Final answer:
The volume of the solid formed by rotating the area between y=x^3 and y=\sqrt{x} about the x-axis is calculated using the disk method, utilizing calculus to integrate the difference in squared radius functions across the interval from x=0 to x=1.
Step-by-step explanation:
The question involves finding the volume of a solid of revolution, where the solid is obtained by rotating the region bounded by y=x^3 and y=\sqrt{x} about the x-axis. To visualize the region, we first sketch the curves, noting that they intersect where x^3 = \sqrt{x}, which implies x^4 = x and thus x=1 is the only non-trivial solution since we only consider positive x values (the curves also intersect at x=0).
To find the volume of the solid, we use the disk method, where a typical disk has a volume of \pi*(outer radius)^2 - \pi*(inner radius)^2 per infinitesimally thin slice. We integrate this volume from x = 0 to x = 1. The outer radius is given by y=\sqrt{x} and the inner radius by y=x^3 for this region. Thus, the volume V is given by:
\[V = \pi \int_{0}^{1} (\sqrt{x})^2 - (x^3)^2 dx\]
Solving this integral yields the volume of the solid of revolution. This approach utilizes the principles of calculus and geometric interpretation of integrals to determine the volume.
When it comes to evaluating the integral, note that this will involve subtracting the inner volume from the outer volume, which comprises the areas between the curves and the axis of rotation, before taking the integral across the bounds of intersection.