/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 Find the flux of \(\mathbf{F}\) ... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 27

Find the flux of \(\mathbf{F}\) across the surface \(\sigma\) by expressing \(\sigma\) parametrically.\(\mathbf{F}(x, y, z)=\mathbf{i}+\mathbf{j}+\mathbf{k} ;\) the surface \(\sigma\) is the portion of the cone \(z=\sqrt{x^{2}+y^{2}}\) between the planes \(z=1\) and \(z=2\), oriented by downward unit normals.

Short Answer

Expert verified
The flux is \(\frac{14\pi}{3}\).

Step by step solution

01

Parametrize the Surface

To parametrize the surface \(\sigma\), use cylindrical coordinates. Here, \(x = r\cos\theta\), \(y = r\sin\theta\), and from the equation of the cone, \(z = r\). Therefore, the parametric equations are \(\mathbf{r}(r, \theta) = (r\cos\theta, r\sin\theta, r)\), where \(1 \leq r \leq 2\) and \(0 \leq \theta < 2\pi\).
02

Compute the Normal Vector

Calculate the partial derivatives \(\mathbf{r}_r\) and \(\mathbf{r}_{\theta}\), where \(\mathbf{r}_r = (\cos\theta, \sin\theta, 1)\) and \(\mathbf{r}_{\theta} = (-r\sin\theta, r\cos\theta, 0)\). The cross product \(\mathbf{n} = \mathbf{r}_r \times \mathbf{r}_{\theta}\) gives the normal vector. Compute it to find \(\mathbf{n} = (-r\cos\theta, -r\sin\theta, r)\).
03

Ensure Normal Vector Correct Orientation

The oriented surface has downward normals, thus check the orientation of \(\mathbf{n}\). Since \(\mathbf{n}\cdot (0,0,-1) > 0\), the current orientation is downward, which is correct for this problem.
04

Evaluate Dot Product \(\mathbf{F} \cdot \mathbf{n}\)

The vector field \(\mathbf{F} = (1, 1, 1)\). Compute the dot product with the normal vector: \[\mathbf{F} \cdot \mathbf{n} = (1, 1, 1) \cdot (-r\cos\theta, -r\sin\theta, r) = -r\cos\theta - r\sin\theta + r\].
05

Set Up the Double Integral for Flux

The flux \(\Phi\) through surface \(\sigma\) is given by the surface integral: \[\Phi = \int_0^{2\pi}\int_1^2 (-r\cos\theta - r\sin\theta + r)r\, dr\, d\theta\]. Simplifying the integrand, we have \(-r^2\cos\theta - r^2\sin\theta + r^2\).
06

Integrate with Respect to \(r\) and \(\theta\)

First, integrate with respect to \(r\) from 1 to 2: \(\int_1^2 (-r^2\cos\theta - r^2\sin\theta + r^2) \,dr\). Then integrate with respect to \(\theta\) from 0 to \(2\pi\). Symmetry of \(\cos\theta\) and \(\sin\theta\) over \([0, 2\pi]\) gives zero for those terms. The remaining integral simplifies to \(\Phi = 2\pi \int_1^2 r^2 \,dr = 2\pi\left[\frac{r^3}{3}\right]_1^2\). Evaluate to get \(\Phi = 2\pi\left(\frac{8}{3} - \frac{1}{3}\right) = \frac{14\pi}{3}\).

