The Pringle FinaleVerify Stokes's Theorem one more time for the vector field F(x, y, z) = (z,x, y), with a Pringle region of integration:s= { «x,y.2)|2 =x* - y* and+:Your task is to parameterize both the surface S and its boundary curve C, and then directly compute both sides ofStokes's Theorem to verify that they are equal:(V xds =F.dr

given surface S=x,y,z|z=x2-y2 and x24+y2≤1 to prove - ∬s∇×F·dS=∮CF·dr where Fx,y,z=z,x,y

Answered: Verify Stokes's Theorem one more time…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

The Pringle Finale
Verify Stokes's Theorem one more time for the vector field F(x, y, z) = (z,x, y), with a Pringle region of integration:
s= { «x,y.2)|2 =x* - y* and+:
Your task is to parameterize both the surface S and its boundary curve C, and then directly compute both sides of
Stokes's Theorem to verify that they are equal:
(V x
ds =
F.dr

Expert Solution

Step 1

given surface

$S = \{(x, y, z) | z = x^{2} - y^{2} a n d \frac{x^{2}}{4} + y^{2} \leq 1\}$

to prove -

$\iint_{s} (\nabla \times F) \cdot d S = \oint_{C} F \cdot d r$

where $F (x, y, z) = (z, x, y)$

Step 2

according to given condition

$\frac{x^{2}}{4} + y^{2} \leq 1 \Rightarrow x^{2} + 4 y^{2} \leq 4$

let $2 y = v$

$\Rightarrow x^{2} + v^{2} \leq 4$

let

$x = 2 \cos θ y = 2 \sin θ$

as given

$F (x, y, z) = (z, x, y)$

now, we will calculate

$\nabla \times F = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z & x & y \end{matrix}|$

$\Rightarrow \nabla \times F = \hat{i} (\frac{\partial}{\partial y} y - \frac{\partial}{\partial z} x) - \hat{j} (\frac{\partial}{\partial x} y - \frac{\partial}{\partial z} z) + \hat{k} (\frac{\partial}{\partial x} x - \frac{\partial}{\partial y} z) = \hat{i} + \hat{j} + \hat{k}$

Step 3

as we know

$\iint_{S} (\nabla \times F) \cdot d S = \iint_{R} (\nabla \times F) \cdot \hat{n} d A$

according to question , base of surface is in x-y plane

$\Rightarrow \hat{n} = \hat{k}$

we have

$\iint_{S} (\nabla \times F) \cdot d S = \iint_{R} (\nabla \times F) \cdot \hat{k} d A$

substitute the value of $\nabla \times F = \hat{i} + \hat{j} + \hat{k}$

$\iint_{R} (\nabla \times F) \cdot \hat{k} d A = \iint_{R} (\hat{i} + \hat{j} + \hat{k}) \cdot \hat{k} r d r d θ$

we have the region

$0 < r < 2 0 < θ < 2 π$

integral becomes

$\iint_{S} (\nabla \times F) \cdot d S = \int_{0}^{2 π} \int_{0}^{2} r d r d θ = \int_{0}^{2 π} {[\frac{r^{2}}{2}]}_{0}^{2} d θ = \int_{0}^{2 π} 2 d θ = 2 {[θ]}_{0}^{2 π} = 4 π$

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