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Here is a sketch of the region to be rotated and the line to be rotated around.

Notice that the volume will be exactly the same if we were to move everything up by 3 units, but with the advantage of rotating around the x axis. So we want to find the volume of the region under $\displaystyle \begin{align*} y = 12 - x^2 \end{align*}$ between $\displaystyle \begin{align*} x= 0 \end{align*}$ and $\displaystyle \begin{align*} x = 3 \end{align*}$ rotated around the x axis, and then subtract the volume of the region under $\displaystyle \begin{align*} y = 3 \end{align*}$ between $\displaystyle \begin{align*} x = 0 \end{align*}$ and $\displaystyle \begin{align*} x = 3 \end{align*}$ rotated around the x axis. Thus the volume we want is

$\displaystyle \begin{align*} V &= \int_0^3{ \pi\,\left( 12 - x^2 \right) ^2\,\mathrm{d}x } - \int_0^3{ \pi\,\left( 3 \right) ^2\,\mathrm{d}x } \\ &= \pi\int_0^3{ \left[ \left( 12 - x^2 \right) ^2 - 3^2 \right]\,\mathrm{d}x } \\ &= \pi\int_0^3{ \left( 144 - 24\,x^2 + x^4 - 9 \right) \,\mathrm{d}x } \\ &= \pi\int_0^3{ \left( 135 - 24\,x^2 + x^4 \right) \,\mathrm{d}x } \\ &= \pi\,\left[ 135\,x - 8\,x^3 + \frac{x^5}{5} \right] _0^3 \\ &= \pi\,\left\{ \left[ 135\,\left( 3 \right) - 8 \,\left( 3 \right) ^3 + \frac{3^5}{5} \right] - \left[ 135\,\left( 0 \right) - 8 \,\left( 0 \right) ^3 + \frac{0^5}{5} \right] \right\} \\ &= \pi\,\left( 405 - 216 + \frac{243}{5} - 0 \right) \\ &= \frac{1188\,\pi}{5}\,\textrm{units}^3 \end{align*}$