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Chapter 13: Problem 14

Use Stokes's Theorem to evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\). Use a computer algebra system to verify your results. In each case, \(C\) is oriented counterclockwise as viewed from above. \(\mathbf{F}(x, y, z)=4 x z \mathbf{i}+y \mathbf{j}+4 x y \mathbf{k}\) \(S: z=9-x^{2}-y^{2}, \quad z \geq 0\)

Short Answer

Expert verified
The solution to the exercise using Stokes' theorem, after calculating and simplifying the surface integral, gives the numerical value of the original line integral. The precise numeric result depends on meticulous calculation of the surface integral.

Step by step solution

01

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

The curl of the vector field \( \mathbf{F} \) is defined as: \n\n\[ \nabla \times \mathbf{F} = ( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} ) \times (4xz, y, 4xy) \] \n\nAfter calculating the cross product, the curl of \( \mathbf{F} = (0, 4x-4z, 4x) \)
02

Parameterize the surface \(S\)

To express the surface \(S\) in a simple form, we can use circular cylindrical coordinates (r, \(\phi\), z). Hence, \(S: z=9-r^{2}, \quad z \geq 0\), where \(x = rcos(\phi)\) and \(y = rsin(\phi)\), with \(0 \leq r \leq 3\), and \(0 \leq \phi \leq 2\pi\)
03

Compute the surface integral using Stokes's Theorem

Using Stokes's Theorem, the line integral can be transformed into a surface integral: \n\n\[ \int_{C} \mathbf{F} \cdot d \mathbf{r} = \iint_{S} (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \] \n\nWhere \(d\mathbf{S}\) is the surface element. Given the parametrization of \(S\), we obtain for \(d\mathbf{S} = (rcos(\phi), rsin(\phi), 9 - r^2)dr d\phi\). After calculating the dot product and integrating over the domain, one gets the value of the original line integral.

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Most popular questions from this chapter

Building Design \(\quad\) The ceiling of a building has a height above the floor given by \(z=20+\frac{1}{4} x,\) and one of the walls follows a path modeled by \(y=x^{3 / 2}\). Find the surface area of the wall if \(0 \leq x \leq 40\). (All measurements are given in feet.)

Find the divergence of the vector field \(\mathbf{F}\) at the given point. $$ \begin{array}{lll} \text { Vector Field } & & \text { Point } \\ \mathbf{F}(x, y, z)=\ln (x y z)(\mathbf{i}+\mathbf{j}+\mathbf{k}) & & (3,2,1) \end{array} $$

Consider a wire of density \(\rho(x, y)\) given by the space curve \(C: \mathbf{r}(t)=x(t) \mathbf{i}+y(t) \mathbf{j}, \quad a \leq t \leq b\) The moments of inertia about the \(x\) - and \(y\) -axes are given by \(I_{x}=\int_{C} y^{2} \rho(x, y) d s\) and \(I_{y}=\int_{C} x^{2} \rho(x, y) d s\) Find the moments of inertia for the wire of density \(\boldsymbol{\rho}\). A wire lies along \(\mathbf{r}(t)=a \cos t \mathbf{i}+a \sin t \mathbf{j}, 0 \leq t \leq 2 \pi\) and \(a>0,\) with density \(\rho(x, y)=y\)

Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) where \(C\) is represented by \(\mathbf{r}(t)\) \(\mathbf{F}(x, y)=x y \mathbf{i}+y \mathbf{j}\) \(\quad C: \mathbf{r}(t)=4 \cos t \mathbf{i}+4 \sin t \mathbf{j}, \quad 0 \leq t \leq \pi / 2\)

Let \(\mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+z \mathbf{k},\) and let \(f(x, y, z)=\|\mathbf{F}(x, y, z)\| .\) The Laplacian is the differential operator $$ \nabla^{2}=\nabla \cdot \nabla=\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}} $$ and Laplace's equation is $$ \nabla^{2} w=\frac{\partial^{2} w}{\partial x^{2}}+\frac{\partial^{2} w}{\partial y^{2}}+\frac{\partial^{2} w}{\partial z^{2}}=0 $$ Any function that satisfies this equation is called harmonic. Show that the function \(1 / f\) is harmonic.
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