Question

ind the volume of the solid that is obtained when the region under the
curve y = 4 − x^2/6 is revolved around the y -axis between y = 0 and y = 4 .
using discs,
v = ∫[0,4] πr^2 dy
where r = x = √(24-6y)
v = π∫[0,4] (24-6y) dy
= π(24y-3y^2) [0,4]
= 48π

using shells,

v = ∫[0,√24] 2πrh dx
where r=x and h=y=4-x^2/6
v = 2π∫[0,√24] x(4-x^2/6) dx
= 2x^2 - x^4/24 [0,√24]
= 48π

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