Answered: Verifying Stokes’ Theorem Confirm that…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter8: Applications Of Trigonometry

Section8.3: Vectors

Problem 60E

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Question

Verifying Stokes’ Theorem Confirm that Stokes’ Theorem holds for
the vector field F = ⟨z - y, x, -x⟩, where S is the hemisphere x² + y² + z² = 4, for z ≥ 0, and C is the circle x² + y² = 4 oriented counterclockwise.

Expert Solution

Step 1

To Verify: Stoke's theorem for the vector field $F = <z - y, x, - x>$ where surface is $S : x^{2} + y^{2} + z^{2} = 4$ for $z \geq 0$ and the it's projection $x^{2} + y^{2} = 4$ is oriented counter clockwise.

Step 2

Stoke's theorem:

Let F be a vector field that is defined in a neighborhood of a surface S. then,

$\int_{C} F \cdot d r = \iint_{S} (Curl F) \cdot n d S$ .

The given vector field is $F = <z - y, x, - x>$ and the surface is $S : x^{2} + y^{2} + z^{2} = 4$ for $z \geq 0$ .

First we will evaluate L.H.S of (i):

parametrize $x^{2} + y^{2} = 4$ ,

Put $x = 2 \cos t, y = 2 \sin t$ .

$r (t) = <2 \cos t, 2 \sin t, 0>$ .

then, $F (r (t)) = <0 - 2 \sin t, 2 \cos t, - 2 \cos t>$ ,

$\begin{array}{rcl} \int F \cdot d r & = & \int_{0}^{2 π} F (r (t)) r' (t) d t \\ = & \int_{0}^{2 π} <- 2 \sin t, 2 \cos t, - 2 \cos t> \cdot <- 2 \sin t, 2 \cos t, 0> d t \\ = & \int_{0}^{2 π} (4 \sin^{2} t + 4 \cos^{2} t) d t \\ = & 4 \int_{0}^{2 π} d t \\ = & 4 {[t]}_{0}^{2 π} = 4 \times 2 π \\ = & 8 π \end{array}$

So, $\int_{C} F \cdot d r = 8 π$ ...................(ii)

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