1. **State the problem:** We need to find the work done by the force field $\mathbf{F} = (y^2, 2xy)$ in moving a particle along the curve $y = x^2$ from the point $(0,0)$ to $(1,1)$. 2. **Recall the formula for work done by a force field along a curve:** $$ W = \int_C \mathbf{F} \cdot d\mathbf{r} $$ where $C$ is the path of the particle. 3. **Parameterize the curve:** Since $y = x^2$, let $x = t$, then $y = t^2$ with $t$ going from 0 to 1. 4. **Express $\mathbf{F}$ and $d\mathbf{r}$ in terms of $t$:** $$ \mathbf{F}(t) = (y^2, 2xy) = ((t^2)^2, 2 \cdot t \cdot t^2) = (t^4, 2t^3) $$ $$ d\mathbf{r} = (dx, dy) = (dt, d(t^2)) = (dt, 2t dt) $$ 5. **Compute the dot product $\mathbf{F} \cdot d\mathbf{r}$:** $$ \mathbf{F} \cdot d\mathbf{r} = t^4 \cdot dt + 2t^3 \cdot 2t dt = t^4 dt + 4t^4 dt = 5t^4 dt $$ 6. **Set up and evaluate the integral:** $$ W = \int_0^1 5t^4 dt = 5 \int_0^1 t^4 dt = 5 \left[ \frac{t^5}{5} \right]_0^1 = 5 \cdot \frac{1}{5} = 1 $$ **Final answer:** The work done by the force field is $1$.