Answer
$$\,V = 8\pi $$
Work Step by Step
$$\eqalign{
& {\text{From the graph we have }} \cr
& x = \sqrt {1 + y} \,\,{\text{and the }}y{\text{ - axis on the interval }}\left[ { - 1,3} \right] \cr
& {\text{The volume of the solid can be calculated using}} \cr
& \,\,\,\,V = \int_c^d {\pi {{\left[ {u\left( y \right)} \right]}^2}dy} \cr
& {\text{Thus}}{\text{,}} \cr
& \,\,\,\,V = \int_{ - 1}^3 {\pi {{\left( {\sqrt {1 + y} } \right)}^2}dy} \cr
& \,\,\,\,V = \pi \int_{ - 1}^3 {\left( {1 + y} \right)dy} \cr
& \,\,\,\,V = \pi \left[ {y + \frac{1}{2}{y^2}} \right]_{ - 1}^3 \cr
& {\text{Evaluating}} \cr
& \,\,\,\,V = \pi \left[ {3 + \frac{1}{2}{{\left( 3 \right)}^2}} \right] - \pi \left[ { - 1 + \frac{1}{2}{{\left( { - 1} \right)}^2}} \right] \cr
& \,\,\,\,V = \pi \left[ {\frac{{15}}{2}} \right] - \pi \left[ { - \frac{1}{2}} \right] \cr
& \,\,\,\,V = 8\pi \cr} $$