \iint_S \vec{F} \cdot d\vec{S} = \iiint_V (abla \cdot \vec{F}) \, dV \]
The divergence measures how much the vector field spreads out from a given point. If a vector field has a positive divergence, it indicates that more of the field is "flowing out" than "flowing in." Conversely, a negative divergence signals more inflows toward a point.
Understanding and applying the Divergence Theorem is crucial when dealing with closed surfaces like spheres, as demonstrated in the problem, turning surface computations into more manageable volume computations.
\[ \iint_S \vec{F} \cdot d\vec{S} = \iiint_V (abla \cdot \vec{F}) \, dV \]
The divergence measures how much the vector field spreads out from a given point. If a vector field has a positive divergence, it indicates that more of the field is "flowing out" than "flowing in." Conversely, a negative divergence signals more inflows toward a point.
Understanding and applying the Divergence Theorem is crucial when dealing with closed surfaces like spheres, as demonstrated in the problem, turning surface computations into more manageable volume computations.
Vector Fields
A vector field assigns a vector to every point in space. It is a way of specifying a direction and magnitude for regions in mathematical spaces. In our exercise, the vector field is defined as \( \vec{F} = -5 \vec{r} \), where \( \vec{r} \) is the position vector \( (x, y, z) \). Therefore, \( \vec{F} \) explicitly becomes \( (-5x, -5y, -5z) \).
Key characteristics of vector fields include:
Key characteristics of vector fields include:
- Directionality: The vector points in a specific direction at each point.
- Magnitude: This is represented by the length of the arrow at each point.
Surface Integral
A surface integral extends the concept of integrating over a curve to integrating over a surface. Surface integrals are used when you want to calculate the flux of a vector field through a surface. In our task, the surface is a sphere centered at the origin.
The equation of the sphere is \( x^2 + y^2 + z^2 = 4 \), which describes all points equidistant from a central point (the origin, in this case) by a radius of 2. Surface integrals involve summing up vector field values over each differential element of the surface area, represented as \( \iint_S \vec{F} \cdot d\vec{S} \).
When using the Divergence Theorem, we convert this surface integral into a volume integral, making the calculation process much simpler. This is especially useful for surfaces that enclose a volume, like a sphere, where checking each differential surface element can be quite complex.
The equation of the sphere is \( x^2 + y^2 + z^2 = 4 \), which describes all points equidistant from a central point (the origin, in this case) by a radius of 2. Surface integrals involve summing up vector field values over each differential element of the surface area, represented as \( \iint_S \vec{F} \cdot d\vec{S} \).
When using the Divergence Theorem, we convert this surface integral into a volume integral, making the calculation process much simpler. This is especially useful for surfaces that enclose a volume, like a sphere, where checking each differential surface element can be quite complex.
Volume Integral
Volume integrals allow us to calculate values over three-dimensional spaces. In our problem, after establishing the divergence of the vector field, we proceed with a volume integral to simplify the task at hand.
The divergence calculated was \(-15\) for the vector field \( \vec{F} \). Thus, the volume integral equation becomes:
\[ \iiint_V (-15) \, dV \]The sphere's volume helps determine the bounds of the integral, calculated using its volume formula \( \frac{4}{3} \pi r^3 \), resulting in:
\[ \frac{4}{3} \pi (2)^3 = \frac{32}{3} \pi \]Multiplying the divergence by this volume gives the flux through the sphere's surface. This straightforward calculation simplifies what would otherwise be a cumbersome surface computation. Understanding volume integrals is essential when applying them in contexts like calculating mass, charge, or any property distributed throughout a region.
The divergence calculated was \(-15\) for the vector field \( \vec{F} \). Thus, the volume integral equation becomes:
\[ \iiint_V (-15) \, dV \]The sphere's volume helps determine the bounds of the integral, calculated using its volume formula \( \frac{4}{3} \pi r^3 \), resulting in:
\[ \frac{4}{3} \pi (2)^3 = \frac{32}{3} \pi \]Multiplying the divergence by this volume gives the flux through the sphere's surface. This straightforward calculation simplifies what would otherwise be a cumbersome surface computation. Understanding volume integrals is essential when applying them in contexts like calculating mass, charge, or any property distributed throughout a region.