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How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by #y=x^2#, x = 2, x = 7, y = 0 revolved about the x=8?

Answer 1

#3565pi/6#

shell method: volume = #int_a^b2pirhdx#

In this problem, #a=2# and #b=7# (from the boundaries x=2 and x=7).

r is the distance from a certain x value in the interval [2,7] to the axis of rotation, which is x=8. Since r must be a positive distance, it is #8-x#.

h is the height of each shell (parallel to axis of rotation). in this case it is the distance between the upper and lower bounds of the region, or #x^2-0=x^2#

plugging in: volume = #int_2^(7)2pi(8-x)(x^2)dx# #=2piint_2^7(-x^3+8x^2)dx# #=2pi * [F(7) - F(2)]#, where #F(x) = -1/4x^4+8/3x^3# (fundamental theorem of calculus). #=2pi * [(-1/4(7)^4+8/3(7)^3) - (-1/4(2)^4+8/3(2)^3)]#

simplify: #=2pi * [-2401/4+2744/3+16/4-64/3]# #=2pi * [-2385/4+2680/3]# #=2pi * [3565/12]# #= 3565pi/6#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.