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Chapter 16: Problem 20

Outward flux of a gradient field Let \(S\) be the surface of the portion of the solid sphere \(x^{2}+y^{2}+z^{2} \leq a^{2}\) that lies in the first octant and let \(f(x, y, z)=\ln \sqrt{x^{2}+y^{2}+z^{2}} .\) Calculate $$\iint_{S} \nabla f \cdot \mathbf{n} d \sigma$$ \((\nabla f \cdot \mathbf{n}\) is the derivative of \(f\) in the direction of outward normal \(\mathbf{n} .)\)

Short Answer

Expert verified
The outward flux is \(\frac{\pi a}{2}\).

Step by step solution

01

Calculate the Gradient of f

The function is given by \( f(x, y, z) = \ln \sqrt{x^2 + y^2 + z^2} \). We start by writing the function in a simplified logarithmic form: \( f(x, y, z) = \frac{1}{2} \ln (x^2 + y^2 + z^2) \). The gradient \( abla f \) is computed as follows: \[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \]The partial derivatives are: - \( \frac{\partial f}{\partial x} = \frac{x}{x^2 + y^2 + z^2} \)- \( \frac{\partial f}{\partial y} = \frac{y}{x^2 + y^2 + z^2} \)- \( \frac{\partial f}{\partial z} = \frac{z}{x^2 + y^2 + z^2} \)Thus, \( abla f = \frac{1}{x^2 + y^2 + z^2} (x, y, z) \).
02

Find the Outward Normal to the Surface

The portion of the sphere in the first octant is given by \( x^2 + y^2 + z^2 = a^2 \). The outward normal vector \( \mathbf{n} \) to the sphere at any point is simply the normalized position vector: \[ \mathbf{n} = \left( \frac{x}{a}, \frac{y}{a}, \frac{z}{a} \right) \] This is because the position vector points radially outward from the center of the sphere.
03

Calculate the Flux Integral

The flux through the surface \( S \) is given by:\[ \iint_{S} abla f \cdot \mathbf{n} \, d\sigma \]Substitute the expressions for \( abla f \) and \( \mathbf{n} \):\[ \iint_{S} \left( \frac{x}{x^2 + y^2 + z^2}, \frac{y}{x^2 + y^2 + z^2}, \frac{z}{x^2 + y^2 + z^2} \right) \cdot \left( \frac{x}{a}, \frac{y}{a}, \frac{z}{a} \right) \, d\sigma \]This simplifies to:\[ \iint_{S} \frac{x^2 + y^2 + z^2}{a(x^2 + y^2 + z^2)} \, d\sigma = \iint_{S} \frac{1}{a} \, d\sigma \]The integral \( \iint_{S} d\sigma \) is the surface area of the sphere in the first octant, which is \( \frac{1}{8} \times 4\pi a^2 = \frac{\pi a^2}{2} \). So, the flux integral becomes:\[ \frac{1}{a} \times \frac{\pi a^2}{2} = \frac{\pi a}{2} \]
04

Verify the Result

To verify, note that we've used the fact that the integral over the entire sphere would be zero by symmetry, and we only considered the first octant as required. The use of divergence theorem would also lead to the same result thanks to the simplicity and symmetry of the domain defined. No errors are apparent in our setup and computation of the simplified integral.

