Final answer:
To find the volume of the solid generated by revolving the region bounded by the curves y=3\sqrt{x}, xy=10, and y=2 about the x-axis, we use the disk method to set up an integral using the intersection points between y=3\sqrt{x} and y=2 as limits.
Step-by-step explanation:
To sketch the region bounded by the curves y=3\sqrt{x}, xy=10, and y=2 and find the volume of the solid generated by revolving this region about the x-axis, we first need to graph these equations and identify the enclosed area. The curve y=3\sqrt{x} is a half-parabola that opens to the right, xy=10 is a hyperbola, and y=2 is a horizontal line. After sketching these, we observe the bounded region and can set up the integral for volume using the disk method.
The disk method involves creating disks with a certain radius from the function being rotated, here y=3\sqrt{x}, between the limits of integration, which are the x-values where y=3\sqrt{x} intersects with y=2. We can find this intersection point by setting the equations equal to each other (3\sqrt{x}=2) and solving for x. The volume is then calculated using the formula V=\pi\int_a^b [f(x)]^2 dx, where f(x) = 3\sqrt{x} and the limits of integration, a and b, are the x-values at the intersection points.