CCan i get some help on how to solve this?

Area of a surface of revolution for y=f(x). Let f(x) be a nonnegative smooth function (smooth means continuously differentiable) over the interval [a,b]. Then, the area of the surface of revolution formed by revolving the graph of y=f(x) about the x-axis is given by S=ab2f(x)1+[f(x)]2dx Part 1. Setup the integral that will give the area of the surface generated by revolving the curve f(x)=2x3 over the interval [4,6] about the x-axis. S= Part 2. Calculate the area of the surface of revolution described above. Round answer to three decimal places. S= units squared