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The path [tex]C[/tex] is parameterized by

[tex]\mathbf r(t)=\langle x(t),y(t)\rangle=\langle t^3,t\rangle[/tex]

with [tex]0\le t\le4[/tex].

We have

[tex]\dfrac{\mathrm d\mathbf r}{\mathrm dt}=\langle3t^2,1\rangle[/tex]
[tex]\left\|\dfrac{\mathrm d\mathbf r}{\mathrm dt}\right\|=\sqrt{9t^4+1}[/tex]

So the line integral is

[tex]\displaystyle\int_Cy^3\,\mathrm dS=\int_{t=0}^{t=4}t^3\sqrt{9t^4+1}\,\mathrm dt[/tex]

Substitute [tex]u=9t^4+1[/tex], so that [tex]\mathrm du=36t^3\,\mathrm dt[/tex].

[tex]=\displaystyle\frac1{36}\int_{u=1}^{u=2305}\sqrt u\,\mathrm du[/tex]
[tex]=\dfrac{2305^{3/2}-1}{54}[/tex]
[tex]\approx2049.31[/tex]

∫C) y³ × ds = 7962624,02 surface units

∫C) y³ × ds .               where  C    x = t³     y = t      with    0 ≤ t ≤ 4

∫C) y³ × ds   = ∫∫R) F(r(t) × dr

Then  F (r(t)) =   y³ . = t³ .          r ( t³ , t )  then .    dr/dt = (3×t² , 1 )

|| dr/dt|| = √ ( 3×t²)² + (1)²

|| dr/dt|| = √9×t⁴ + 1

∫C) y³ × ds   = ∫∫R) F(r(t) × dr . =  ∫₀⁴ t³ ×√9×t⁴ + 1    ×dt

To solve the integral we make a change of variables, we call

9×t⁴ + 1  = v .  then .    36×t³ ×dt = dv

Now as we change variables we need to change the limits of integration as follows:If   9×t⁴ + 1 = v .       then when t = 0   v = 1

And when  t = 4 .                then . v = 9×(4)⁴ + 1 . = 2305

The integral takes that form

∫₁²³⁰⁵ [ 1/36)× √v × dv . =   (1/36) ×∫₁²³⁰⁵ (v)¹/² dv  

= (1/36)×(2/3) × v³/² |₁²³⁰⁵ = (2/108)× √v³|₁²³⁰⁵ =

= (2/108) [ √( 9×t⁴ + 1 )³|₀⁴ =  (2/108) [ √( 9⁴×4¹⁶ + 1  - √(9)⁴×0 + 1)

=  (2/108) [√( (9⁴×4¹⁶ + 1 ) . + 1 ]

Neglecting  in √( (9⁴×4¹⁶ + 1 )  1 unit .  we get 9²×4⁸ = 429981696

= (2/108) [429981697]

= 7962624,02

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