The function $f$ is defined by $$f(x)=3(1+x)^{0.5}\cos\left(\frac{\pi x}{6}\right)$$ for $0\leq x\leq3$. The function $g$ is continuous and decreasing for $0\leq x\le3$ with $g(3)=0$.
The figure above on the left shows the graphs of $f$ and $g$ and the regions $R$ and $S$. $R$ is the region bounded by the graph of $g$ and the $x$- and $y$-axes. Region $R$ has area $3.24125$. $S$ is the region bounded by the $y$-axis and the graphs of $f$ and $g$.
The figure above on the right shows the graph of $y=(g(x))^2$ and the region $T$. $T$ is the region bounded by the graph of $y=(g(x))^2$ and the $x$- and $y$-axes. Region $T$ has area $5.32021$.
I'm given these three functions and three areas, and I need to find the volume of the shape created when $S$ is revolved around $y = -3$.
I can rotate around the $x$-axis fine, as I am given $(g(x))^2$, so I can plug in the area of $T$ and find the integral from $0$ to $3$ of $(f(x))^2$, however, I don't know how to revolve around $y = -3$, as it would involve $(g(x)+3)^2$, not $(g(x))^2$, and since I don't have $g(x)$, I cannot find the integral.