/*! 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 13 Evaluate the line integral \(\oi... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the line integral \(\oint_{C} \mathbf{F} \cdot\) dr by evaluating the surface integral in Stokes" Theorem with an appropriate choice of S. Assume C has a counterclockwise orientation. \(\mathbf{F}=\left\langle x^{2}-z^{2}, y, 2 x z\right\rangle ; C\) is the boundary of the plane \(z=4-x-y\) in the first octant.

Short Answer

Expert verified
Using Stokes' Theorem, we found that the value of the line integral $$\oint_{C} \mathbf{F} \cdot d\mathbf{r}$$ for the given vector field $$\mathbf{F} = \left\langle x^{2}-z^{2}, y, 2 x z\right\rangle$$ around a curve in the first octant and the surface S defined by the plane $$z = 4-x-y$$ is equal to $$64$$.

Step by step solution

01

Calculate the curl of the vector field

Compute the curl of the given vector field \(\mathbf{F}=\left\langle x^{2}-z^{2}, y, 2 x z\right\rangle\): $$\nabla \times \mathbf{F} = \left\langle\frac{\partial(2xz)}{\partial y} - \frac{\partial y}{\partial z}, \frac{\partial(x^2-z^2)}{\partial z} - \frac{\partial(2xz)}{\partial x}, \frac{\partial y}{\partial x} - \frac{\partial(x^2-z^2)}{\partial y}\right\rangle = \left\langle 0, -2z-2x, 0 \right\rangle.$$
02

Parametrize the surface S

The surface S is defined as the plane \(z = 4-x-y\) in the first octant. We can parametrize this plane as $$\mathbf{r}(x, y) = \left\langle x, y, 4-x-y\right\rangle,$$ where \(x\) and \(y\) are both non-negative and their sum does not exceed 4.
03

Find the normal vector \(d\mathbf{S}\)

In order to evaluate the surface integral, we need the normal vector \(d\mathbf{S}\). We can obtain this using the cross product of the partial derivatives of \(\mathbf{r}(x, y)\) with respect to \(x\) and \(y\): $$\frac{\partial \mathbf{r}}{\partial x} = \left\langle 1, 0, -1\right\rangle \quad \text{and} \quad \frac{\partial \mathbf{r}}{\partial y} = \left\langle 0, 1, -1\right\rangle.$$ Now, compute the cross product: $$d\mathbf{S} = \frac{\partial \mathbf{r}}{\partial x} \times \frac{\partial \mathbf{r}}{\partial y} = \left\langle -1, -1, 1\right\rangle dx dy.$$
04

Evaluate the surface integral

Now use Stokes' Theorem to evaluate the line integral as a surface integral: $$\oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{S} (\nabla \times \mathbf{F}) \cdot d\mathbf{S}.$$ Plug in the results from Steps 1 and 3 and compute the integral: $$\oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{S} \left\langle 0, -2z-2x, 0 \right\rangle \cdot \left\langle-1, -1, 1\right\rangle dx dy \\ = \iint_{S} (2x+2z) dx dy.$$ To find the limits of integration, we can use the bounds of the first octant and the given plane: $$0 \le x \le 4-y, \quad 0 \le y \le 4.$$ Evaluate the surface integral: $$\oint_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{0}^{4} \int_{0}^{4-y} (2x+2(4-x-y)) dx dy = \int_{0}^{4} \int_{0}^{4-y} 8 dy dx.$$ The result of the integral is: $$\oint_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{0}^{4} (32-8y) dy = 64.$$

