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Chapter 9: Problem 29

Use Stokes' Theorem to evaluate the integral $$ \int_{C}-y^{3} d x+x^{3} d y-z^{3} d z $$ for \(C\) the (positively oriented) curve of intersection between the cylinder \(x^{2}+y^{2}=1\) and the plane \(x+y+z=1\)

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Step by step solution

01

Find the Vector Field

The vector field \(\mathbf{F}\) can be written as: \[\mathbf{F} = -y^{3} \mathbf{i} + x^{3} \mathbf{j} - z^{3} \mathbf{k}\] By weighting the differential elements \(dx\), \(dy\), and \(dz\) by the physical quantities \(-y^3\), \(x^3\), and \(-z^3\) respectively.
02

Curl of the Vector Field

The curl of the field \(\mathbf{F}\) is calculated using \(\nabla \times \mathbf{F}\). The curl results to: \[\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ -y^{3} & x^{3} & -z^{3} \end{vmatrix} = (3z^2\mathbf{i} - 0\mathbf{j} - 3x^{2}\mathbf{k}) = 3z^{2}\mathbf{i} - 3x^{2}\mathbf{k}\]
03

Application of Stokes' Theorem

Integrate the flux of the curl over a disk \(D\) that lies in the plane \(z = 1 - x - y\), and is enclosed by the given curve \(C\): \[\int_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_D (\nabla \times \mathbf{F}) \cdot d\mathbf{S}\] where the surface \(S\) is positively oriented towards the upward direction \(n = \mathbf{k}\), the disk \(D\) has polar coordinates \(x = r\cos(\theta)\), \(y = r\sin(\theta)\), and \(z = 1 - r\). So our double integral simplifies into: \[\iint_D (\nabla \times \mathbf{F}) \cdot d\mathbf{S} =\iint_D [3r^2(1-r)\mathbf{i} - 3r^2\cos^2(\theta)\mathbf{k} ] \cdot \mathbf{k} \, rdrd\theta = -3 \int^{2\pi}_{0} \int^1_0 r^4 \cos^2(\theta) \, dr d\theta\]
04

Solve the Integral

Since the integrand does not depend on \(r\), we get \[-3 \int^{2\pi}_{0} \cos^2(\theta) d\theta \int^1_0 r^4 \, dr = -3 \cdot \frac{1}{2} \cdot \frac{1}{5} = -3/10 \] Note here that the integral of \(r^4\) dr from 0 to 1 is \(1/5\), and the integral of \(\cos^2(\theta)\) d\theta from 0 to \(2\pi\) is \(\pi\), hence you get the factor of \(\pi/2\).

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Most popular questions from this chapter

For cylindrical coordinates, $$ \begin{aligned} x &=r \cos \theta \\ y &=r \sin \theta \\ z &=z \end{aligned} $$ find the scale factors and derive the following expressions: $$ \begin{gathered} \nabla f=\frac{\partial f}{\partial r} \hat{\mathbf{e}}_{r}+\frac{1}{r} \frac{\partial f}{\partial \theta} \hat{\mathbf{e}}_{\theta}+\frac{\partial f}{\partial z} \hat{\mathbf{e}}_{z} \\ \nabla \cdot \mathbf{F}=\frac{1}{r} \frac{\partial\left(r F_{r}\right)}{\partial r}+\frac{1}{r} \frac{\partial F_{\theta}}{\partial \theta}+\frac{\partial F_{z}}{\partial z} \\ \nabla \times \mathbf{F}=\left(\frac{1}{r} \frac{\partial F_{z}}{\partial \theta}-\frac{\partial F_{\theta}}{\partial z}\right) \hat{\mathbf{e}}_{r}+\left(\frac{\partial F_{r}}{\partial z}-\frac{\partial F_{z}}{\partial r}\right) \hat{\mathbf{e}}_{\theta}+\frac{1}{r}\left(\frac{\partial\left(r F_{\theta}\right)}{\partial r}-\frac{\partial F_{r}}{\partial \theta}\right) \\\ \nabla^{2} f=\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial f}{\partial r}\right)+\frac{1}{r^{2}} \frac{\partial^{2} f}{\partial \theta^{2}}+\frac{\partial^{2} f}{\partial z^{2}}. \end{gathered} $$

Let \(C\) be a closed curve and \(D\) the enclosed region. Prove the identity $$ \int_{C} \phi \nabla \phi \cdot \mathbf{n} d s=\int_{D}\left(\phi \nabla^{2} \phi+\nabla \phi \cdot \nabla \phi\right) d A. $$

Prove the following vector identities: a. \((\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=(\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d})-(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \cdot \mathbf{c})\) b. \((\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{d})=(\mathbf{a} \cdot \mathbf{b} \times \mathbf{d}) \mathbf{c}-(\mathbf{a} \cdot \mathbf{b} \times \mathbf{c}) \mathbf{d}\).

Consider the integral \(\int_{C} y^{2} d x-2 x^{2} d y\). Evaluate this integral for the following curves: a. \(C\) is a straight line from \((0,2)\) to \((1,1)\). b. \(C\) is the parabolic curve \(y=x^{2}\) from \((0,0)\) to \((2,4)\). c. \(C\) is the circular path from \((1,0)\) to \((0,1)\) in a clockwise direction.

For the given vector field, find the divergence and curl of the field. a. \(\mathbf{F}=x \mathbf{i}+y \mathbf{j}\) b. \(\mathbf{F}=\frac{y}{r} \mathbf{i}-\frac{x}{r} \mathbf{j}\), for \(r=\sqrt{x^{2}+y^{2}}\). c. \(\mathbf{F}=x^{2} y \mathbf{i}+z \mathbf{j}+x y z \mathbf{k} .\)
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