Finite Volume, Infinite Surface Area

Take the reciprocal curve on the interval from 1 to ∞ and rotate it about the horizontal axis. The solid it sweeps out has finite volume and infinite surface area. You could fill it with paint and never manage to coat its inner wall.

Setup: r(X) = 1/xfor x≧1
Volume by disks

slice perpendicular to the axis. Each cross section is a disk of radius r(X) so the volume integral runs form 1 to ∞ as an improper integral.

V π*(∫_1^∞)(r(X)*d(x))= π*(∫_1^∞)(1/(x^2)*d(x))

V = (lim_b)(∞) π*(∫_1^b)(1/(x^2)*d(x)) = (lim_b)(∞)*π*(1-1/b)=π

The volume is exactly π,so it is finite.

Surface Area

surface area of revolution carries the arc length factor and that is what changes the outcome

(r^′)*X=1/(x^2)

S = 2*π*(∫_1^∞)(r(X)√(,(1+(r^′)*(X_)^2)))*d(x) = 2*π*(∫_1^∞)(1/x√(,(1+1/(x^4))))*d(x)

For every x≧1the square root exceeds 1, so the integrand is bounded below by 1/x

S > 2*π*(∫_1^∞)(1/x*d(x))= 2*π*(lim_b)(∞)*ln(b)= ∞

The harmonic integral diverges, so the surface area does too. So it is infinite.

y=1/x and y=−1/x on [1,10]