Find the area of the surface generated when y=4x and x=1 is revolved about the y-axis.

Answer 1

No calculus need for this one. It's just a cylinder with a cone cut out.

r=1
h=4

v = πr^2h - π/3 r^2h = 2π/3 r^2h = 8π/3

Ok ok. If you want to use calculus, then with shells,

v = ∫[0,1] 2πrh dx
where r=x and h=y=4x
v = ∫[0,1] 2πx*4x dx = 8π/3

using discs (washers) you get

v = ∫[0,4] π(R^2-r^2) dy
where R=1 and r=x=y/4
∫[0,4] π(1-y^2/16) dy = 8π/3

Answer 2

We have to use the Surface area of revolution formula

integral(f(x) aqrt(1+f'(x)^2))