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Compute the flux of \(\vec{F}\) across \(\mathcal{S}\). (If \(\mathcal{S}\) is not a closed surface, choose \(\vec{n}\) so that it has a positive z-component, unless otherwise indicated.) \(\mathcal{S}\) is the unit sphere; \(\vec{F}=\langle y-z, z-x, x-y\rangle\)

Short Answer

Expert verified
The flux of \(\vec{F}\) across the unit sphere \(\mathcal{S}\) is zero.

Step by step solution

01

Identify the Surface and Vector Field

The surface \(\mathcal{S}\) is given as the unit sphere, described by the equation \(x^2 + y^2 + z^2 = 1\). The vector field is \(\vec{F} = \langle y-z, z-x, x-y\rangle\). Our task is to compute the flux of \(\vec{F}\) across this surface.
02

Recall the Flux Integral Formula

To compute the flux of the vector field \(\vec{F}\) across a closed surface \(\mathcal{S}\), we use the surface integral \(\iint_{\mathcal{S}} \vec{F} \cdot \vec{n} \, dS\), where \(\vec{n}\) is the outward-pointing unit normal vector and \(dS\) is the differential area element.
03

Use the Divergence Theorem

Because \(\mathcal{S}\) is a closed surface, we can apply the divergence theorem, which relates the flux through a closed surface to a volume integral over the region enclosed by the surface. The divergence theorem states: \[\iint_{\mathcal{S}} \vec{F} \cdot \vec{n} \, dS = \iiint_{V} (abla \cdot \vec{F}) \, dV\]\ where \(V\) is the volume inside \(\mathcal{S}\).
04

Compute the Divergence of \(\vec{F}\)

Calculate the divergence of \(\vec{F}\). We have: \[abla \cdot \vec{F} = \frac{\partial}{\partial x}(y-z) + \frac{\partial}{\partial y}(z-x) + \frac{\partial}{\partial z}(x-y)\] Compute each term: \[\frac{\partial}{\partial x}(y-z) = 0,\ \frac{\partial}{\partial y}(z-x) = 0,\ \frac{\partial}{\partial z}(x-y) = 0\] Thus, \(abla \cdot \vec{F} = 0\).
05

Evaluate the Volume Integral

Since \(abla \cdot \vec{F} = 0\), the volume integral over the volume enclosed by \(\mathcal{S}\) is zero: \[\iiint_{V} (abla \cdot \vec{F}) \, dV = \iiint_{V} 0 \, dV = 0\]
06

Conclude the Flux Calculation

By the divergence theorem, since the volume integral is zero, the flux of \(\vec{F}\) across the surface \(\mathcal{S}\) is also zero: \[\iint_{\mathcal{S}} \vec{F} \cdot \vec{n} \, dS = 0\]

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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 in vector calculus that links the flow of a vector field through a closed surface to the behavior of the vector inside the volume. It acts like a bridge between surface integrals and triple integrals by transforming a complicated surface calculation into a potentially simpler volume calculation.

Here's how it works: If you have a closed surface, denoted by \( \mathcal{S} \), enclosing a volume \( V \), the theorem states that the total flux of a vector field \( \vec{F} \) across \( \mathcal{S} \) can be computed as a volume integral of the divergence of \( \vec{F} \) over \( V \). The formula is:

\[ \iint_{\mathcal{S}} \vec{F} \cdot \vec{n} \, dS = \iiint_{V} (abla \cdot \vec{F}) \, dV \]

In simpler terms, the theorem helps by eliminating the need to compute complex surface integrals directly if the divergence of the vector field is easier to calculate.

For instance, in our exercise, the divergence of \( \vec{F} = \langle y-z, z-x, x-y \rangle \) was found to be zero. When the divergence is zero, the triple integral becomes zero, simplifying the problem significantly and showing that the net flux through the unit sphere is zero.
Surface Integrals
Surface integrals are an extension of line integrals to higher dimensions. They allow us to calculate the total flux or flow across a surface in 3D space. When dealing with a vector field and a surface, the surface integral computes how much of the vector field is passing through the surface.

