Use the cylindrical shell method to find the volume of the solid obtained by rotating the region bounded by the curve y=sin(3x^2) and the x-axis, 0≤x≤√π/3, about the y-axis.
AAAaannndd the bot gets it wrong yet again!
The negative result should have raised a red flag, since the area is under one arch of the curve.
v = ∫[0,√π/3] 2πrh dx
where r = x and h = y = sin(3x^2)
v = ∫[0,√π/3] 2πxsin(3x^2) dx = π/6
V = 2π∫_0^√π/3 y*r dr
V = 2π∫_0^√π/3 sin(3x^2)*x dx
V = 2π[-(1/3)cos(3x^2)]_0^√π/3
V = 2π[-(1/3)cos(π/3) - (-1/3)]
V = 2π[-(1/3)cos(π/3) + (1/3)]
V = 2π[(2/3)cos(π/3)]
V = (4π/3)cos(π/3)
V = (4π/3)(-1/2)
V = -2π/3