Problem :
Let $A$ denote a region bounded by $\frac{x^2}{4}+\frac{y^2}{9}=1$.
Find the volume of the solid of revolution formed by revolving the region $A$ about the line $3x + 4y = 12$.
I tried to apply Pappus–Guldinus theorem but I can't use this in this case since the line $3x + 4y = 12$ intersects the region $A$.
Then, how can I compute the volume of the solid in this case?
I asked a question about volume of the solid before and it could be solved by rotating the xy-plane.
But I think I can't apply that to this problem.
Is there any general method to rotate about $y=mx + n$?
It's not a full answer, but i think, you can apply the Pappus–Guldinus theorem.
To be honest, I don't know, if there the line intersects the region. But here it's not a problem, since the upper right part of the ellipse is fully contained in the lower left part of the ellipse, if we reflect it on the line.
Hence, instead of $A$, we only rotate the lower left part.
Let $\ell:=\{(x,y)\mid 3x+4y=12\}$ and $A:=\{(x,y)\mid\frac{x^2}{4}+\frac{y^2}{9}\leq 1\}$. Then $\partial A\cap \ell=\{((0,3), (\frac 85, \frac 95)\}.$ We can parametrize $$A=\{(2r\cos t, 3r\sin t) \mid 0\leq t\leq 2\pi,\ 0\leq r\leq 1\}$$ Now we cut the sector $POQ$ away. Let $t_0:=\frac\pi2$ (corresponds to $Q$) and $t_1:=2\pi+\arccos(\frac 45)$ (corresponds to $P$). We get $$\tilde{A}=\{(2r\cos t, 3r\sin t) \mid \frac\pi2\leq t\leq 2\pi+\arccos(\frac 45),\ 0\leq r\leq 1\}\cup\operatorname{conv}\{(0,0),(0,3),(\frac 85, \frac 95)\}.$$ Now, you just have to compute the new centroid and the new area. Hopefully, it helps.