Volumes with infinite integrands Find the volume of the described solid of revolution or state that it does not exist.The region bounded by f(x) = (4 ? x)?1/3 and the x-axis on the interval [0, 4) is revolved about the y-axis.

Problem 39EVolumes with infinite integrands Find the volume of the described solid of revolution or state that it does not exist.The region bounded by f(x) = (4 x)1/3 and the x-axis on the interval [0, 4) is revolved about the y-axis.Answer; Step-1; The shell method ; If R is the region under the curve y = f(x) on the interval [a , b] , then the volume of the solid obtained by revolving R about the y-axis is V = 2x f(x) dx.Step-2 Now , we have to find out the volume of the described solid under the region bounded by f(x) = and the x-axis on the interval [0 , 4) is revolved about the y-axis. Consider , y = f(x) = The visual representation of the interval is given below; | | axis of revolution\n | 1\/(4 - x)^(1\/3)\n | x-axis\n(axes not equally scaled) | Therefore , the volume of the described solid under the region bounded by f(x) = and the x-axis on the interval [0 , 4) is revolved about the y-axis is ; Volume (V) = 2x f(x) dx , since by the above formula. = 2x dx , since f(x) = . = dx , since = ……..(1) To evaluate this integral , let us use substitution method . Step-3 ; Consider , (4 -x) = , then = t , and 4 - = x If x = 0 , the lower limit of t = = = . If x = 4 , the upper limit of t = = = Differentiate both sides with respect to x , then ; ( 0 - 1) = 3 , since = n -dx = 3dt ……………(2)Step-4 ; From , (1) and (2) , the volume of the integral can be written as ; V = dx = , since from (2). = 2 (-3t)(4 - )dt = 6 dt , since f(x)dx = 6 = = 6(, since ( 5.0396842 - 2.0158737) 6(3.0238105) 56 .997485 Therefore , volume(V) = dx 56 .997485 Thus , the volume of the solid is ; .