Hello, FLTR!
Your integral is slightly off . . .
The region R is bounded by the graphs of $\displaystyle x2y=3$ and $\displaystyle x=y^2$.
Set up (but do not evaluate) the integral that gives the volume of the solid
obtained by rotating R around the line $\displaystyle x = 1$.
Did you make a sketch? Code:

:  /
:  ...o (9,3)
:  ..*::::/
:  .*:::::::/
: *::::::::/
+*::::/
1: *::::/
:  o (1,1)
:  / *
: / *
/ 
We will integrate with respect to $\displaystyle y$ and use "washers".
The two functions are: .$\displaystyle \begin{array}{cc}x\:=\:2y + 3 \\ x = y^2\end{array}$
The "outer radius" is: .$\displaystyle r_o\:=\2y + 3) (1) \:=\:2y + 4$
The "inner radius" is: .$\displaystyle r_i\:=\:y^2  (1) \:=\:y^2 + 1$
. . And $\displaystyle y$ ranges from $\displaystyle 1$ to $\displaystyle 3.$
. . . . $\displaystyle V \;=\;\pi\int^3_{1} \bigg[(2y + 4)^2  (y^2 + 1)^2\bigg]\,dy$