Finding the volume obtained of a curve after revolving it around $y=-2$, using Mathematica gives $$V=\pi \int_{0}^{2} \left(4+x^{2} \right)^{2}dx=\dfrac{896 \pi }{15}$$ However, in the solution manual it is, $$R=x^{2}+2-(-2)=x^{2}+4$$ $$r=0-(-2)=2$$ $$V=\pi \int_{0}^{2} \left[(x^{2}+4)^{2} -2^{2} \right]dx=\dfrac{776\pi}{15}$$
I don't understand why did Mathematica take only the outer radius? I am confused now to which one is correct.
Your radius is $y$, since you are rotating around the $x$-axis (y=0), right?. It is sufficient to compute \begin{eqnarray} V &=& \pi \int_0^2 y(x)^2 \,dx \\ &=& \pi \int_0^2 (x^2+2)^2 \,dx \\ &=& \pi \left(\frac{1}{5}2^5 + \frac{2}{3}2^3 + 2 \cdot2\right)\\ &=& \pi \frac{96 + 90 + 60}{15} \\ &=& \pi \frac{246}{15} \\ \end{eqnarray} Am I understanding you correctly?
Edit:
If it revolves about the line $y=-2$, you need to add the volume of the additionally created cylinder with radius $r = 2$ and height $h = 4$ ($V_C = \pi r^2h = 16 \pi$). This yields a total volume of \begin{eqnarray} V_{tot}&=& \pi\left(\frac{246}{15}+16\right) = \pi\left(\frac{246+240}{15}\right) =\pi\frac{486}{15} \end{eqnarray}