Evaluate the line integral ∫C27x2yzds with respect to s along thecurve C:x=t,y=t3,z=23t3(0≤t≤10

Step 1The line integral is&nbsp;∫c&nbsp;27x2yzdsx=t,y=t3,z=23t3,(0≤t≤1)&nbsp;dx&nbsp;&nbsp;dt&nbsp;=1,&nbsp;dy&nbsp;&nbsp;dt&nbsp;=3t2,&nbsp;dz&nbsp;&nbsp;dt&nbsp;=2t2Step 2At first we findds=(&nbsp;dx&nbsp;&nbsp;dt&nbsp;)2+(&nbsp;dy&nbsp;&nbsp;dt&nbsp;)2+(&nbsp;dz&nbsp;&nbsp;dt&nbsp;)2()=1+9t4()+4t4=1+13t4()the limits for integration would be 0 and 1now,∫Cf(x,y)ds=∫0127(t)2(t3)(23t3)1+9t4+4t4()at=18∫01t81+13t4()&nbsp;dt&nbsp;=6.243258

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ezpimpin6988ok1n

Answered question

2022-04-04

Evaluate the line integral

\int_{C} 27 x^{2} y z d s

with respect to s along the
curve

C : x = t, y = t^{3}, z = \frac{2}{3} t^{3} (0 \leq t \leq 10

Answer & Explanation

Nathanial Carey

Beginner2022-04-05Added 12 answers

Step 1
The line integral is $\int_{c}^{} 27 x^{2} y z d s$
$x = t, y = t^{3}, z = \frac{2}{3} t^{3},$
$(0 \leq t \leq 1)$
$\frac{d x}{d t} = 1, \frac{d y}{d t} = 3 t^{2}, \frac{d z}{d t} = 2 t^{2}$
Step 2
At first we find
$d s = \sqrt[()]{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2} + {(\frac{d z}{d t})}^{2}}$
$= \sqrt[()]{1 + 9 t^{4}} + 4 t^{4}$
$= \sqrt[()]{1 + 13 t^{4}}$
the limits for integration would be 0 and 1
now,
$\int_{C} f (x, y) d s = \int_{0}^{1} 27 {(t)}^{2} (t^{3}) (\frac{2}{3} t^{3}) \sqrt[()]{1 + 9 t^{4} + 4 t^{4}} a t$
$= 18 \int_{0}^{1} t^{8} \sqrt[()]{1 + 13 t^{4}} d t$
$= 6.243258$

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