1. **Problem statement:** Find the volume of the solid bounded between the planes perpendicular to the x-axis at $x=0$ and $x=1$, where the cross sections perpendicular to the x-axis are circular disks. The diameter of each disk runs from the parabola $y=x^2$ to the parabola $y=\sqrt{x}$. 2. **Formula and explanation:** The volume $V$ of the solid can be found by integrating the area of the cross-sectional disks along the x-axis from 0 to 1. The diameter $d(x)$ of each disk is the vertical distance between the two curves: $$d(x) = \sqrt{x} - x^2$$ The radius $r(x)$ of each disk is half the diameter: $$r(x) = \frac{d(x)}{2} = \frac{\sqrt{x} - x^2}{2}$$ The area $A(x)$ of each circular cross section is: $$A(x) = \pi r(x)^2 = \pi \left(\frac{\sqrt{x} - x^2}{2}\right)^2 = \pi \frac{(\sqrt{x} - x^2)^2}{4}$$ 3. **Set up the integral for volume:** $$V = \int_0^1 A(x) \, dx = \int_0^1 \pi \frac{(\sqrt{x} - x^2)^2}{4} \, dx = \frac{\pi}{4} \int_0^1 (\sqrt{x} - x^2)^2 \, dx$$ 4. **Expand the integrand:** $$(\sqrt{x} - x^2)^2 = (x^{1/2} - x^2)^2 = x - 2x^{5/2} + x^4$$ 5. **Rewrite the integral:** $$V = \frac{\pi}{4} \int_0^1 \left(x - 2x^{5/2} + x^4\right) dx$$ 6. **Integrate term-by-term:** $$\int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}$$ $$\int_0^1 2x^{5/2} \, dx = 2 \left[ \frac{x^{7/2}}{7/2} \right]_0^1 = 2 \cdot \frac{2}{7} = \frac{4}{7}$$ $$\int_0^1 x^4 \, dx = \left[ \frac{x^5}{5} \right]_0^1 = \frac{1}{5}$$ 7. **Combine the results:** $$V = \frac{\pi}{4} \left( \frac{1}{2} - \frac{4}{7} + \frac{1}{5} \right) = \frac{\pi}{4} \left( \frac{35}{70} - \frac{40}{70} + \frac{14}{70} \right) = \frac{\pi}{4} \cdot \frac{9}{70} = \frac{9\pi}{280}$$ **Final answer:** $$\boxed{\frac{9\pi}{280}}$$