You should use the following formula to evaluate the volume of solid ormed by revolving the region bounded by the graph of `y=(x-3)^2` , x axis and `x_1 = 0` , `x_2` , such that:

`V = pi*int_0^(x_2) y^2 dx`

Since the problem does not provide the upper limit of integration, yields:

`V = pi*int_0^(x_2) ((x-3)^2)^2 dx`

`V = pi*int_0^(x_2) (x-3)^4 dx`

`V = pi*(x - 3)^5/5|_0^(x_2)`

Using the fundamental theorem of calculus, yields:

`V = pi*(x_2 - 3)^5/5 - pi*(0-3)^5/5`

`V = pi*((x_2 - 3)^5 + 243)/5`

**Hence, evaluating the volume of solid formed by revolving the region bounded by the given curves, yields **`V = pi*((x_2 - 3)^5 + 243)/5.`