Answer
$\displaystyle{V=\frac{471\pi }{14}}\\ $
Work Step by Step
$\displaystyle{A\left(x\right)=\pi \left(x^3+3\right)^2-\pi \left(1+3\right)^2}\\
\displaystyle{A\left(x\right)=\pi \left(7-6x^3-x^6\right)}\\$
$\displaystyle{V=\int_{1}^2A\left(x\right)\ dx}\\
\displaystyle{V=\int_{1}^2\pi \left(7-6x^3-x^6\right)\ dx}\\
\displaystyle{V=\pi \int_{1}^27-6x^3-x^6\ dx}\\
\displaystyle{V=\pi\left[7x-\frac{3}{2}x^4-\frac{1}{7}x^7\right]_{1}^2}\\
\displaystyle{V=\pi\left(\left(7(2)-\frac{3}{2}(2)^4-\frac{1}{7}(2)^7\right)-\left(7(1)-\frac{3}{2}(1)^4-\frac{1}{7}(1)^7\right)\right)}\\
\displaystyle{V=\frac{471\pi }{14}}\\ $