Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=1/x^2, y=0, x=4, x=5; about y=−2

3 answers

the shell method seems best

the corners of the cross section are
(4,0), (5,0), (5,.04), (4,.0625)
not needed for solution, but helps to visualize

the radius of the shell is x+2

volume = 2π * r * L * thickness
= 2π * (x + 2) * (1 / x^2) * dx

integration limits are 4 to 5

have at it
the problem with the above scenario is that the shells have thickness dy, since the region is rotated around a horizontal axis.

v = ∫[1/25,1/16] 2πrh dy
where r=y+2 and h=x-4=(1/√y - 4)
v = ∫[1/25,1/16] 2π(y+2)(1/√y - 4) dy = 4913π/120000

Hmmm. Let's check that using discs of thickness dx:

v = ∫[4,5] π(R^2-r^2) dx
where r=51/25 and R=y+2=1/x^2+2
v = ∫[4,5] π((1/x^2+2)^2-(51/25)^2) dx = 4913π/120000
oops...got my axes swapped

thanks, Cal