plz hlp me

Area of a surface of revolution for y = f(z). Let f(a) be a nonnegative smooth function (smooth means continuously differentiable) over the interval Ja, b). Then, the area of the surface of revolution formed by revolving the graph of y - f() about the x-axis is given by 27f (2) /1+ [f' (z)12 da Part 1. Setup the integral that will give the area of the surface generated by revolving the curve f(a) = over the interval [3, 7) about the a-axis. = Part 2. Calculate the area of the surface of revolution described above. Round answer to three decimal places. units squared