The region is roughly trapezoid shaped, with vertices at (0,1)(0,2)(2,1)(5,2)
With shells, we have to integrate on x, since the shell thickness is dx.
From x=0-2, we just have a rectangle (which, revolved is just a cylinder of radius 2, height 1), and for x=2-5, we have shells of height 2-y.
Since x=1+y^2, y = √(x-1)
v = 2π*2*1 + ∫[2,5] 2πrh dx
where r = x and h = 2-√(x-1)
v = 4π + 2π∫[2,5] x(2-√(x-1)) dx
= 4π + 2π(59/15)
= 178/15 π
just to check, we can use discs, and we have
v = ∫[1,2] πr^2 dy
where r = x
v = π∫[1,2] (y^2+1)^2 dy
= 178/15 π
use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis. sketch the region and a typical shell.
x=1+y^2, x=0, y=1, y=2
3 answers
just one more question how did you get the y=squareroot of x-1?? so confuse
x = 1+y^2
x-1 = y^2
√(x-1) = y
Algebra I, man, Algebra I.
x-1 = y^2
√(x-1) = y
Algebra I, man, Algebra I.