Final answer:
The correct answer is option b. Work = ∫[0,1] π(1 - x²) dy.
Step-by-step explanation:
The student's question pertains to calculating the work required to pump water out of a solid formed by revolving a region around the y-axis. This solid of revolution is based on the region between y = x², x = 0, and y = 1. When considering the work done, we need to account for the displacement of the water and gravitational force. Since the water is being moved over the top of the tank, the distance each slice of water must be moved is essentially the radius of the tank minus the y-value of the slice since y-values are the same as the depth in this context. The force required to move a slice of water is equal to its weight, which is the volume of the slice times the density of water times gravity.
The correct integral to represent the work done in pumping the water is based on the volume of the slices of water, which depends on the area of the cross-section of the solid perpendicularly sliced against the y-axis and the distance it must be moved.
Therefore, to obtain the area of a cross-section slice, we have Area = πx², where x = √y. This area is multiplied by the distance it must be pumped over the tank, which is 1 - y because the tank top is at y = 1. Hence, the force exerted on each differential slice is its weight, which can be expressed as density × gravity × Area × dy.
When integrating, we're effectively summing all the infinitesimal elements of work for these slices. The expression for work is then Work = π ∫[0,1] (1 - y)y dy, which simplifies to Work = π ∫[0,1] (y - y²) dy since x = √y. Therefore, the correct integral for finding the work done by pumping the water over the tank is:
Work = π ∫[0,1] (1 - x²) dy, which matches option b.