1. **Stating the problem:** Calculate the double integral $$I = \iint_D x^2 \, dx \, dy$$ where the domain $$D = \{(x,y) \in \mathbb{R}^2 \mid x \leq 1, y \geq 0, y^2 \leq x \}$$. 2. **Understanding the domain:** The domain is defined by: - $$x \leq 1$$ - $$y \geq 0$$ - $$y^2 \leq x$$ which implies $$x \geq y^2$$. So for each fixed $$y \geq 0$$, $$x$$ ranges from $$x = y^2$$ to $$x = 1$$. 3. **Setting up the integral:** Since $$y \geq 0$$ and $$x$$ ranges between $$y^2$$ and $$1$$, we need to find the range of $$y$$. From $$x \leq 1$$ and $$x \geq y^2$$, the maximum $$y$$ satisfies $$y^2 \leq 1$$, so $$y \in [0,1]$$. 4. **Writing the integral with limits:** $$I = \int_0^1 \int_{y^2}^1 x^2 \, dx \, dy$$. 5. **Integrate with respect to $$x$$:** $$\int_{y^2}^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_{x=y^2}^{x=1} = \frac{1^3}{3} - \frac{(y^2)^3}{3} = \frac{1}{3} - \frac{y^6}{3} = \frac{1 - y^6}{3}$$. 6. **Substitute back and integrate with respect to $$y$$:** $$I = \int_0^1 \frac{1 - y^6}{3} \, dy = \frac{1}{3} \int_0^1 (1 - y^6) \, dy = \frac{1}{3} \left[ y - \frac{y^7}{7} \right]_0^1 = \frac{1}{3} \left(1 - \frac{1}{7} \right) = \frac{1}{3} \times \frac{6}{7} = \frac{2}{7}$$. 7. **Final answer:** $$\boxed{\frac{2}{7}}$$