Evaluate the definite integral that represents the area of the surface generated by revolving the curve on the indicated interval about the x-axis. y 6 2 xStep 1 Write the original equation: y = 6 2 x Since y has a continuous derivative on the interval [1, 3], then the area of the surface of the revolution is given below. S = 2x dx Here, the distance between the x-axis and the graph of fis r(x) = f(x) Step 2 Differentiate the function y = f(x) : + 1 with respect to x. 6 2 x 1 f '( x) = 2 2 2x Therefore, 2 1 + [f '(x) ]2 = 1+ 2 2 N 2x 1 = + 2 2 2 2xStep 3 Now substitute the values of 1 + [f '(x)]2 and r(x) in S = 2n (x)V 1+ [f '(x)]2 dx to obtain the area of the surface of revolution. Therefore, S = 21 dx 2 x 2 5 1 X dx N 12 X 3 4X 1 3 = 21 8 X 4 X 5 X 1 = 0