1. **State the problem:** Evaluate the double integral $$\int_0^1 \int_y^{\sqrt{y}} (x^2 + y) \, dx \, dy.$$\n\n2. **Understand the integral:** The integral is over $x$ from $y$ to $\sqrt{y}$ and over $y$ from 0 to 1. The integrand is $x^2 + y$.\n\n3. **Integrate with respect to $x$ first:** Use the formula $$\int (x^2 + y) \, dx = \int x^2 \, dx + \int y \, dx = \frac{x^3}{3} + yx + C.$$\n\n4. **Evaluate the inner integral:**\n$$\int_y^{\sqrt{y}} (x^2 + y) \, dx = \left[ \frac{x^3}{3} + yx \right]_y^{\sqrt{y}} = \left( \frac{(\sqrt{y})^3}{3} + y \sqrt{y} \right) - \left( \frac{y^3}{3} + y \cdot y \right).$$\n\n5. **Simplify the terms:**\nNote that $(\sqrt{y})^3 = y^{3/2}$ and $y \sqrt{y} = y^{3/2}$. So,\n$$\frac{y^{3/2}}{3} + y^{3/2} = \frac{y^{3/2}}{3} + \frac{3y^{3/2}}{3} = \frac{4y^{3/2}}{3}.$$\nAlso, $\frac{y^3}{3} + y^2 = \frac{y^3}{3} + \frac{3y^2}{3} = \frac{y^3 + 3y^2}{3}.$\n\n6. **Rewrite the inner integral result:**\n$$\int_y^{\sqrt{y}} (x^2 + y) \, dx = \frac{4y^{3/2}}{3} - \frac{y^3 + 3y^2}{3} = \frac{4y^{3/2} - y^3 - 3y^2}{3}.$$\n\n7. **Now integrate with respect to $y$ from 0 to 1:**\n$$\int_0^1 \frac{4y^{3/2} - y^3 - 3y^2}{3} \, dy = \frac{1}{3} \int_0^1 (4y^{3/2} - y^3 - 3y^2) \, dy.$$\n\n8. **Integrate each term separately:**\n$$\int_0^1 4y^{3/2} \, dy = 4 \cdot \frac{y^{5/2}}{5/2} \Big|_0^1 = 4 \cdot \frac{2}{5} = \frac{8}{5}.$$\n$$\int_0^1 y^3 \, dy = \frac{y^4}{4} \Big|_0^1 = \frac{1}{4}.$$\n$$\int_0^1 3y^2 \, dy = 3 \cdot \frac{y^3}{3} \Big|_0^1 = 1.$$\n\n9. **Combine the results:**\n$$\frac{1}{3} \left( \frac{8}{5} - \frac{1}{4} - 1 \right) = \frac{1}{3} \left( \frac{8}{5} - \frac{1}{4} - \frac{4}{4} \right) = \frac{1}{3} \left( \frac{8}{5} - \frac{5}{4} \right).$$\n\n10. **Find common denominator and simplify:**\n$$\frac{8}{5} - \frac{5}{4} = \frac{32}{20} - \frac{25}{20} = \frac{7}{20}.$$\n\n11. **Final answer:**\n$$\frac{1}{3} \cdot \frac{7}{20} = \frac{7}{60}.$$\n\n**Therefore, the value of the double integral is** $\boxed{\frac{7}{60}}$.