Final answer:
The volume of the region formed by rotating the area under the curve y = x^2, bounded by the x-axis and the line x = 3 around the y-axis, is 40.5π cubic units, which does not match any of the provided options.
Step-by-step explanation:
The question asks us to find the volume of the region formed by rotating the area under the curve y = x^2, from the x-axis to the line x = 3, around the y-axis. To solve this, we'll use the method of disks or washers, integrating with respect to y because we are rotating around the y-axis.
First, we solve the equation y = x^2 for x, yielding x = √y. The radius of a typical disk is x, and therefore √y, and the height of each disk is an infinitesimal change in y, or dy. The volume of a thin disk is found by the formula πr^2h, where r is √y and h is dy in this case. We integrate this from y=0 to y=9 (since x goes from 0 to 3 and x^2=9 when x=3) to find the volume of the entire solid.
The integral becomes V = ∫_0^9 π(√y)^2 dy = π∫_0^9 y dy, which evaluates to 1/2 π y^2 from 0 to 9. This gives us 1/2 π(81-0) which simplifies to 40.5π. So, the volume of the region is 40.5π cubic units.
None of the options (a, b, c, d) given correspond to the correct answer, so it seems like there may be an error in the question or the options provided.