We are given the vector field $\mathbf{F} = (y^2, x^2)$ and the curve $\mathcal{C}$ defined by $y = x - 1$ for $1 \leq x \leq 2$, oriented from left to right. The parametrization of $\mathcal{C}$ is given by:

$\mathbf{r}(t) = (t, t - 1)$ where $t$ ranges from 1 to 2, corresponding to the interval for $x$ in the original curve.

Step 1: Compute $\mathbf{F}(\mathbf{r}(t))$

First, we need to evaluate the vector field $\mathbf{F}$ at the point $\mathbf{r}(t)$. Since $\mathbf{F}(x, y) = (y^2, x^2)$, we substitute $x = t$ and $y = t - 1$ into $\mathbf{F}$:

$\mathbf{F}(\mathbf{r}(t)) = ((t - 1)^2, t^2)$

Step 2: Compute $\mathbf{r}'(t)$

Now, we differentiate $\mathbf{r}(t) = (t, t - 1)$ with respect to $t$:

$\mathbf{r}'(t) = \left( \frac{d}{dt}(t), \frac{d}{dt}(t - 1) \right) = (1, 1)$

Step 3: Compute the dot product $\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)$

We now compute the dot product of $\mathbf{F}(\mathbf{r}(t))$ and $\mathbf{r}'(t)$. We have:

$\mathbf{F}(\mathbf{r}(t)) = ((t - 1)^2, t^2)$ $\mathbf{r}'(t) = (1, 1)$

The dot product is:

$\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = (t - 1)^2 \cdot 1 + t^2 \cdot 1 = (t - 1)^2 + t^2$

Simplifying:

$(t - 1)^2 + t^2 = (t^2 - 2t + 1) + t^2 = 2t^2 - 2t + 1$

Step 4: Set up the integral $\int_{\mathcal{C}} \mathbf{F} \cdot d\mathbf{r}$

We now set up the line integral $\int_{\mathcal{C}} \mathbf{F} \cdot d\mathbf{r}$, which can be written as:

$\int_{1}^{2} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt$

Using the result from Step 3, the integral becomes:

$\int_{1}^{2} (2t^2 - 2t + 1) \, dt$

Step 5: Evaluate the integral

Now, we evaluate the integral:

$\int_{1}^{2} (2t^2 - 2t + 1) \, dt$

First, integrate each term:

$\int 2t^2 \, dt = \frac{2t^3}{3}, \quad \int -2t \, dt = -t^2, \quad \int 1 \, dt = t$

Evaluating from $t = 1$ to $t = 2$:

$\left[ \frac{2t^3}{3} - t^2 + t \right]_{1}^{2}$

At $t = 2$:

$\frac{2(2)^3}{3} - (2)^2 + (2) = \frac{16}{3} - 4 + 2 = \frac{16}{3} - 2$

At $t = 1$:

$\frac{2(1)^3}{3} - (1)^2 + (1) = \frac{2}{3} - 1 + 1 = \frac{2}{3}$

Now subtract the values:

$\left( \frac{16}{3} - 2 \right) - \frac{2}{3} = \frac{16}{3} - \frac{6}{3} - \frac{2}{3} = \frac{8}{3}$

Thus, the value of the line integral is:

$\int_{\mathcal{C}} \mathbf{F} \cdot d\mathbf{r} = \frac{8}{3}$

Final Answer:

The value of the line integral is $\frac{8}{3}$.

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Would you like a more detailed breakdown of any specific part of the calculation?

Here are some related questions you might find interesting:

1. How would the result change if the curve $\mathcal{C}$ was oriented in the opposite direction (right to left)?
2. How do we compute line integrals of scalar fields instead of vector fields?
3. What happens to the line integral if the vector field $\mathbf{F}$ is conservative?
4. How can the parametrization of a curve affect the computation of a line integral?
5. What is the physical interpretation of this line integral in terms of work done by a force?

Tip: Always pay attention to the parametrization and bounds when setting up a line integral, as they can affect the direction and limits of integration.