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

int_(C)y^(3)ds,quad C:x=t^(3),quad y=t,quad0 <= t <= 2

Evaluate the line integral, where $C$ is the given curve. $\newline$ $\int_{C} y^{3} d s, \quad C: x=t^{3}, \quad y=t, \quad 0 \leq t \leq 2$

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Find derivatives of sine and cosine functions

Full solution

Q. Evaluate the line integral, where

C

is the given curve.

\newline

\int_{C} y^{3} d s, \quad C: x=t^{3}, \quad y=t, \quad 0 \leq t \leq 2

Rephrase Problem: Step $1$ : Rephrase the problem. $\newline$ question_prompt: Evaluate the line integral of $y^3$ along the curve $C$ , where $C$ is defined by $x = t^3$ and $y = t$ from $t = 0$ to $t = 2$ .
Parametrize Curve: Step $2$ : Parametrize the curve. $x = t^3$ , $y = t$ . The curve $C$ is parametrized by these equations.
Find Arc Length: Step $3$ : Find $ds$ , the arc length differential. $\newline$ $ds = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$ $\newline$ $= \sqrt{(3t^2)^2 + (1)^2} dt$ $\newline$ $= \sqrt{9t^4 + 1} dt$ .
Substitute $y$ and $ds$ : Step $4$ : Substitute $y$ and $ds$ into the integral. $\int_{0}^{2} t^3 \sqrt{9t^4 + 1} \, dt.$
Evaluate Integral: Step $5$ : Evaluate the integral. $\newline$ This step involves actual integration, which might require numerical methods or special functions since the integral of $t^3 \sqrt{9t^4 + 1} \, dt$ is not straightforward to solve analytically.

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