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Key Concepts

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

Line Integral
The line integral measures the work done by a vector field along a path. Imagine walking along a curve, while affected by forces such as wind or gravity. A line integral sums up these forces' effects along your path. It's essential in physics for calculating things like potential energy or circulation of fields. A line integral of a vector field \( \mathbf{F} \) over a curve \( C \), denoted as \( \oint_C \mathbf{F} \cdot d\mathbf{r} \), requires understanding both the direction and the magnitude of the vector field along the curve. It's expressed as: \[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\, dt \] where \( \mathbf{r}(t) \) is a parameterization of the path \( C \).Stokes' Theorem offers an efficient way to evaluate line integrals by converting them into surface integrals, which can be simpler to work with depending on the context.
Surface Integral
Surface integrals extend the concept of line integrals to surfaces. They measure flux, or the flow of a vector field across a surface. Surface integrals compute this across curvy, complex surfaces using infinitesimally small bits, added up to give a total sum.For a surface \( S \) parametrized as \( \mathbf{r}(u, v) \), the surface integral of a vector field \( \mathbf{F} \) is:\[ \iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_D \mathbf{F}(\mathbf{r}(u, v)) \cdot (\mathbf{r}_u \times \mathbf{r}_v) \, dudv \]where \( \mathbf{r}_u \times \mathbf{r}_v \) defines an oriented surface segment.In our problem, Stokes' Theorem relates the line integral \( \oint_C \mathbf{F} \cdot dr \) around the boundary \( C \), with the surface integral of the curl of \( \mathbf{F} \) across \( S \). This theorem helps simplify complex calculations by shifting them into a different dimensional space.
Curl of a Vector Field
The curl of a vector field indicates the field's rotational or swirling behavior. Imagine swirling water in a cup; the curl represents how fast, and in what direction it's turning. In mathematics, the curl of a vector field \( \mathbf{F} = \langle P, Q, R \rangle \) is given by the determinant:\[ abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \]It serves a crucial role in Stokes' Theorem, showing the connection between line integrals and surface integrals.In our exercise, the curl \( abla \times \mathbf{F} \) observed as \( \langle 0, -2z-2x, 0 \rangle \) captures the local rotational strength perpendicular to our surface.
Parametrization of a Surface
Parametrization describes a surface using functions of two parameters, typically \( u \) and \( v \). Think of it as assigning coordinates that simplify complex shapes into a more manageable domain. This method is pivotal for evaluating integrals over surfaces.In our exercise, the plane \( z = 4-x-y \) needs to be parametrized. A suitable parametrization is \( \mathbf{r}(x, y) = \langle x, y, 4-x-y \rangle \) within defined bounds. This form allows transformation and calculation of vector field interactions over the surface easily.Effective parametrization can greatly simplify understanding and solving integrals over a given surface. It involves setting constraints that define the region accurately, in this case, \( 0 \leq x \leq 4-y, \quad 0 \leq y \leq 4 \), in the first octant where constraints ensure our calculations remain within defined intersections.

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Most popular questions from this chapter

Suppose a solid object in \(\mathbb{R}^{3}\) has a temperature distribution given by \(T(x, y, z) .\) The heat flow vector field in the object is \(\mathbf{F}=-k \nabla T,\) where the conductivity \(k>0\) is a property of the material. Note that the heat flow vector points in the direction opposite to that of the gradient, which is the direction of greatest temperature decrease. The divergence of the heat flow vector is \(\nabla \cdot \mathbf{F}=-k \nabla \cdot \nabla T=-k \nabla^{2} T(\text {the Laplacian of } T) .\) Compute the heat flow vector field and its divergence for the following temperature distributions. $$T(x, y, z)=100(1+\sqrt{x^{2}+y^{2}+z^{2}})$$

Prove that for a real number \(p,\) with \(\mathbf{r}=\langle x, y, z\rangle, \nabla\left(\frac{1}{|\mathbf{r}|^{p}}\right)=\frac{-p \mathbf{r}}{|\mathbf{r}|^{p+2}}\).

Green's First Identity Prove Green's First Identity for twice differentiable scalar-valued functions \(u\) and \(v\) defined on a region \(D:\) $$\iiint_{D}\left(u \nabla^{2} v+\nabla u \cdot \nabla v\right) d V=\iint_{S} u \nabla v \cdot \mathbf{n} d S$$where \(\nabla^{2} v=\nabla \cdot \nabla v .\) You may apply Gauss' Formula in Exercise 48 to \(\mathbf{F}=\nabla v\) or apply the Divergence Theorem to \(\mathbf{F}=u \nabla v\)

Flux across concentric spheres Consider the radial fields \(\mathbf{F}=\frac{\langle x, y, z\rangle}{\left(x^{2}+y^{2}+z^{2}\right)^{p / 2}}=\frac{\mathbf{r}}{|\mathbf{r}|^{p}},\) where \(p\) is a real number. Let \(S\) consist of the spheres \(A\) and \(B\) centered at the origin with radii \(0

Conditions for Green's Theorem Consider the radial field \(\mathbf{F}=\langle f, g\rangle=\frac{\langle x, y\rangle}{\sqrt{x^{2}+y^{2}}}=\frac{\mathbf{r}}{|\mathbf{r}|}\) a. Explain why the conditions of Green's Theorem do not apply to F on a region that includes the origin. b. Let \(R\) be the unit disk centered at the origin and compute \(\iint_{R}\left(\frac{\partial f}{\partial x}+\frac{\partial g}{\partial y}\right) d A\) c. Evaluate the line integral in the flux form of Green's Theorem on the boundary of \(R\) d. Do the results of parts (b) and (c) agree? Explain.

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