Example Find the line integral $${\int_{C}(x^3 + y^4)ds}$$, where $C$ is a line connecting a point $(0,0,0)$ and a point $(1,1,1)$.

Answer We parametrize the line connecting a point $(0,0,0)$ and a point $(1,1,1)$. Then

$$\left\{\begin{array}{l} x = t\\ y = t\\ z = t \end{array}\right. \ \ 0 \leq t \leq 1$$

Then the line $C$ is represented by $\boldsymbol{r}(t) = t\:\boldsymbol{i} + t\:\boldsymbol{j} + t\:\boldsymbol{k}$. The arc length $s$ of $C$ is

$$s(t) = \int_{0}^{t}\vert\frac{d \boldsymbol{r}}{dt}\vert dt = \in... ...}^{t} \vert\boldsymbol{i} + \boldsymbol{j} + \boldsymbol{k}\vert dt = \sqrt{3}t$$

Thus, $ds = \sqrt{3}dt$ and we have

$$\int_{C}(x^3 + y^4)ds = \int_{0}^{1} (t^3 + t^4) \sqrt{3}dt = \sqrt{3}(\frac{1}{4} + \frac{1}{5}) = \frac{9\sqrt{3}}{20}$$

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