shells are just cylinders, so
v = ∫[0,1] 2πrh dy
where r = 1-y and h = 4-x = 4-4∛y
v = 8π∫[0,1] (1-y)(1-∛y) dy
= 8π∫[0,1] 1 - y^(1/3) - y + y^(4/3) dy
= 8π (y - 3/4 y^(3/4) - 1/2 y^2 + 3/7 y^(3/7) [0,1]
= 8π (1 - 3/4 - 1/2 + 3/7)
= 8π (5/28)
= 10π/7
Just for a check, try using discs (washers)
Now, we have
v = ∫[0,4] π(R^2-r^2) dx
where R=1 and r = 1-y = 1-x^3/64
v = π∫[0,4] (1-(1-x^3/64)^2) dx
= π∫[0,4] (x^3/32 - x^6/4096) dx
= π (x^4/128 - x^7/26872) [0,4]
= 10π/7
Consider the given curves to do the following.
64 y = x^3, y = 0, x = 4
Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about y = 1.
1 answer
1 answer