calculus
posted by nicko .
find the volume formed by revolving the parabola y=4x^2 in the first quadrant:
a.) about x=0
b.) about x=2

a)
after making the diagram, take horizontal discs, so
volume = π[integral] x^2 dy from 0 to 4
= π[integral](4y)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 = 4x^2, x=2 and y=4
we need the radius or the x of that part, which is 2  √(4y)
but we will need x^2 which is
4  2√(4y) + 4y
= 8  y  2(4y)^(1/2)
this will have to be integrated to give
8y  y^2/2 + (4/3)(4y)^(3/2)
see if you can finish it from here.