1. **State the problem:** Evaluate the line integral $$\int_C y^2 \, dx + x^2 y \, dy$$ where $C$ is the rectangle with vertices $(0,0)$, $(5,0)$, $(5,4)$, and $(0,4)$ oriented counterclockwise. 2. **Recall Green's Theorem:** Green's Theorem relates a line integral around a simple closed curve $C$ to a double integral over the region $D$ enclosed by $C$: $$\int_C P \, dx + Q \, dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA$$ where $P = y^2$ and $Q = x^2 y$. 3. **Compute partial derivatives:** $$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (x^2 y) = 2xy$$ $$\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (y^2) = 2y$$ 4. **Set up the double integral:** $$\iint_D (2xy - 2y) \, dA = \iint_D 2y(x - 1) \, dA$$ 5. **Describe the region $D$:** The rectangle has $x$ from 0 to 5 and $y$ from 0 to 4. 6. **Write the integral with limits:** $$\int_0^5 \int_0^4 2y(x - 1) \, dy \, dx$$ 7. **Integrate with respect to $y$ first:** $$\int_0^5 \left[ 2(x - 1) \int_0^4 y \, dy \right] dx = \int_0^5 2(x - 1) \left[ \frac{y^2}{2} \right]_0^4 dx = \int_0^5 2(x - 1) \times 8 \, dx = \int_0^5 16(x - 1) \, dx$$ 8. **Integrate with respect to $x$:** $$\int_0^5 16(x - 1) \, dx = 16 \int_0^5 (x - 1) \, dx = 16 \left[ \frac{x^2}{2} - x \right]_0^5 = 16 \left( \frac{25}{2} - 5 \right) = 16 \times \frac{15}{2} = 120$$ **Final answer:** $$\boxed{120}$$