Find the volume of the solid generated by revolving the region bounded by the graphs about the line $x=6$.
The graphs are $$y=x^2$$ $$x=4$$ Setting up the integral using the shell method gives the equation $$2\pi \int_0^{16}\left(6-x^2\right)\left(2 \right)dx$$
After multiplying out we get $$2\pi \int_0^{16}12-2x^2 dx$$ I haven't integrated yet because I wanna make sure I have the correct integral first. The way I understand it is, you take the graph you're revolving and subtract it from the line you're revolving around. This being the axis of rotation.
In this case it would be $6-x^2$. Then doing the same for $x=4$, this being $2$ from, $6-4$.
Am I correct in my understanding, is there an easier way to think of this and if I am wrong how much of this do I have wrong?
Using cylindrical shell method
$$2\pi \int_0^4\left(6-x\right)\left(x^2 \right)dx=128\pi $$