Answer
$V = \pi \ln 5$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{{\sqrt {x + 1} }},{\text{ }}y = 0,{\text{ }}x = 0,{\text{ }}x = 4 \cr
& {\text{Let }}R\left( x \right) = f\left( x \right) \cr
& {\text{Apply the disk method}} \cr
& V = \pi \int_a^b {{{\left[ {R\left( x \right)} \right]}^2}} dx \cr
& {\text{So}},{\text{ the volume of the solid of revolution is}} \cr
& V = \pi \int_0^4 {{{\left[ {\frac{1}{{\sqrt {x + 1} }}} \right]}^2}} dx \cr
& V = \pi \int_0^4 {\frac{1}{{x + 1}}} dx \cr
& {\text{Integrate}} \cr
& V = \pi \left[ {\ln \left| {x + 1} \right|} \right]_0^4 \cr
& V = \pi \left[ {\ln \left| {4 + 1} \right| - \ln \left| {0 + 1} \right|} \right] \cr
& V = \pi \ln 5 \cr} $$
