I don't know what the cylindrical shells means, but I would just do
Pi ( Integral from 0 to 4 of (4 -(4-y)) dy)
I'll explain how I got that if you want.
my teacher gave this to me on a test that i failed i would like to know how to do this kind of problem in case it appears in the future
a region bounded by y=(-x^2)+4, y=4, and x=2 is revolved around the y-axis. use cylindrical shells to find the volume of the solid generated.
hope someone can help.
Okay, so a graph of the three functions gives an enclosed region that kind of looks like a right triangle, with the right angle in the top right, and the hypotenuse being a curve, in the first quadrant.
If you are to revolve this around the y-axis you will get a disc shape because of the space between the area and the axis.
If you remember how to find the area of the disc, it is the outer radius squared, minus the inner radius squared, times Pi.
Since you are revolving around the y-axis, your change is going to be along the y axis; hence dy. To integrate with respect to y, you need to change your equation to x=
This gives you x= +- sqrt[4-y] (Sorry I do not know how to use math tags)
You only need the positive portion of this graph since we only need to worry about y from 0 to 4 as bounded by the fuctions.
When you imagine the disc that will be generated. x=2 becomes your outer radius, and x = sqrt[4-y] your inner radius.
You integrate starting at y = 0 because this is where the graphs of the two previous functions intersect. You end at 4 because your third function bounds the area here.
This all gives you the integral I gave you before.
Tell me if anything doesn't make sense.
EDIT:
nevermind.. didn't know there is a shell method