using shells,
v = ∫[0,1] 2πrh dy
where r = 1-y and h = √y - y^2
v = 2π∫[0,1] (1-y)(√y - y^2) dy = 11π/30
using washers,
v = ∫[0,1] π(R^2-r^2) dx
where R = 1-x^2 and r = 1-√x
v = π∫[0,1] ((1-x^2)^2-(1-√x)^2) dx = 11π/30
Find the volume of the solid obtained by rotating the region bounded by y=x^2, y=sqrt(x), about y=1.
I have to use either the disk or washer method and i believe i use washer. i set up the integral to be pi[(1-x^3)^2 -(1-sqrt(x))^2] from 0 to 1 for my limits. i worked it out by hand and got 31pi/21 but when i checked my work using a calculator it said approx. 1.495998. Am i setting up the integral correct?
1 answer
1 answer