How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by #y = 1/x^4#, y = 0, x = 1, x = 4 revolved about the x=-4?

1 Answer
Dec 16, 2017

#(57pi)/16#

Explanation:

the formula for the shell method is #int_a^b2pirhdx#

#a# and #b# are the x-bounds, which are x=1 and x=4, so #a=1# and #b=4#.

#r# is the distance from a certain x-value in the interval #[1,4]# and the axis of rotation, which is x=-4. #r=x-(-4)=x+4#

#h# is the height of the cylinder at a certain x-value in the interval #[1,4]#, which is #1/x^4-0=1/x^4# (because #1/x^4# is always greater than #0# and h must be positive).

plugging it all in: volume #=int_1^4(2pi(x+4)(1/x^4))dx#
you should get: #(57pi)/16#