Question

What is the surface area of the solid created by revolving `f(x)=xsqrt(x+1)` for `x in [0,1]` around the x-axis?

Answer 1

`(7 pi) /12`

Explanation:

Small element of width `Delta x` has volume `Delta V = pi f^2(x) Delta x` so

`V = pi \int_0^1 \ (x(sqrt(x+1))^2 \ dx`

`= pi \int_0^1 \ x^2(x+1) \ dx = pi \int_0^1 \ x^3+ x^2 \ dx`

`= pi [ (x^4)/4+ (x^3)/3]_0^1`

`= (7 pi) /12`