   Chapter 16, Problem 3P

Chapter
Section
Textbook Problem

Let C be a simple closed piecewise-smooth space curve that lies in a plane with unit normal vector n = ⟨a, b, c⟩ and has positive orientation with respect to n. Show that the plane area enclosed by C is 1 2 ∫C (bz - cy) dx + (cx − az) dy + (ay − bx) dz

To determine

To show: The plane area enclosed by C is 12C(bzcy)dx+(cxaz)dy+(aybx)dz .

Explanation

Given data:

Simple closed piecewise-smooth space curve that lies in a plane with unit normal vector n=a,b,c and has positive orientation with respect to n.

Formula used:

Write the expression for Stokes’ Theorem.

ScurlFdS=CFdr (1)

Write the required differential and integration formulae to evaluate the given integral.

ddxxn=nxn1[f(x)]ndx=[f(x)]n+1n+1

Write the expression for curlF .

curlF=|ijkxyzPQR| (2)

Consider the expression as follows.

F(x,y,z)=12(bzcy)i+12(cxaz)j+12(aybx)k (3)

Write the expression for F(x,y,z) .

F(x,y,z)=Pi+Qj+Rk (4)

Compare equation (3) and (4).

P=12(bzcy)Q=12(cxaz)R=12(aybx)

Substitute 12(bzcy) for P , 12(cxaz) for Q and 12(aybx) for R in equation (2),

curlF=|ijkxyz12(bzcy)12(cxaz)12(aybx)|={i[[12(aybx)]y[12(cxaz)]z]j[[12(aybx)]x[12(bzcy)]z

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