Here is the region we need to rotate:

And here is the volume generated when we rotate the region around the `y`-axis:

The volume generated when revolving the curve bounded by `y=x^3`, `x=0` and `y=4` around the `y`-axis.

We first must express *x *in terms of* y*, so that we can apply the volume of solid of revolution formula.

If *y* = *x*^{3} then *x* = *y*^{1/3}

The formula requires *x*^{2}, and on squaring we have *x*^{2} = *y*^{2/3}

`text[Vol] = pi int_c^d x^2 dy`

`=pi int_0^4 y^[2//3] dy`

`=pi [(3y^[5//3])/(5)]_0^4`

`=(3pi)/(5)[y^[5//3]]_0^4`

`=(3pi)/(5)[10.079-0]`

`=19.0\ text[units]^3`

Please support IntMath!