math
posted by sara .
find the volume of the solid by rotating y=1x^2, y=0 around the xaxis

using discs, of thickness dx
v = ∫[1,1] πr^2 dx
where r = y = 1x^2
So,
v = π∫[1,1] (1x^2)^2 dx = 16/15 π
Using shells, of thickness dy, we have to account for the two branches of the parabola, so it's easier just to use symmetry and get
v = 2∫[0,1] 2πrh dy
where r = y and h = x = √(1y)
v = 4π∫[0,1] y√(1y) dy = 16/15 π 
pi y^2 dx from 1 to +1
same as 2 pi y^2 dx from 0 to 1
2 pi (12 x^2 + x^4) dx from 0 to 1
2 pi (10)  4 pi (1^3/30) + 2 pi (1^5/50)
2 pi  4 pi/3 + 2 pi/5
(pi/15) ( 30  20 + 6)
(16/15) pi
Check my arithmetic !