Answer
$$\frac{{36}}{{105}}\pi $$
Work Step by Step
$$\eqalign{
& 2\pi \int_{ - 1}^0 {{x^2}\sqrt {x + 1} dy} \cr
& {\text{Integrate by substitution}} \cr
& {\text{Let }}u = x + 1,{\text{ }}x = u - 1,{\text{ }}dx = du \cr
& {\text{The new limits of integration are:}} \cr
& x = 0 \to u = 1 \cr
& x = - 1 \to u = 0 \cr
& {\text{Substituting}} \cr
& 2\pi \int_{ - 1}^0 {{x^2}\sqrt {x + 1} dy} = 2\pi \int_0^1 {{{\left( {u - 1} \right)}^2}{u^{1/2}}du} \cr
& = 2\pi \int_0^1 {\left( {{u^2} - 2u + 1} \right){u^{1/2}}du} \cr
& = 2\pi \int_0^1 {\left( {{u^{5/2}} - 2{u^{3/2}} + {u^{1/2}}} \right)du} \cr
& = 2\pi \left[ {\frac{2}{7}{u^{7/2}} - \frac{4}{5}{u^{5/2}} + \frac{2}{3}{u^{3/2}}} \right]_0^1 \cr
& = 2\pi \left[ {\frac{2}{7}{{\left( 1 \right)}^{7/2}} - \frac{4}{5}{{\left( 1 \right)}^{5/2}} + \frac{2}{3}{{\left( 1 \right)}^{3/2}}} \right] - 2\pi \left[ 0 \right] \cr
& {\text{Simplifying}} \cr
& = 2\pi \left( {\frac{{16}}{{105}}} \right) \cr
& = \frac{{36}}{{105}}\pi \cr} $$
