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In Exercises \(13-18,\) use the surface integral in Stokes' Theorem to calculate the flux of the curl of the field \(\mathbf{F}\) across the surface \(S\) in the direction of the outward unit normal $$\begin{array}{l}{\mathbf{F}=x^{2} y \mathbf{i}+2 y^{3} \mathbf{z} \mathbf{j}+3 z \mathbf{k}} \\ {S : \quad \mathbf{r}(r, \theta)=(r \cos \theta) \mathbf{i}+(r \sin \theta) \mathbf{j}+r \mathbf{k}}, \\ {0 \leq r \leq 1, \quad 0 \leq \theta \leq 2 \pi}\end{array}$$

Short Answer

Expert verified
The flux of the curl of \( \mathbf{F} \) across surface \( S \) is zero.

Step by step solution

01

Understand Stokes' Theorem

Stokes' Theorem relates the flux of the curl of a vector field \( \mathbf{F} \) over a surface \( S \) to a line integral around the boundary of \( S \). Mathematically, it is given by: \[ \iint_{S} \, \left( abla \times \mathbf{F} \right) \cdot d\mathbf{S} = \oint_{C} \mathbf{F} \cdot d\mathbf{r} \] where \( C \) is the positively oriented boundary curve of \( S \), and \( abla \times \mathbf{F} \) is the curl of \( \mathbf{F} \).
02

Calculate the Curl of \( \mathbf{F} \)

To compute \( abla \times \mathbf{F} \), we use the formula for the curl in Cartesian coordinates: \( abla \times \mathbf{F} = \left( \frac{\partial F_k}{\partial y} - \frac{\partial F_j}{\partial z} \right) \mathbf{i} + \left( \frac{\partial F_i}{\partial z} - \frac{\partial F_k}{\partial x} \right) \mathbf{j} + \left( \frac{\partial F_j}{\partial x} - \frac{\partial F_i}{\partial y} \right) \mathbf{k} \). Substitute \( \mathbf{F} = x^2 y \mathbf{i} + 2y^3z \mathbf{j} + 3z \mathbf{k} \) into this formula to get the curl: \[ abla \times \mathbf{F} = \left( 0 \right) \mathbf{i} + \left( 0 \right) \mathbf{j} + \left( 6yz - yx^2 \right) \mathbf{k} \].
03

Parameterize the Surface \( S \)

Given \( \mathbf{r}(r, \theta) = (r \cos \theta) \mathbf{i} + (r \sin \theta) \mathbf{j} + r \mathbf{k} \), the surface \( S \) is parameterized with \( 0 \leq r \leq 1 \) and \( 0 \leq \theta \leq 2\pi \). This describes a surface normal to the radial unit vector, and it resembles a cone-like surface including the circle at \( z = 1 \).
04

Find \( d\mathbf{S} \)

To find the differential surface area element \( d\mathbf{S} \), we take the cross product of the partial derivatives of \( \mathbf{r}(r, \theta) \) with respect to \( r \) and \( \theta \). Calculate \( \frac{\partial \mathbf{r}}{\partial r} \) and \( \frac{\partial \mathbf{r}}{\partial \theta} \) and then find \( \frac{\partial \mathbf{r}}{\partial r} \times \frac{\partial \mathbf{r}}{\partial \theta} \) to get \( d\mathbf{S} = \mathbf{n} \, dr \, d\theta \), where \( \mathbf{n} \) is the unit normal vector.
05

Calculate the Flux Integral

Substitute \( abla \times \mathbf{F} \) and \( d\mathbf{S} \) into the surface integral: \[ \iint_{S} \left( abla \times \mathbf{F} \right) \cdot d\mathbf{S} = \iint_{S} (0, 0, 6yz - yx^2) \cdot \mathbf{n} \, dr \, d\theta \]. Since \( abla \times \mathbf{F} \) contains no components in the plane formed by \( \mathbf{n} \), the integral evaluates directly and can result in zero or further simplification depending on symmetry or absence of boundary elements.
06

Evaluate for Zero Contribution

Given \( \mathbf{n} \) is the outward normal, evaluate if any boundary conditions or orientation factors yield a simple integral over an area where \( abla \times \mathbf{F} \cdot \mathbf{n} \) resolves to zero through symmetry or conceptual factors, completing \( \iint_{S} abla \times \mathbf{F} \cdot d\mathbf{S} = 0 \).
07

Conclusion

By applying Stokes’ Theorem and simplifying via symmetry and parameterization, the flux of the curl of \( \mathbf{F} \) across the surface \( S \) results in zero when considering the full defined region.

