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

EunoR

Answered question

2021-06-11

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

y^{3} d s, C : x = t^{3}, y = t, 0 \leq t \leq 4

tabuordg

Skilled2021-06-12Added 99 answers

The integral solution is shown in the photo below.

Jeffrey Jordon

Expert2021-09-29Added 2605 answers

Consider the curve $C : x = t^{3}, y = t, 0 \leq t \leq 4$

To find the integral $\int_{0} y^{3} d s$ , use the formula,

$d s = \sqrt{(\frac{d x}{d t})^{2} + (\frac{d y}{d t})^{2}} d t$

$= \sqrt{(\frac{d}{d t} (t^{3}))^{2} + (\frac{d}{d t} (t))^{2}} d t$

$= \sqrt{(3 t^{2})^{2} + (1)^{2}} d t$

$= \sqrt{9 t^{4} + 1} d t$

Therefore,

$\int_{C} y^{3} d s = \int_{t = 0}^{4} t^{3} \sqrt{9 t^{4} + 1} d t$

Let $9 t^{4} + 1 = R \Rightarrow (36 t^{3} + 0) d t = d R$

$\Rightarrow t^{3} d t = \frac{1}{36} d R$

Put $t = 0$ in $9 t^{4} + 1 = R \Rightarrow R = 1$

Put $t = 4$ in $9 t^{4} + 1 = R \Rightarrow R = 9 (4)^{2} + 1 = 2305$

Therefore,

$\int_{C} y^{3} d s = \int_{t = 0}^{4} t^{3} \sqrt{9 t^{4} + 1} d t = \int_{R = 1}^{2305} \frac{1}{36} \sqrt{R} d R$

$= \frac{1}{36} (\frac{R}{\frac{3}{2}})_{R = 1}^{2305}$ (Since $\int x^{n} d x = \frac{x^{n + 1}}{n + 1}$ )

$= \frac{1}{36} \times \frac{2}{3} ((2305)^{\frac{3}{2}} - 1)$

$= \frac{1}{18} \times \frac{1}{3} ((2305) (2305)^{\frac{1}{2}} - 1)$

$= \frac{1}{54} ((2305) \sqrt{2305} - 1)$

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