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Chapter 16: Problem 32

Find the outward flux of the field \(\mathbf{F}=x z \mathbf{i}+y z \mathbf{j}+\mathbf{k}\) across the surface of the upper cap cut from the solid sphere \(x^{2}+y^{2}+z^{2} \leq 25\) by the plane \(z=3\)

Short Answer

Expert verified
The outward flux is \( 16\pi \).

Step by step solution

01

Understand the Problem Context

We need to find the outward flux of the vector field \( \mathbf{F}=xz\mathbf{i}+yz\mathbf{j}+\mathbf{k} \) across the surface of the upper cap of a sphere. The sphere is defined by the equation \( x^2 + y^2 + z^2 \leq 25 \) and is intersected by the plane \( z=3 \). The surface we are focused on is thus the spherical cap where the sphere is cut by the plane.
02

Determine Surface Parameters

The center of the sphere is at the origin, and its radius is 5, indicated by the equation \(x^2 + y^2 + z^2 = 25\). The plane \(z = 3\) cuts this sphere into an upper cap. The equation of the cap surface is obtained by substituting \(z=3\) into the sphere's equation, resulting in \(x^2 + y^2 = 16\). This describes a circle of radius 4 in the z=3 plane.
03

Set Up the Surface Integral for Flux

The outward flux of \( \mathbf{F} \) across a surface \( S \) is given by the surface integral \( \iint_S \mathbf{F} \cdot d\mathbf{S} \). We can use the differential surface area vector \( d\mathbf{S} = \mathbf{n} \, dS \), where \( \mathbf{n} \) is the unit normal to the surface. For the cap at \(z = 3\), the outward normal is \(\mathbf{n} = \mathbf{k} \).
04

Perform the Integration

The surface integral becomes \( \int \int_S (xz\mathbf{i}+yz\mathbf{j}+\mathbf{k}) \cdot \mathbf{k} \, dS = \int \int_S 1 \, dS \) over the circular region \(x^2 + y^2 \le 16\) in the plane \(z = 3\). The integral simplifies to finding the area of the circle, which is \( \pi (4)^2 = 16\pi \). Thus, the flux across the cap is \( 16\pi \).

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Key Concepts

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

flux calculation
Flux calculation in vector calculus refers to computing how much of a vector field passes through a given surface. Imagine a vector field as a flow of water, and the surface as a net. Calculating flux helps us understand how much water is passing through the net. In mathematical terms, flux is given by the surface integral of the field over the surface.
Conceptually, flux can tell us if a vector field is diverging (moving away from a point) or converging (moving towards a point).
  • A positive flux indicates more field lines leaving through the surface than entering.
  • A negative flux means more field lines are entering than leaving.
  • A zero flux suggests a balance, with as many lines entering as leaving.
In practice, finding the flux involves integrating the vector field dot product with the differential surface area over the entire surface in question.
For our example, we calculate the flux of the vector field \( \mathbf{F}=xz\mathbf{i}+yz\mathbf{j}+\mathbf{k} \) across the surface, highlighting the outward flow through this upper cap.
surface integrals
Surface integrals help in examining how vector fields interact with surfaces in 3D space. Think of surface integrals as an extension of line integrals into more dimensions, where instead of a path, you are working over surfaces. In simpler terms, while line integrals are like sliding a ruler along a path, surface integrals
are like covering an entire surface with a mat.
Surface integrals for flux are expressed using the formula:\[ \iint_S \mathbf{F} \cdot d\mathbf{S} \]where \( \mathbf{F} \) is the vector field, and \( d\mathbf{S} \) is the differential surface area vector.
  • \( d\mathbf{S} \) contains information about the orientation and magnitude of the surface.
  • It generally includes a unit normal vector to the surface (\( \mathbf{n} \)) and an infinitesimally small piece of area (\( dS \)).
In our case, since the cap is at a constant \( z = 3 \), the unit normal vector \( \mathbf{n} \) is simply \( \mathbf{k} \), making the calculation more straightforward. We focus on the area of a circular cap defined by a plane intersecting a sphere, simplifying it to a circle in 2D for integration. Understanding surface integrals is key when working with flux calculations, as it allows a deeper comprehension of how fields pass through different surfaces.
vector fields
Vector fields are collections of vectors associated with every point in space, often used to represent various physical quantities like magnetic fields, fluid flows, or force directions.
In terms of mathematics, a vector field is a function that assigns a vector to each point in a subset of space.
  • The function \( \mathbf{F}=xz\mathbf{i}+yz\mathbf{j}+\mathbf{k} \) in the exercise represents such a vector field in 3D space.
  • The vector \( xz\mathbf{i} \) represents the component in the direction of the x-axis.
  • The vector \( yz\mathbf{j} \) is for the y-direction.
  • \( \mathbf{k} \) is simply in the z-direction, representing the simplest uniform field in one direction.
Vector fields can often reveal important insights about the nature of flow, directionality, and changes in quantities across space.
The provided vector field helps us understand how the components influence flux calculations, as they interact with given surfaces. Understanding vector fields equips you with the capabilities to model and solve real-world problems involving distributions of forces, velocities, or energies throughout a region.

