Answer
$$V = \frac{{128}}{3}\pi $$
Work Step by Step
$$\eqalign{
& {\text{Calculate the volume using the Shell method about the }}y{\text{ - axis}} \cr
& V = \int_a^b {2\pi x\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& {\text{Let }}f\left( x \right) = \frac{1}{{\sqrt {x + 4} }}{\text{ and }}g\left( x \right) = 0,\,\,\,{\text{on the interval }}\left[ {0,12} \right] \cr
& V = \int_0^{12} {2\pi x\left[ {\frac{1}{{\sqrt {x + 4} }}} \right]} dx \cr
& V = 2\pi \int_0^{12} {\frac{x}{{\sqrt {x + 4} }}} dx \cr
& {\text{Integrate by tables using the formula 94}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int {\frac{x}{{\sqrt {ax + b} }}} dx = \frac{2}{{3{a^2}}}\left( {ax - 2b} \right)\sqrt {ax + b} + C \cr
& V = 2\pi \left[ {\frac{2}{3}\left( {x - 8} \right)\sqrt {x + 4} } \right]_0^{12} \cr
& V = 2\pi \left[ {\frac{2}{3}\left( {12 - 8} \right)\sqrt {12 + 4} } \right] - 2\pi \left[ {\frac{2}{3}\left( {0 - 8} \right)\sqrt {0 + 4} } \right] \cr
& V = 2\pi \left( {\frac{{32}}{3}} \right) - 2\pi \left( { - \frac{{32}}{3}} \right) \cr
& V = \frac{{128}}{3}\pi \cr} $$
