posted by nicko .
find the volume formed by revolving the parabola y=4-x^2 in the first quadrant:
a.) about x=0
b.) about x=2
after making the diagram, take horizontal discs, so
volume = π[integral] x^2 dy from 0 to 4
= π[integral](4-y)dy from 0 to 4
= π [4y - y^2/2] from 0 to 4
= π( 16 - 8 - 0) = 8π
b) since you only want to rotate the part in the first quadrant, look at your diagram.
I would find the volume of the cylinder with radius 2 and height of 4, the "hollow out" the part contained by y = 4-x^2, x=2 and y=4
we need the radius or the x of that part, which is 2 - √(4-y)
but we will need x^2 which is
4 - 2√(4-y) + 4-y
= 8 - y - 2(4-y)^(1/2)
this will have to be integrated to give
8y - y^2/2 + (4/3)(4-y)^(3/2)
see if you can finish it from here.