# 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?

Jun 23, 2016

$\frac{7 \pi}{12}$

#### Explanation:

Small element of width $\Delta x$ has volume $\Delta V = \pi {f}^{2} \left(x\right) \Delta x$ so

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

$= \pi \setminus {\int}_{0}^{1} \setminus {x}^{2} \left(x + 1\right) \setminus \mathrm{dx} = \pi \setminus {\int}_{0}^{1} \setminus {x}^{3} + {x}^{2} \setminus \mathrm{dx}$

$= \pi {\left[\frac{{x}^{4}}{4} + \frac{{x}^{3}}{3}\right]}_{0}^{1}$

$= \frac{7 \pi}{12}$