Problem 16
Question
In Exercises 9-20, use the Divergence Theorem to find the outward flux of \(\mathbf{F}\) across the boundary of the region \(D .\) Sphere \(\quad \mathbf{F}=x^{3} \mathbf{i}+y^{3} \mathbf{j}+z^{3} \mathbf{k}\) \(D :\) The solid sphere \(x^{2}+y^{2}+z^{2} \leq a^{2}\)
Step-by-Step Solution
Verified Answer
The divergence of \( \mathbf{F} \) is \( 3(x^2+y^2+z^2) \).
1Step 1: Understanding the Divergence Theorem
The Divergence Theorem states \( \iint_{\partial D} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_{D} abla \cdot \mathbf{F} \, dV \), where \( \mathbf{F} \) is a vector field, \( \partial D \) is the boundary of the region \( D \), and \( abla \cdot \mathbf{F} \) is the divergence of \( \mathbf{F} \). This theorem allows us to convert a surface integral into a volume integral.
2Step 2: Find the Divergence of \( \mathbf{F} \)
Given \( \mathbf{F} = x^{3} \mathbf{i} + y^{3} \mathbf{j} + z^{3} \mathbf{k} \), we find the divergence: \( abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^{3}) + \frac{\partial}{\partial y}(y^{3}) + \frac{\partial}{\partial z}(z^{3}) = 3x^{2} + 3y^{2} + 3z^{2} \).
Key Concepts
Flux CalculationSurface IntegralVolume IntegralVector Field Divergence
Flux Calculation
Flux is a fundamental concept in vector calculus used to compute the flow of a vector field across a surface. Consider a vector field \( \mathbf{F} \), which represents the flow within a space. The flux of \( \mathbf{F} \) through a surface is essentially the amount of \( \mathbf{F} \) passing through that surface along a particular direction. In this sphere exercise, the task is to determine the outward flux of \( \mathbf{F} \) across the boundary of the region \( D \). By leveraging the Divergence Theorem, the problem simplifies from a complex surface integral to a more manageable volume integral. This shift to volume significantly eases the calculation process in this context. It requires finding the divergence of \( \mathbf{F} \) to integrate over the sphere's volume.
Surface Integral
A surface integral extends the concept of a line integral to higher dimensions. It is used to calculate the total field passing through a 3D surface. For our sphere example, a surface integral of vector field \( \mathbf{F} \) would consider the entirety of the sphere's surface \( \partial D \).The surface integral \( \iint_{\partial D} \mathbf{F} \cdot \mathbf{n} \, dS \) involves integrating the dot product of \( \mathbf{F} \) and the unit normal vector \( \mathbf{n} \) across the boundary surface. This calculation directly measures the flow of \( \mathbf{F} \) through \( \partial D \). However, instead of performing this direct computation, the Divergence Theorem allows an alternative approach by converting the surface integral into an equivalent, and often simpler, volume integral.
Volume Integral
Volume integrals extend integrals into three-dimensional space, helping to evaluate a quantity over a volume in space. The Divergence Theorem assists in transforming a surface integral into this volume integral \( \iiint_{D} abla \cdot \mathbf{F} \, dV \). Here, you integrate the divergence of \( \mathbf{F} \) over the region \( D \) enclosed by the sphere.In our exercise, you've already derived the divergence of \( \mathbf{F} \) as \( 3x^2 + 3y^2 + 3z^2 \). This expression is further integrated over the sphere defined by \( x^2 + y^2 + z^2 \leq a^2 \), greatly simplifying the task of computing the flux through the surface by focusing instead on the volume.
Vector Field Divergence
Divergence is a measure of how much a vector field spreads out from a given point, and it is represented as \( abla \cdot \mathbf{F} \). For any vector field \( \mathbf{F} \), calculating the divergence involves taking partial derivatives of its components with respect to their respective variables and summing the results.In this problem, \( \mathbf{F} = x^3 \mathbf{i} + y^3 \mathbf{j} + z^3 \mathbf{k} \). The divergence is found as follows:
- Partial derivative of \( x^3 \) with respect to \( x \) gives \( 3x^2 \).
- Partial derivative of \( y^3 \) with respect to \( y \) gives \( 3y^2 \).
- Partial derivative of \( z^3 \) with respect to \( z \) gives \( 3z^2 \).
Other exercises in this chapter
Problem 15
The paths of integration for Exercises 15 and 16 Integrate \(f ( x , y , z ) = x + \sqrt { y } - z ^ { 2 }\) over the path from \(( 0,0,0 )\) to \(( 1,1,1 )\) (
View solution Problem 15
Let \(S\) be the cylinder \(x^{2}+y^{2}=a^{2}, 0 \leq z \leq h,\) together with its top, \(x^{2}+y^{2} \leq a^{2}, z=h .\) Let \(\mathbf{F}=-y \mathbf{i}+x \mat
View solution Problem 16
In Exercises \(1-16,\) find a parametrization of the surface. (There are many correct ways to do these, so your answers may not be the same as those in the back
View solution Problem 16
Integrate \(G(x, y, z)=x\) over the surface given by $$z=x^{2}+y \quad \text {for} \quad 0 \leq x \leq 1, \quad-1 \leq y \leq 1$$
View solution