## Problem on direct computation of a line integral

Directly compute $\int_C \mathbf{F} \cdot d\mathbf{r}$ where $\mathbf{F}(x,y) = x y \ \mathbf{i} + \mathbf{j}$ and $C$ is the curve connecting $(0,0)$ to $(1,1)$ along $y=x^2$.
• ## Solution

Recall that:

#### Step 1 - Parameterize the curve

Before parameterizing the curve, we draw it. The function $y=x^2$ from $(0,0)$ to $(1,1)$ looks like:
Recall that:
Because the curve $C$ is given by the function $y=x^2$, we choose the parameter $t$ and let $x(t) = t$. Thus, $y(t) = t^2$.
Our parameterization of the curve $C$ is thus $$\mathbf{r}(t) = x(t) \ \mathbf{i} + y(t) \ \mathbf{j} = t \ \mathbf{i} + t^2 \ \mathbf{j}, \tag{1}$$ for $0 \leq t \leq 1$.

#### Step 2 - Express everything in terms of the parameter

We now express the integral exclusively in terms of $t$.
Expressing $\mathbf{F}$ in terms of $t$: $$F(x(t),y(t)) = x(t) y(t) \ \mathbf{i} + \mathbf{j} = t\cdot t^2 \ \mathbf{i}+ \mathbf{j} = t^3 \ \mathbf{i} + \mathbf{j}.$$
Expressing $d\mathbf{r}$ in terms of $t$, we differentiate (1): $$d\mathbf{r}(t) = (\mathbf{i} + 2 t \ \mathbf{j}) \ dt.$$
Expressing $\mathbf{F} \cdot d\mathbf{r}$ in terms of $t$:$$\mathbf{F} \cdot d\mathbf{r} = (t^3 + 2 t) dt.$$
Noting that the bounds of the parameter $t$ are $0$ and $1$, we can rewrite the integral as:$$\int_C \mathbf{F}\cdot d\mathbf{r} = \int_0^1( t^3 + 2t) dt.$$

#### Step 3 - Evaluate the 1d integral

Evaluating the integral in $t$, we see that \begin{align} \int_C \mathbf{F}\cdot \mathbf{r}& = \int_0^1 (t^3 + 2t )dt \\
&= \frac{1}{4} t^4 + t^2 \Bigg |_0^1\\
&= \frac{5}{4}.
\end{align}
We conclude that $$\int_C \mathbf{F} \cdot d\mathbf{r}= \frac{5}{4}.$$