Let R be the region bounded by the following curves. Let S be the solid generate

Question

Let R be the region bounded by the following curves. Let S be the solid generated when R is revolved about the given axis. If possible, find the volume of S by both the disk/washer and shell methods. Check that your results agree. y = x - x^4, y = 0: revolved about the y-axis. Set up the integral(s) that gives the volume of the solid as a single integral if possible using the disk/washer method, if it is possible to do so. Use increasing limits of integration. Select the correct choice below and fill in any answer boxes within integral ()dy = integral ()dy (Type exact answers.) integral ()dy (Type exact answers.) It is not possible to use the disk/washer method to find the volume of the solid. Set up the integral(s) that gives the volume of the solid as a single integral if possible using the shell method. If it is possible to do so. Use increasing limits of integration. Select the correct choice below and fill in any answer boxes within your choice. integral ()dx = integral ()dx (Type exact answers.) integral ()dx (Type exact answers.) It is not possible to use the shell method to find the volume of the solid. The volume of the solid is (Type an exact answer.)

Explanation / Answer

C) it is not possible to use the disk/washer method to find the volume of the solid

for curve y=x-x4, if is not possible to write x as a function of y

given curves y=x-x4, y=0

when curves intersect

x-x4=0

=>x=0, 1

for volume about y axis :

using method of cylindrical shells:
radius of shell=x
height of shell =(x-x4)-0=(x-x4)

volume=[0 to 1]2x[(x-x4)] dx

volume=[0 to 1]2(x2-x5) dx

volume=[0 to 1]2((1/3)x3-(1/6)x6)

volume=2((1/3)13-(1/6)16) -2((1/3)03-(1/6)06)

volume=2((1/3)-(1/6)) -0

volume=2(1/6)

volume=(/3)

the volume of the solid is (/3)

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