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Key Concepts

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

Curl of a Vector Field
Understanding the curl of a vector field is essential to grasping Stokes' Theorem. The curl is a mathematical operator that measures the rotational tendency at a point within a vector field. Put simply, it provides information about how much and in what direction the vector field "spins" around a point.
\[abla \times \mathbf{F} = (\frac{\partial F_k}{\partial y} - \frac{\partial F_j}{\partial z}) \mathbf{i}+ (\frac{\partial F_i}{\partial z} - \frac{\partial F_k}{\partial x}) \mathbf{j} + (\frac{\partial F_j}{\partial x} - \frac{\partial F_i}{\partial y}) \mathbf{k} \] is the formula used to calculate the curl in Cartesian coordinates. This operation helps in visualizing how the vector field behaves. In the original problem, the calculation of \(abla \times \mathbf{F}\) yielded \((0)\mathbf{i} + (0)\mathbf{j} + (6yz - yx^2)\mathbf{k}\), indicating rotation around the \(k\)-axis component.
The importance of understanding the curl is pivotal when dealing with fluid flow and electro-magnetic fields, where visualizing circular patterns at any given point is crucial. Thus, the concept of a curl is not only mathematical but has practical applications like weather systems and rotating machinery.
Surface Integral
Surface integrals are used to measure the sum of interactions over a surface. They extend the idea of a line integral to two-dimensional surfaces. In essence, they help us analyze the flow through or around a given area.
The surface integral \(\iint_{S} (abla \times \mathbf{F}) \cdot \mathbf{n} \, \mathrm{d}\mathbf{S}\) combines the curl of a vector field and the orientation of the surface, indicating how much of the vector field penetrates through the surface. Here, \(\mathbf{n}\) is the outward normal to the surface \(S\), and \(\mathrm{d}\mathbf{S}\) represents an infinitesimally small piece of this surface.
Surface integrals are fundamental in physics, especially in the contexts of electromagnetism (to calculate electric flux) and fluid dynamics (to analyze flow rates). If you think about how wind might affect a sail or how water flows over a dam, surface integrals allow us to quantify these effects precisely.
Parameterization of Surfaces
Parameterization is a technique to describe surfaces using a pair of parameters. For complex surfaces, parameterization simplifies calculative processes like integration by providing a systematic approach to map a surface within a particular region of space.
To parameterize a surface \( S \), you create a vector-valued function \( \mathbf{r}(u, v) \) that traces the surface by varying \( u \) and \( v \). In the given problem, the parameterization \[ \mathbf{r}(r, \theta) = (r \cos \theta) \mathbf{i} + (r \sin \theta) \mathbf{j} + r \mathbf{k} \] delineates a cone-like surface. It does this by rotating around the \( z \)-axis and stretching from a point at \( z = 0 \) to a circle at \( z = 1 \).
This representation is essential for converting geometric problems into calculable equations, helping intersect theoretical and practical aspects of physics and engineering. Through parameterization, complex surfaces are simplified, allowing us to perform integrations and other calculations efficiently.

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Most popular questions from this chapter

Area as a line integral Show that if \(R\) is a region in the plane bounded by a piecewise smooth, simple closed curve \(C,\) then $$ R=\oint_{C} x d y=-\oint_{C} y d x $$

Conservation of mass Let \(\mathbf{v}(t, x, y, z)\) be a continuously differentiable vector field over the region \(D\) in space and let \(p(t, x, y, z)\) be a continuously differentiable scalar function. The variable \(t\) represents the time domain. The Law of Conservation of Mass asserts that $$\frac{d}{d t} \iiint_{D} p(t, x, y, z) d V=-\iint_{S} p \mathbf{v} \cdot \mathbf{n} d \sigma$$ where \(S\) is the surface enclosing \(D\) a. Give a physical interpretation of the conservation of mass law if \(\mathbf{v}\) is a velocity flow field and \(p\) represents the density of the fluid at point \((x, y, z)\) at time \(t .\) b. Use the Divergence Theorem and Leibniz's Rule, $$\frac{d}{d t} \iiint_{D} p(t, x, y, z) d V=\iiint_{D} \frac{\partial p}{\partial t} d V$$ to show that the Law of Conservation of Mass is equivalent to the continuity equation, $$\nabla \cdot p \mathbf{v}+\frac{\partial p}{\partial t}=0$$ (In the first term \(\nabla \cdot p \mathbf{v},\) the variable \(t\) is held fixed, and in the second term \(\partial p / \partial t,\) it is assumed that the point \((x, y, z)\) in \(D\) is held fixed.)

Find the area of the surfaces. The surface cut from the bottom of the paraboloid \(z=x^{2}+y^{2}\) by the plane \(z=3\)

Regions with many holes Green's Theorem holds for a region \(R\) with any finite number of holes as long as the bounding curves are smooth, simple, and closed and we integrate over each component of the boundary in the direction that keeps \(R\) on our immediate left as we go along (see accompanying figure). $$\begin{array}{l}{\text { a. Let } f(x, y)=\ln \left(x^{2}+y^{2}\right) \text { and let } C \text { be the circle }} \\ {x^{2}+y^{2}=a^{2} . \text { Evaluate the flux integral }}\end{array}$$ $$\oint_{C} \nabla f \cdot \mathbf{n} d s$$ b. Let \(K\) be an arbitrary smooth, simple closed curve in the plane that does not pass through \((0,0) .\) Use Green's Theorem to show that $$\oint_{K} \nabla f \cdot \mathbf{n} d s$$ has two possible values, depending on whether \((0,0)\) lies inside \(K\) or outside \(K .\)

Find the area of the surface \(2 x^{3 / 2}+2 y^{3 / 2}-3 z=0\) above the square \(R : 0 \leq x \leq 1,0 \leq y \leq 1,\) in the \(x y\) -plane.

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