The mathematical expression for a surface integral of a vector field \( \vec{F} \) across a surface \( \mathcal{S} \) is:
\[ \iint_{\mathcal{S}} \vec{F} \cdot \vec{n} \, dS \]

Here:
  • \( \vec{n} \) is the unit normal vector, indicating the direction perpendicular to the surface.
  • \( dS \) is the differential element of the surface area.

This integral sums up the component of \( \vec{F} \) that is normal to the surface across the entire surface.

If the surface is closed, like a sphere, and using the Divergence Theorem can simplify the calculation by converting it into a volume integral. This approach is beneficial when the vector field is complex but has useful properties like a zero divergence.
Vector Fields
A vector field is a function that assigns a vector to every point in space. These vectors can represent quantities like velocity, force, or electromagnetic fields.

In our exercise, the given vector field is \( \vec{F} = \langle y-z, z-x, x-y \rangle \).
This field assigns a vector direction and magnitude based on the values of \( x, y, \) and \( z \).

Understanding vector fields requires exploring their properties, such as divergence and curl.
  • Divergence: Measures the "outflow" of a vector field from an infinitesimal volume around a point. In the exercise, we calculated that \( abla \cdot \vec{F} = 0 \), signaling no net change in volume.
  • Curl: Describes the rotation or "twirl" of the field around a point. Although not directly calculated here, it's another vital property of vector fields.

Knowing these properties helps determine how the field behaves when interacting with various surfaces. A zero divergence within a closed surface can indicate no net flow through the surface, as seen in this situation where the flux through the unit sphere was zero.

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Most popular questions from this chapter

In Exercises \(11-14,\) a planar curve \(C\) is given along with a surface \(f\) that is defined over \(C\). Set up the line integral \(\int_{C} f(s) d s,\) then approximate its value using technology. $$ \begin{aligned} &C \text { is the portion of } y=x^{3} \text { on }[-1,1] ; \text { the surface is } f(x, y)=\\\ &2 x+3 y+5 \end{aligned} $$

Set up the double integral that finds the surface area \(S\) of the given surface \(\mathcal{S},\) then use technology to approximate its value. \(\mathcal{S}\) is the hyperbolic paraboloid \(z=x^{2}-y^{2}\) over the circular disk of radius 1 centered at the origin.

A vector field \(\vec{F}\) and a curve \(C\) are given. Evaluate \(\int_{C} \vec{F} \cdot \vec{n} d s,\) the flux of \(\vec{F}\) over \(C\). \(\vec{F}=\langle y, 0\rangle ; C\) is the line segment from (0,0) to \((0,1) .\)

In Exercises \(7-12,\) a vector field \(\vec{F}\) and a curve \(C\) are given. Evaluate \(\int_{c} \vec{F} \cdot d \vec{r} .\) $$ \begin{aligned} &\vec{F}=\langle y+z, x+z, x+y\rangle ; C \text { is the hellx } \vec{r}(t)=\\\ &\langle\cos t, \sin t, t /(2 \pi)\rangle \text { on } 0 \leq t \leq 2 \pi \end{aligned} $$

In Exercises \(9-12,\) a closed curve \(C\) that is the boundary of a surface \(S\) is given along with a vector field \(\vec{F}\). Verify Stokes' Theorem on \(C ;\) that is, show \(\oint_{c} \vec{F} \cdot d \vec{r}=\iint_{S}(\operatorname{curl} \vec{F}) \cdot \vec{n} d S\). $$ \begin{aligned} &C \text { is the curve parametrized by } \vec{r}(t)=\left\langle\cos t, \sin t, e^{-1}\right\rangle\\\ &\text { and } \mathcal{S} \text { is the portion of } z=e^{-x^{2}-y^{2}} \text { enclosed by } c ; \vec{F}=\\\ &\langle-y, x, 1\rangle . \end{aligned} $$

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