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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?

1 Answer

Jun 23, 2016

# (7 pi) /12#

Explanation:

Desmos

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#

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