Calculate volume of a solid rotating around the y-axis

[Nivelo]

Sorry if i made any language errors, English isn't my first language.
Question: The limited area in the plane is created when the space between the line y=1 and the graph to the function f(x)=3*x/(x^2+1) rotates around the y-axis. Calculate the volume of the solid.I want to sum up all the circular discs that make up the body in order to get the volume. One disc has the
base area: pi*radius^2 * the height dy. Since its rotating around the y-axis i assume i first need to find the inverse by solving x from y since the radius is the distance from the y-axis and therefore x. I assume the limits of the x-axis is where f(x)=1 and when i solve this i get 4 points, 2 left of the y-axis and 2 right, which i listed below. Since i find the inverse and integrate over y i want the inverse limits. I get the lower limit to y=1 and the upper to y=3/2 (from finding maximum). The problem is i get a very complicated integral which makes me think its wrong and it also not the same as the correct answer.
I get the integral I=2*pi*integral from 1 to 3/2 of (3+sqrt(9-4*y^2)/(2*y))^2)dy)

My question is if my limits are correct and if i need to integrate on both sides of the y-axis or if i can mulitply by 2 instead.
I am a bit lost on this question and would appreciate some guidance in general. Can u give me some steps and data I've missed?

Limits x
X1=(-3/2)-sqrt(5/2)
X2=(-3/2)+sqrt(5/2
X3=(3/2)-sqrt(5/2
X4=(3/2)+sqrt(5/2

Any help is appreciated, thanks in advance!

 

[skeeter]

recommend the method of cylindrical shells ...

shell height as a function of x

$h(x) = f(x)-1$

shell circumference as a function of x

$C(x) = 2\pi \cdot x$

shell thickness = $dx$

$\displaystyle dV = h(x) \cdot C(x) \cdot dx \implies V = \int_a^b h(x) \cdot C(x) \cdot dx$

determine the limits of integration, $a$ and $b$ ...

$\dfrac{3x}{x^2+1} = 1 \implies x = \dfrac{3 \pm \sqrt{5}}{2} \implies a = \dfrac{3 - \sqrt{5}}{2} \text{ and } b = \dfrac{3 + \sqrt{5}}{2}$

\(\displaystyle V = 2\pi \int_a^b \left(\dfrac{3x}{x^2+1}-1\right) \cdot x \, dx =2\pi \int_a^b 3 - \dfrac{3}{x^2+1}-x \, dx \)