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Key Concepts

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

Parametrization
To find the flux of a vector field across a surface, the first step is to parametrize the surface. Parametrization translates a complex surface into simpler variables, often using coordinates like cylindrical or spherical. In this exercise, we parametrize a conical surface using cylindrical coordinates. The cone is described by the equation \(z = \sqrt{x^2 + y^2}\). Using cylindrical coordinates, we set \(x = r\cos\theta\), \(y = r\sin\theta\), and consequently, \(z = r\). This adapts the problem into a new set of parameters \((r, \theta)\), where \(r\) ranges from 1 to 2, and \(\theta\) ranges from \(0\) to \(2\pi\).This transformation not only simplifies calculations but also clearly defines the boundaries for calculating the integral, making it easier to compute the surface integral that follows.
Surface Integral
The surface integral is central in flux calculations. A surface integral computes the integral over a curved surface, translating the vector field and the surface into a single comprehensible quantity. To evaluate it, we first determine the normal vector to the surface. Differentiating the parametric equations \( \mathbf{r}(r, \theta) = (r\cos\theta, r\sin\theta, r)\), we find the partial derivatives \( \mathbf{r}_r = (\cos\theta, \sin\theta, 1)\) and \( \mathbf{r}_{\theta} = (-r\sin\theta, r\cos\theta, 0)\). The cross product of these vectors gives the normal vector \( \mathbf{n} = (-r\cos\theta, -r\sin\theta, r)\).The flux through the surface is then found by evaluating the surface integral of the dot product of the vector field and this normal vector over the entire surface. This results in an integral involving \(r\) and \(\theta\), which represents the combined effect of \(\mathbf{F}\) across the surface \(\sigma\). In our exercise, the surface integral is effectively calculated to yield a total flux of \( \frac{14\pi}{3} \).
Vector Field
Vector fields are functions that assign a vector to each point in space. In this exercise, the vector field \(\mathbf{F}\) is constant and given by \(\mathbf{F}(x, y, z) = (1, 1, 1)\). Here, each vector in the field represents a force moving in the positive direction of all three coordinate axes, with equal magnitude. Understanding the vector field's behavior is crucial in flux calculations, as it describes how a fluid or force "flows" through different points of the surface.The key interaction in flux calculation involves the dot product \(\mathbf{F}\cdot\mathbf{n}\), where \(\mathbf{n}\) is the normal vector to the surface. The result of this dot product shows how much of \(\mathbf{F}\) penetrates through the surface at each point. In this step-by-step process, the dot product simplifies to \(-r\cos\theta - r\sin\theta + r\), showing the relationship between the vector field and the surface's orientation, which is essential for determining the net flux.

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Most popular questions from this chapter

In each part, evaluate the integral $$ \int_{C} y d x+z d y-x d z $$ along the stated curve. (a) The line segment from \((0,0,0)\) to \((1,1,1)\).(b) The twisted cubic \(x=t, y=t^{2}, z=t^{3}\) from \((0,0,0)\) to \((1,1,1)\). (c) The helix \(x=\cos \pi t, y=\sin \pi t, z=t\) from \((1,0,0)\) to \((-1,0,1)\).

Let \(\mathbf{F}(x, y, z)=a \mathbf{i}+b \mathbf{j}+c \mathbf{k}\) be a constant vector field and let \(\sigma\) be the surface of a solid \(G\). Use the Divergence Theorem to show that the flux of \(\mathbf{F}\) across \(\sigma\) is zero. Give an informal physical explanation of this result.

(a) Let \(\sigma\) denote the surface of a solid \(G\) with \(\mathbf{n}\) the outward unit normal vector field to \(\sigma\). Assume that \(\mathbf{F}\) is a vector field with continuous first-order partial derivatives on \(\sigma\). Prove that $$ \iint_{\sigma}(\operatorname{curl} \mathbf{F}) \cdot \mathbf{n} d S=0 $$ [Hint: Let \(C\) denote a simple closed curve on \(\sigma\) that separates the surface into two subsurfaces \(\sigma_{1}\) and \(\sigma_{2}\) that share \(C\) as their common boundary. Apply Stokes' Theorem to \(\sigma_{1}\) and to \(\sigma_{2}\) and add the results.] (b) The vector field \(\operatorname{curl}(\mathbf{F})\) is called the curl field of F. In words, interpret the formula in part (a) as a statement about the flux of the curl field.

Evaluate the line integral with respect to \(s\) along the curve \(C\).$$ \begin{aligned} &\int_{C} 3 x^{2} y z d s \\ &C: x=t, y=t^{2}, z=\frac{2}{3} t^{3} \quad(0 \leq t \leq 1) \end{aligned} $$

Let \(k\) be a constant, \(\mathbf{F}=\mathbf{F}(x, y, z), \mathbf{G}=\mathbf{G}(x, y, z)\), and \(\phi=\phi(x, y, z)\). Prove the following identities, assuming that all derivatives involved exist and are continuous.\(\operatorname{div}(\operatorname{curl} \mathbf{F})=0\)
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