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
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 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
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
thanks, Cal