Problem: Consider the region $R$ bounded by $y=f(x)>0$, $x=1$, and $x=2$. If the volume of a solid formed by rotating $R$ about the $y$-axis is $\pi$, and the volume formed by rotating $R$ about the line $x=-3$ is $6\pi$, find the area of $R$.
I am not sure how to go about this using either the shell method or washer method. Also, I am not familiar with the proper math tools on this site, I apologize for that.
On
This problem is literally trivial when you use Pappus's ($2^{nd}$) Centroid Theorem: the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid $C$, i.e., $2\pi C$. The bottom line is that the volume is given simply by $V=2\pi CA$.
Thus, $$ V_1=2\pi CA=\pi\\ V_2=2\pi (C+3)A=6\pi\\ C=\frac{1}{2A}\\ A=\frac{5}{6} $$
where $C$ is the centroid of the area $A$ relative to $y=0$. Notice that there are no integrals and no reference to the function $f(x)$.
This is how it works
For rotation about ($x = 0$)-axis $$ V_{x=0} = \int_0^{2\pi}\left(\int_1^2 f(r)rdr\right)d\phi = 2\pi\int_1^2 f(r)rdr = \pi $$ That means $$ \int_1^2 f(r)rdr = 1/2 \tag{*} $$ For rotation about ($x = -3$)-axis $$ V_{x=-3} = \int_0^{2\pi}\left(\int_4^5 f(r)rdr\right)d\phi = 2\pi\int_4^5 f(r-3)rdr = 6\pi $$ That is $$ \int_4^5 f(r-3)rdr = 3 $$ For the last integral change variables $r = x +3$ so it becomes $$ 3 = \int_4^5 f(r-3)rdr = \int_1^2 f(x)(x+3)dx = \int_1^2 f(x)xdx + 3\int_1^2 f(x)dx $$ the area $R$ is the $\int_1^2 f(x)dx$ and from Eq.($*$) you can get $\int_1^2 f(x)xdx$. Substitute these in the last equation you get $$ 3 = \frac{1}{2} + 3R \Rightarrow R = \frac{5}{6} $$