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Magazine Chapter 16: Problem 25
Evaluate the surface integral \(\int_{S} \mathbf{F} \cdot d \mathbf{S}\) for the given vector field \(\mathbf{F}\) and the oriented surface \(S .\) In other words, find the flux of \(\mathbf{F}\) across \(S .\) For closed surfaces, use the positive (outward) orientation. \(\mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+z^{2} \mathbf{k}, \quad S\) is the sphere with radius 1 and center the origin
Short Answer
Expert verified
The flux is \( \frac{11\pi}{6} \).
Step by step solution
01
Understand the Problem
We are given a vector field \( \mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+z^{2} \mathbf{k} \) and the surface \( S \), which is a sphere of radius 1 centered at the origin. We need to calculate the flux of \( \mathbf{F} \) across this closed surface using the outward orientation.
02
Recognize the Surface Type
The surface \( S \) is a sphere, which is a closed surface. For closed surfaces, we can apply the Divergence Theorem, which states that the surface integral of a vector field \( \mathbf{F} \) across a closed surface is equal to the volume integral of the divergence of \( \mathbf{F} \) over the region \( V \) inside the surface.
03
Apply the Divergence Theorem
According to the Divergence Theorem, the flux can be calculated as: \[ \int_{S} \mathbf{F} \cdot d \mathbf{S} = \int_{V} (abla \cdot \mathbf{F}) \ dV \]where \( V \) is the volume inside the sphere.
04
Compute the Divergence of \( \mathbf{F} \)
First, find the divergence of the vector field \( \mathbf{F} \). The divergence is given by the formula: \[ abla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \]For our vector field, \( \mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+z^{2} \mathbf{k} \), we have:\[ abla \cdot \mathbf{F} = \frac{\partial }{\partial x}(x) + \frac{\partial }{\partial y}(y) + \frac{\partial }{\partial z}(z^2) = 1 + 1 + 2z = 2 + 2z \]
05
Set Up the Volume Integral
Substitute the divergence calculated into the volume integral. The volume integral now becomes: \[ \int_{V} (2 + 2z) \ dV \]The volume \( V \) is the interior of a sphere with radius 1, for which it's convenient to use spherical coordinates.
06
Convert to Spherical Coordinates
In spherical coordinates, the element of volume \( dV \) becomes \( \rho^2 \sin \phi \, d\rho \, d\theta \, d\phi \). The spherical coordinates \( (\rho, \theta, \phi) \) range, respectively, from 0 to 1, 0 to \( 2\pi \), and 0 to \( \pi \) for a full sphere.
07
Calculate the Integral
The integral becomes: \[ \int_{0}^{1} \int_{0}^{\pi} \int_{0}^{2\pi} (2 + 2\rho \cos \phi)(\rho^2 \sin \phi) \, d\theta \, d\phi \, d\rho \]After computing: \[ = 4\pi \int_{0}^{1} \left( \rho^2 \right) d\rho \ + 2\pi \int_{0}^{1} \left( \rho^3 \right) d\rho \] \[ = 4\pi \cdot \frac{1}{3} + 2\pi \cdot \frac{1}{4} \] \[ = \frac{4\pi}{3} + \frac{2\pi}{4} = \frac{11\pi}{6} \]
08
Final Evaluation
The calculation of the integrals gives the result of \( \frac{11\pi}{6} \). Therefore, the flux of \( \mathbf{F} \) across the sphere is \( \frac{11\pi}{6} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Flux Across a Surface
Flux is an important concept in vector calculus that measures how much of a vector field passes through a given surface. In this exercise, we are interested in finding the flux of a vector field \( \mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+z^{2} \mathbf{k} \) across a sphere.
The flux across a surface is given by the surface integral of the vector field over the surface:
The flux across a surface is given by the surface integral of the vector field over the surface:
- It shows how much of the field lines cross the surface and in which direction.
- An outward flux indicates that more field lines are exiting the surface than entering.
Sphere Geometry
The geometry of a sphere is fundamental in calculations involving spherical surfaces. A sphere is a perfectly round geometrical object in three-dimensional space, like the surface of a round ball. In the context of this exercise, the sphere:
- Has a unit radius.
- Is centered at the origin \( (0, 0, 0) \).
Spherical Coordinates
Spherical coordinates are a system of coordinates that is very useful when dealing with spherical geometries.
Instead of the usual Cartesian coordinates \((x, y, z)\), spherical coordinates \((\rho, \theta, \phi)\) consist of:
Instead of the usual Cartesian coordinates \((x, y, z)\), spherical coordinates \((\rho, \theta, \phi)\) consist of:
- \( \rho \): the radial distance from the origin.
- \( \theta \): the azimuthal angle, which ranges from 0 to \(2\pi\).
- \( \phi \): the polar angle, which ranges from 0 to \(\pi\).
Surface Integrals
Surface integrals extend the concept of integration to surfaces, allowing calculations involving vector fields across an area in space. In this exercise:
This reduces the complexity since, in a closed surface scenario, it changes a two-dimensional integration problem into a three-dimensional one, often making computations more manageable. This theorem is particularly useful for spherical and closed surfaces, as in the given exercise.
- The surface of interest is the sphere centered at the origin with radius 1.
- The surface integral of the vector field \( \mathbf{F} \) over the sphere gives us the flux through the sphere.
This reduces the complexity since, in a closed surface scenario, it changes a two-dimensional integration problem into a three-dimensional one, often making computations more manageable. This theorem is particularly useful for spherical and closed surfaces, as in the given exercise.
