Find the volume generated by revolving the region enclosed above by the x-axis and the graph of the function

f(x) = x^2 -1 about the y-axis.

I cant make sense of it. What is it asking me to do? What are the integral bounds?

Makes no sense to me neither.

f(x) = x^2 - 1 is a parabola opening upwards, so when we revolve it about the y-axis we have it open-ended above.
We need a closed region to revolve.
Did you mean the region below the x-axis ?

The region is bounded above by the x-axis. That is, it is the region below the axis.

Since it is symmetric about the y-axis, we really only need to rotate the region from 0 to 1.

So, using discs,

v = ∫[0,1] πr^2 dy
where r = x, so r^2 = x^2 = y+1
v = ∫[0,1] π(y+1) dy

Thanks. Would that just be answer, or would I need to multiply the integral by 2?

Oh wait, shouldnt the limits be [0,-1] not [0,1]? Also is the answer then pi/2?

To understand what the question is asking you to do, let's break it down:

1. "The region enclosed above the x-axis and the graph of the function f(x) = x^2 - 1": This refers to the portion of the curve of the function that lies above the x-axis.

2. "Revolving... about the y-axis": Imagine taking the region described in step 1 and rotating it around the y-axis, which is the vertical line passing through the origin (0,0).

Now, to find the volume generated, you need to use the method of cylindrical shells and integrate over the region.

To determine the integral bounds, you need to find the x-values where the region starts and ends. In this case, notice that the function f(x) = x^2 - 1 crosses the x-axis when x = -1 and x = 1. Therefore, the bounds of integration for this problem will be -1 ≤ x ≤ 1.

In summary, you need to find the volume of the solid generated by rotating the region between the x-axis and the curve y = x^2 - 1 about the y-axis, using the method of cylindrical shells, with the bounds of integration being -1 and 1.