Shells:
It's PI, not PIE for goodness sake.
This is a question on one of my past exams that I'm trying to do...
the question says to find the volume of the regions bounded by y=x^2-2 and
y=x
rotated about x=2
I was thinking that the radius equalls (2-x) [I'm not entirely sure about this]
and the height equals (x-(x^2-2))
all of them multiplied together with PIE under the integral from -1 to 2
is this correct?
to avoid confusion, remember that we always consider the distance from the y-axis to the bounded region to be x (that is when we are revolving about a vertical line of course). the line x = 2 is to the right of the region, so the distance from the y-axis some point in the region is x, and the distance from that point to the line x = 2 is (2 - x). the (2 - x) is the radius of our revolution, so that's what we want.
Actually, it does allow it...to a point. I have posted them before. But, with this one, I had to resize the picture. When I did, it wouldn't animate. I used to post animated graphs, but it seems lately it won't handle them unless I resize, then it doesn't work.