Asked by Melissa
                Find the area of the surface generated when y=4x and x=1 is revolved about the y-axis.
            
            
        Answers
                    Answered by
            Steve
            
    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
    
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
                    Answered by
            Melissa
            
    We have to use the Surface area of revolution formula 
integral(f(x) aqrt(1+f'(x)^2))
    
integral(f(x) aqrt(1+f'(x)^2))
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