Evaluate the line integral after converting it to an ordinary integral with respect to the parameter.∫C(y−z)ds&nbsp;C is the helix&nbsp;r(t)=&lt;3cos⁡t,3sin⁡t,t&gt;, for&nbsp;0≤t≤2π

∫C(y−z)dsr(t)=&lt;3cos⁡t,3sin⁡t,t&gt;0≤t≤2πr′(t)=&lt;−3sin⁡t,3cos⁡t+1&gt;|r′(t)|=9sin2t+9cos2t+1|r′(t)|=9(cos2t+sin2t)+1=10=∫02π(3sin⁡t−t)10dt10∫02π(3sin⁡t−t)dtIntegrate=10[−3cos⁡t−t22]02π=10[−3cos⁡2π−(2π)22+3cos⁡(0)]=10[−3−2π2+3]=−2π210

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Travis Fogle

Answered question

2021-11-12

Evaluate the line integral after converting it to an ordinary integral with respect to the parameter.
$\int_{C} (y - z) d s$
C is the helix $r (t) = < 3 \cos t, 3 \sin t, t >$ , for $0 \leq t \leq 2 π$

Answer & Explanation

Steacensen69

Beginner2021-11-13Added 15 answers

\int_{C} (y - z) d s

r (t) = < 3 \cos t, 3 \sin t, t >

0 \leq t \leq 2 π

r^{'} (t) = < - 3 \sin t, 3 \cos t + 1 >

| r^{'} (t) | = \sqrt{9 \sin^{2} t + 9 \cos^{2} t + 1}

| r^{'} (t) | = \sqrt{9 (\cos^{2} t + \sin^{2} t) + 1} = \sqrt{10}

= \int_{0}^{2 π} (3 \sin t - t) \sqrt{10} d t

\sqrt{10} \int_{0}^{2 π} (3 \sin t - t) d t

Integrate

= \sqrt{10} {[- 3 \cos t - \frac{t^{2}}{2}]}_{0}^{2 π}

= \sqrt{10} [- 3 \cos 2 π - \frac{{(2 π)}^{2}}{2} + 3 \cos (0)]

= \sqrt{10} [- 3 - 2 π^{2} + 3]

= - 2 π^{2} \sqrt{10}

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