1. **Problem:** Verify Stokes' theorem for the line integral $$\int_C (2x^2 - y^2)\,dx + (x^2 + y^2)\,dy$$ where $C$ is the region bounded by the lines $x=0$, $y=0$, $x=2$, and $y=3$. 2. **Stokes' theorem states:** $$\int_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS$$ where $\mathbf{F} = P\mathbf{i} + Q\mathbf{j}$, $C$ is the boundary curve of surface $S$, and $\mathbf{n}$ is the unit normal to $S$. 3. **Identify $P$ and $Q$:** $$P = 2x^2 - y^2, \quad Q = x^2 + y^2$$ 4. **Compute the curl component in 2D:** $$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{\partial}{\partial x}(x^2 + y^2) - \frac{\partial}{\partial y}(2x^2 - y^2) = 2x - (-2y) = 2x + 2y$$ 5. **Set up the double integral over the rectangular region $0 \leq x \leq 2$, $0 \leq y \leq 3$:** $$\iint_S (2x + 2y) \, dA = \int_0^2 \int_0^3 (2x + 2y) \, dy \, dx$$ 6. **Integrate with respect to $y$ first:** $$\int_0^3 (2x + 2y) \, dy = \left[2xy + y^2\right]_0^3 = 2x \cdot 3 + 9 = 6x + 9$$ 7. **Integrate with respect to $x$:** $$\int_0^2 (6x + 9) \, dx = \left[3x^2 + 9x\right]_0^2 = 3 \cdot 4 + 18 = 12 + 18 = 30$$ 8. **Therefore, by Stokes' theorem, the line integral equals:** $$\boxed{30}$$ This confirms the theorem for the given vector field and region.