Evaluate the line integral, where C is the given curve. ∫y3 ds, C÷x=t3, y=t, 0? t? 3

Step 1f(x, y)=y3r(t)=⟨t3, t⟩r′(t)=⟨3t2, 1⟩ds=|r′(t)|dtds=(3t2)2+12dtds=(9t4)+1dtIntegral y3dsIntegral =∫[0 to 3]f(r(t))⋅ds=∫[0 to 3]t3⋅[(9t4)+1]dtLet (9t4)+1=u⇒9×4t3dt+0=du⇒t3dt=dt36⇒∫3⋅(9t4)+1dt⇒∫udu36⇒(136)u32(32)+c⇒(154)u32+c⇒(154)[(9t4)+1]32+c∫03t3⋅(9t4)+1]dt=∫03(154)[(9t4)+1]32+c⇒(154)[(9×34)+1]32+c−(154)[(9×04)+1]32−c⇒(154)[730]32−(154)⇒(154)[73032−1]⇒365.23

Check the solution to "Evaluate the line integral, where C is the..."

Rui Baldwin

Answered question

2021-06-13

Evaluate the line integral, where C is the given curve.

\int y 3 d s, C \div x = t 3, y = t, 0 ? t ? 3

Answer & Explanation

opsadnojD

Skilled2021-06-14Added 95 answers

Step 1
$f (x, y) = y^{3}$
$r (t) = ⟨ t^{3}, t ⟩$
$r^{'} (t) = ⟨ 3 t^{2}, 1 ⟩$
$d s = | r^{'} (t) | d t$
$d s = \sqrt{(3 t^{2})^{2} + 1^{2}} d t$
$d s = \sqrt{(9 t^{4}) + 1} d t$
Integral $y^{3} d s$
Integral $= \int_{[0 t o 3]} f (r (t)) \cdot d s$
$= \int_{[0 t o 3]} t^{3} \cdot \sqrt{[(9 t^{4}) + 1]} d t$
Let $(9 t^{4}) + 1 = u$
$\Rightarrow 9 \times 4 t^{3} d t + 0 = d u \Rightarrow t^{3} d t = \frac{d t}{36}$
$\Rightarrow \int^{3} \cdot \sqrt{(9 t^{4}) + 1} d t$
$\Rightarrow \int \sqrt{u} \frac{d u}{36}$
$\Rightarrow (\frac{1}{36}) \frac{u^{\frac{3}{2}}}{(\frac{3}{2})} + c$
$\Rightarrow (\frac{1}{54}) u^{\frac{3}{2}} + c$
$\Rightarrow (\frac{1}{54}) [(9 t^{4}) + 1]^{\frac{3}{2}} + c$
$\int_{0}^{3} t^{3} \cdot \sqrt{(9 t^{4}) + 1]} d t = \int_{0}^{3} (\frac{1}{54}) [(9 t^{4}) + 1]^{\frac{3}{2}} + c$
$\Rightarrow (\frac{1}{54}) [(9 \times 3^{4}) + 1]^{\frac{3}{2}} + c - (\frac{1}{54}) [(9 \times 0^{4}) + 1]^{\frac{3}{2}} - c$
$\Rightarrow (\frac{1}{54}) [730]^{\frac{3}{2}} - (\frac{1}{54})$
$\Rightarrow (\frac{1}{54}) [730^{\frac{3}{2}} - 1]$
$\Rightarrow 365.23$

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