Can you tell me if im doing this right?

Find the volume of the region obtained by revolving the area below about the line x=3.
y=x3, x=2, y=0

v=pi[2,0] (x^3)^3dx= pi[2,0] x^9
v=pi/10 x^10(2,0)= 1024/10 pi

what you have tried is the volume when rotating the region about the x-axis.

But, of course, the volume of a disk is pi r^2 dx, not pi r^3 dx

In addition, your radius is 3-x, not y, since you are rotating about the line x=3, not y=0.

So, if you want to use discs, you need to integrate over y, since the discs stack up vertically. Note that the discs have a hole of radius 1 in the center, since the axis of rotation is 1 unit beyond the region being revolved.

v = ∫[0,8] π(R^2-r^2) dy
where R = 3-x = 3-∛y and r=1
v = π∫[0,8] (3-∛y)^2 - 1 dy
= π∫[0,8] (9 - 3∛y + ∛y^2 - 1) dy
= π(8y - 9/4 y^(4/3) + 3/5 y^(5/3))[0,8]
= 56/5 π

In this case, it is easier to use shells of thickness dx, so we have

v = ∫[0,2] 2πrh dx
where r = 3-x and h = y = x^3
v = 2π∫[0,2] (3-x)x^3 dx
= 56/5 π