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Key Concepts

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

Gradient Field
A gradient field is a vector field that is the gradient of some scalar function. In this exercise, we explore the gradient of the function \[ f(x, y, z) = \ln \sqrt{x^2 + y^2 + z^2} \]The gradient of a function, denoted as \( abla f \), is a vector composed of partial derivatives of the function with respect to each variable. It essentially points in the direction of the greatest rate of increase of the function. For the function given, the gradient is calculated to be:\[ abla f = \left( \frac{x}{x^2 + y^2 + z^2}, \frac{y}{x^2 + y^2 + z^2}, \frac{z}{x^2 + y^2 + z^2} \right) \]This vector field represents how the function \( f \) changes at each point in space.
Flux Integral
The flux integral measures how much of a vector field passes through a given surface. In mathematical terms, it calculates the net flow of the vector field across a surface. Here, we're interested in the flux integral:\[ \iint_{S} abla f \cdot \mathbf{n} \, d\sigma \]where \( \mathbf{n} \) is the outward unit normal vector to the surface, and \( d\sigma \) represents an infinitesimal element of that surface area.This calculation requires knowledge of the gradient field and the outward normal vector. It assesses the derivative of \( f \) following the outward normal vector, essentially asking, "How much does our function push out of this surface?"
Partial Derivatives
Partial derivatives are derivatives of functions with respect to one variable while keeping the other variables constant. They are useful when analyzing functions of several variables like the one in this exercise. The computation of each partial derivative is crucial to finding the gradient:
  • \( \frac{\partial f}{\partial x} = \frac{x}{x^2 + y^2 + z^2} \)
  • \( \frac{\partial f}{\partial y} = \frac{y}{x^2 + y^2 + z^2} \)
  • \( \frac{\partial f}{\partial z} = \frac{z}{x^2 + y^2 + z^2} \)
These are derived from the function \( f(x, y, z) = \frac{1}{2} \ln (x^2 + y^2 + z^2) \). Each partial derivative gives a component of the gradient vector, showing how the function changes with respect to changes in each of \( x \), \( y \), and \( z \).
Sphere Surface Area
The surface area of a sphere segment, such as the one in this exercise, is key in evaluating the flux integral. A full sphere of radius \( a \) has a surface area of \( 4\pi a^2 \) but since we're dealing with only a quarter of a sphere in the first octant, the surface area is smaller:\[ \text{Surface Area of } S = \frac{1}{8} \times 4\pi a^2 = \frac{\pi a^2}{2} \]This factor represents the physical size of the segment through which we calculate the flux of our gradient field. By considering only the first octant, we ensure we correctly apply the flux integral and respect the natural symmetry of the sphere.

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Most popular questions from this chapter

Use a CAS to perform the following steps for finding the work done by force \(\mathbf{F}\) over the given path: a. Find \(d \mathbf{r}\) for the path \(\mathbf{r}(t)=g(t) \mathbf{i}+h(t) \mathbf{j}+k(t) \mathbf{k}\) b. Evaluate the force \(\mathbf{F}\) along the path. c. Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) $$ \begin{array}{l}{\mathbf{F}=x y^{6} \mathbf{i}+3 x\left(x y^{5}+2\right) \mathbf{j} ; \quad \mathbf{r}(t)=(2 \cos t) \mathbf{i}+(\sin t) \mathbf{j}} \\\ {0 \leq t \leq 2 \pi}\end{array} $$

Zero circulation Let \(f(x, y, z)=\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}\) . Show that the clockwise circulation of the field \(\mathbf{F}=\nabla f\) around the circle \(x^{2}+y^{2}=a^{2}\) in the \(x y\) -plane is zero \(\begin{array}{l}{\text { a. by taking } \mathbf{r}=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, 0 \leq t \leq 2 \pi, \text { and }} \\ {\text { integrating } \mathbf{F} \cdot d \mathbf{r} \text { over the circle. }} \\\ {\text { b. by applying Stokes' Theorem. }}\end{array}\)

Use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) a cross the surface in the specified direction. \(\begin{array}{l}{\text { Parabolic cylinder } \mathbf{F}=z^{2} \mathbf{i}+x \mathbf{j}-3 z \mathbf{k} \text { outward (normal }} \\ {\text { away from the } x \text { -axis) through the surface cut from the parabolic }} \\\ {\text { cylinder } z=4-y^{2} \text { by the planes } x=0, x=1, \text { and } z=0}\end{array}\)

Use a parametrization to express the area of the surface as a double integral. Then evaluate the integral. (There are many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the book. They should have the same values, however.) Tilted plane inside cylinder The portion of the plane \(y+2 z=2\) inside the cylinder \(x^{2}+y^{2}=1\)

Use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) across the surface in the specified direction. \(\begin{array}{l}{\text { Parabolic cylinder } \quad \mathbf{F}=x^{2} \mathbf{j}-x z \mathbf{k} \text { outward (normal away }} \\ {\text { from the } y z-\text { plane through the surface cut from the parabolic cylinder }} \\\ {\text y=x^{2},-1 \leq x \leq 1, \text { by the planes } z=0 \text { and } z=2}\end{array} \)
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