Volume of revolution

x= 3y^2 and x =3

rotate around x=3

now I should be able to use the shell's method right?

so the upper and lower limit should be 0 to 3 and and the integral is (3-x) (shell's height)

now what is the shell's height? Is it just the function it self but in terms of y? so is it squareroot of x/3 ?

You can always use any method you like...it's just for some problems, one way is "easier" than the other (i.e. one way doesn't cause you to have to break the integral into parts, while the other way does).

In this case, I really don't think one way is better than the other...in fact, I think the washer (disk) method is better.

And no, you don't have the shell method correct, you are trying to apply the washer method (or disk, in this case it's just a disk). If you are using the shell method, your the height should be in the y direction (which is parallel to x = 3).

So let's use the disk method, since I think that will be better.

volume of a disk:

πr²*h
-->

dV = πr²*dh

dh should be parallel to the axis (x = 3), so that means it is dy.

So we have:

dV = πr²dy

r is the distance from the axis (which you correctly said was (3 - x)
-->

dV = π(3 - x)²dy

But wait, we have x and dy...we'll have to change something...but this is easy, just plug in x = 3y²
-->

dV = π(3 - 3y²)²dy

You are going to integrate from y = -1 to y = 1 (solve equation 3y² = 3 --> y² = 1 --> y = ±1)

Now multiply that out (so you can integrate):

dV = π(9 - 18y² + 9y⁴)dy
--> integrate

V = π(9y - 6y³ + 9/5y⁵)

--> plug in y = 1 and y = -1

V = π * { (9 - 6 + 9/5) - (-9 - 18 - 9/5) }
-->

V = π * (2*9 + 2 * 9/5) = 2π*(9 + 9/5)
-->

9 + 9/5 = 45/5 + 9/5 = 54/5
-->

V = 108π/5

Edit:

To check our work, here is the shell method: