1. **Problem Statement:** Evaluate the integral of $f(x,y) = x^2 + y^2$ over the triangular region with vertices $(0,0)$, $(1,0)$, and $(0,1)$. 2. **Formula and Setup:** The integral over a region $D$ is given by $$\iint_D (x^2 + y^2) \, dA.$$ For the triangular region, the limits for $x$ and $y$ are $0 \leq x \leq 1$ and $0 \leq y \leq 1 - x$. 3. **Integral Expression:** $$\int_0^1 \int_0^{1-x} (x^2 + y^2) \, dy \, dx.$$ 4. **Integrate with respect to $y$ first:** $$\int_0^{1-x} (x^2 + y^2) \, dy = x^2 y + \frac{y^3}{3} \Big|_0^{1-x} = x^2 (1-x) + \frac{(1-x)^3}{3}.$$ 5. **Simplify the inner integral:** $$x^2 (1-x) + \frac{(1-x)^3}{3} = x^2 - x^3 + \frac{1 - 3x + 3x^2 - x^3}{3} = x^2 - x^3 + \frac{1}{3} - x + x^2 - \frac{x^3}{3}.$$ 6. **Combine like terms:** $$= \frac{1}{3} - x + 2x^2 - \frac{4x^3}{3}.$$ 7. **Now integrate with respect to $x$ from 0 to 1:** $$\int_0^1 \left( \frac{1}{3} - x + 2x^2 - \frac{4x^3}{3} \right) dx = \left[ \frac{x}{3} - \frac{x^2}{2} + \frac{2x^3}{3} - \frac{x^4}{3} \right]_0^1.$$ 8. **Evaluate at the bounds:** $$\frac{1}{3} - \frac{1}{2} + \frac{2}{3} - \frac{1}{3} = \left( \frac{1}{3} + \frac{2}{3} - \frac{1}{3} \right) - \frac{1}{2} = \frac{2}{3} - \frac{1}{2} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6}.$$ 9. **Final answer:** The value of the integral is $\boxed{\frac{1}{6}}$.