The region specified I understand to be shown above below the x-axis between the red curve y=9x-12x4 and blue line x=3, which meet at (3,-945), where 9×3-12×81=-945. This region is to be rotated around the blue line as axis of rotation. The distance between any point P(x,y) on the curve and the line forms the radius of a circle and a disc is created by the rotating point. The area of the disc is π(3-x)2. If the disc has a thickness dy, its volume is π(3-x)2dy so the volume of the solid created by many stacked discs=-945∫0π(3-x)2dy. Since y=9x-12x4, dy can be replaced by (9-48x3)dx, and the limits changed. When y=0, 9x-12x4=0=3x(3-4x3), so x=0 or ∛¾. The latter is the actual low limit for x while x=3 is the high limit. Because of the scale of the graph it's not easy to see the low limit.
We can now evaluate the integral:
π∛¾∫3(3-x)2(9-48x3)dx=
π∛¾∫3(9-6x+x2)(9-48x3)dx=
π∛¾∫3(81-54x+9x2-432x3+288x4-48x5)dx=
π[81x-27x2+3x3-108x4+288x5/5-8x6]∛¾3=
π[(243-243+81-8748+69984/5-5832)-(81∛¾-27(¾⅔)+9/4-108(¾4/3)+288(¾5/3)-9/2)=
-π(502.2+11.1228)=-513.323π =-1612.65 cubic units approx.
This volume lies below the x-axis, hence the negative sign, so the actual volume of this "pinnacle" is about 1612.65 cubic units.
(A cone with base radius 2 and height 945 has a volume of ⅓(22)×945π=1260π=3958.41 cubic units. Since the cone encloses the pinnacle, the volume of the cone should exceed the volume of the pinnacle---which it clearly does.)