Find the area of the region, R, bounded by the curve y=x^(-2/3), the line x = 1 and the x axis . In addition, find the volume of revolution of this region when rotated 2 pi radians around the x axis.

As we are looking to find the area under a curve we can use integration. The area under a curve with equation y = f(x), bounded by the lines x = a and x = b and the x axis can be expressed as:A = integral (from a to b) f(x) dxThus, the area of the region, R, can be expressed as:R = integral (from 1 to infinity) x^(-2/3) dxR = [3x^(1/3)] (from 1 to infinity)R = (3infinity(1/3))-(31^(1/3)) = infinity - 3 = infinityR = infinityTherefore the region has infinite area, Considering now the volume of revolution, again using integration:The volume of revolution 2pi radians around the x axis of the same region described above can be expressed:V = pi * integral (from a to b) f(x)^2 dxThus, for the curve in question:V = pi * integral (from 1 to infinity) [x^(-2/3)]^2 dx = pi * integral (from 1 to infinity) x^(4/3) dxV = [-3x^(-1/3)] (from 1 to infinity) = (-3infinity^(-1/3))-(-31^(-1/3)) = 0 - (-3) = 3V = 3This interesting problem shows that a region can have infinity area but its revolution can have fininte volume