Answer
$$\frac{{28}}{{15}}\pi $$
Work Step by Step
$$\eqalign{
& 2\pi \int_0^1 {\left( {y + 1} \right)\sqrt {1 - y} dy} \cr
& {\text{Integrate by substitution}} \cr
& {\text{Let }}u = 1 - y,{\text{ }}y = 1 - u,{\text{ }}dy = - du \cr
& {\text{The new limits of integration are:}} \cr
& y = 1 \to u = 0 \cr
& y = 0 \to u = 1 \cr
& {\text{Substituting}} \cr
& 2\pi \int_0^1 {\left( {y + 1} \right)\sqrt {1 - y} dy} = 2\pi \int_1^0 {\left( {1 - u + 1} \right)\sqrt u \left( { - 1} \right)du} \cr
& = 2\pi \int_1^0 {\left( {u - 2} \right){u^{1/2}}du} \cr
& = 2\pi \int_1^0 {\left( {{u^{3/2}} - 2{u^{1/2}}} \right)du} \cr
& = 2\pi \left[ {\frac{2}{5}{u^{3/2}} - \frac{4}{3}{u^{3/2}}} \right]_1^0 \cr
& = 2\pi \left[ {\frac{2}{5}{{\left( 0 \right)}^{3/2}} - \frac{4}{3}{{\left( 0 \right)}^{3/2}}} \right] - 2\pi \left[ {\frac{2}{5}{{\left( 1 \right)}^{3/2}} - \frac{4}{3}{{\left( 1 \right)}^{3/2}}} \right] \cr
& = - 2\pi \left( {\frac{2}{5} - \frac{4}{3}} \right) \cr
& = \frac{{28}}{{15}}\pi \cr} $$
