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 ?
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?
4 answers
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
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?
1 answer