I am stuck with this problem given that $a=3$ and $b=6.$ I was trying to cut with respect to $y$. I tried to do two integrals: from 0 to 6 and 6 to 15 because $y = 9+6$. However, I think my integral is incorrect and I got my answer as $999\pi/2$ which seems dubious.
On
Your curve is currently $y=x^2+6.$ We first translate it $3$ units to the left so we can revolve it around the $y$-axis instead of the line $x=3.$
After this, the easiest way would be to first find the volume of the cylinder when the combined regions A and B are revolved around $x=0$ and then subtract the volume of solid formed due to the lower region (i.e. region B). The latter can be computed by expressing $x^2$ in terms of $y$ and then using the well-studied disc method to find the volume.
We first translate the curve by the vector $-3\choose0$ to yield $$y=(x+3)^2+6$$
Now, we find the volume of the cylinder, with radius $a=3$ units, and height $15$ units. Hence the volume will be $$V_{\text{cylinder}}=\pi \cdot 3^2\cdot 15=135\pi$$
For the volume made by region B, we calculate the volume using the "disc method." We first express $x^2$ in terms of $y$ as follows
$$y-6=(x+3)^2\implies x^2=(\sqrt{y-6}-3)^2=y+3-6\sqrt{y-6}$$
Note that we only include the positive square root since we're dealing with the right side of the parabola's vertex. Our $x$ variable now acts as the radius of one infinitesimally thin disc, so the volume $\pi x^2$ for each disc summed up can be written as the integral
$$V_{\text{region B}}=\pi\int_{x=-3}^{x=0}{x^2}dy$$ $$=\pi\int_{y=6}^{y=15}{(y+3-6\sqrt{y-6})}dy$$
Evaluating this integral yields $$V_{\text{region B}}=\frac{27\pi}{2}$$
Hence, $$V_{\text{region A}}=V_{\text{cylinder}}-V_{\text{region B}}=135\pi-\frac{27\pi}{2}=\boxed{\frac{243\pi}{2}}$$
First find the volume of the cylinder which is formed by rotating the brown region (with both A and B) around x = 3. Then, find the volume of the solid formed by rotating the region B around x = 3. Then subtract this volume from the volume of the cylinder to get your answer.
To find the volume of the solid obtained by rotating B around x = 3, integrate the square of the given function or (x2 + 6)2 and multiply that by pi (this is the disc method).