   Chapter 16.2, Problem 31E

Chapter
Section
Textbook Problem

Find the exact value of ∫C x3y2 z ds, where C is the curve with parametric equations x = e−t cos 4t, y = e−t sin 4t, z = e−t, 0 ⩽ t ⩽ 2π.

To determine

To find: The exact value of Cx3y2zds .

Explanation

Given data:

The parametric equations of curve and its limits are given as follows.

C:x=etcos4t,y=etsin4t,z=et,0t2π

Formula used:

Write the expression to evaluate the line integral for a function f(x,y,z) along the curve C .

Cf(x,y,z)ds=abf(x(t),y(t),z(t))(dxdt)2+(dydt)2+(dzdt)2dt (1)

Here,

a is the lower limit of the curve C and

b is the upper limit of the curve C .

Write the required differential formula to evaluate the given integral.

ddtet=etddt[u(t)v(t)]=u(t)ddt[v(t)]+v(t)ddt[u(t)]ddtcosnt=nsinntddtsinnt=ncosnt

Calculation of expression (x3y2z) :

Substitute etcos4t for x , etsin4t for y , and et for z in the expression (x3y2z) ,

x3y2z=(etcos4t)3(etsin4t)2(et)=(e3tcos34t)(e2tsin24t)(et)=e6tcos34tsin24t

Calculation of term (dxdt)2+(dydt)2+(dzdt)2 :

Substitute etcos4t for x , etsin4t for y , and et for z in the expression (dxdt)2+(dydt)2+(dzdt)2 ,

(dxdt)2+(dydt)2+(dzdt)2=[(ddtetcos4t)2+(ddtetsin4t)2+(ddtet)2]=[(etddtcos4t+cos4tddtet)2+(etddtsin4t+sin4tddtet)2+(et)

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