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Most popular questions from this chapter

If \(S\) is the surface defined by a function \(z=f(x, y)\) that has continuous first partial derivatives throughout a region \(R_{x y}\) in the \(x y\) -plane (Figure \(16.49 ),\) then \(S\) is also the level surface \(F(x, y, z)=0\) of the function \(F(x, y, z)=f(x, y)-z\) . Taking the unit normal to \(R_{x y}\) to be \(\mathbf{p}=\mathbf{k}\) then gives $$ \begin{aligned}|\nabla F|=\left|f_{x} \mathbf{i}+f_{y} \mathbf{j}-\mathbf{k}\right| &=\sqrt{f_{x}^{2}+f_{y}^{2}+1} \\\|\nabla F \cdot \mathbf{p}| &=\left|\left(f_{x} \mathbf{i}+f_{y} \mathbf{j}-\mathbf{k}\right) \cdot \mathbf{k}\right|=|-1|=1 \end{aligned} $$ and $$ \iint_{R_{\mathrm{xy}}} \frac{|\nabla F|}{|\nabla F \cdot \mathbf{p}|} d A=\iint_{R_{\mathrm{xy}}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y $$ Similarly, the area of a smooth surface \(x=f(y, z)\) over a region \(R_{y z}\) in the \(y z\) -plane is $$ A=\iint_{R_{y x}} \sqrt{f_{y}^{2}+f_{z}^{2}+1} d y d z $$ and the area of a smooth \(y=f(x, z)\) over a region \(R_{x z}\) in the \(x z\) -plane is $$ A=\iint_{R_{\mathrm{xz}}} \sqrt{f_{x}^{2}+f_{z}^{2}+1} d x d z $$ Use Equations \((11)-(13)\) to find the area of the surfaces in Exercises \(39-44 .\) The surface cut from the bottom of the paraboloid \(z=x^{2}+y^{2}\) by the plane \(z=3\)

Find the area of the surface cut from the paraboloid \(x^{2}+y+z^{2}=\) 2 by the plane \(y=0\) .

The tangent plane at a point \(P_{0}\left(f\left(u_{0}, v_{0}\right), g\left(u_{0}, v_{0}\right), h\left(u_{0}, v_{0}\right)\right)\) on a parametrized surface \(\mathbf{r}(u, v)=f(u, v) \mathbf{i}+g(u, v) \mathbf{j}+h(u, v) \mathbf{k}\) is the plane through \(P_{0}\) normal to the vector \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right) \times \mathbf{r}_{v}\left(u_{0}, v_{0}\right),\) the cross product of the tangent vectors \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right)\) and \(\mathbf{r}_{v}\left(u_{0}, v_{0}\right)\) at \(P_{0} .\) In Exercises \(49-52,\) find an equation for the plane tangent to the surface at \(P_{0} .\) Then find a Cartesian equation for the surface and sketch the surface and tangent plane together. Parabolic cylinder The parabolic cylinder surface \(\mathbf{r}(x, y)=\) \(x \mathbf{i}+y \mathbf{j}-x^{2} \mathbf{k},-\infty

Gradient of a line integral Suppose that \(\mathbf{F}=\nabla f\) is a conservative vector field and $$ g(x, y, z)=\int_{(0,0,0)}^{(x, y, z)} \mathbf{F} \cdot d \mathbf{r} $$ Show that \(\nabla g=\mathbf{F}\)

Let \(S\) be the portion of the cylinder \(y=e^{x}\) in the first octant that projects parallel to the \(x\) -axis onto the rectangle \(R_{y z} : 1 \leq y \leq 2\) \(0 \leq z \leq 1\) in the \(y z\) -plane (see the accompanying figure). Let \(\mathbf{n}\) be the unit vector normal to \(S\) that points away from the \(y z\) -plane. Find the flux of the field \(\mathbf{F}(x, y, z)=-2 \mathbf{i}+2 y \mathbf{j}+z \mathbf{k}\) across \(S\) in the direction of \(\mathbf{n} .\)
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