Show that when the portion R of the plane under x*y = 9 and to the right of x= 1 is revolved about the x axis?

Question

the volume generated = 81pi cu. unit . But the area of the surface is infinite

Answer 1

Rewrite the equation as y=9/x. If we take one point on that curve and rotate it around the axis, the circle generated has a circumference of 18π/x and an area of 81π/x² (2πr and πr², respectively). Each of these circles is as an infinitesimally thin slice of the entire solid of revolution, and so we can find the surface area and volume of it by integrating the circumference and area of these circles, respectively. Thus the surface area of the entire curve is (n→∞)lim (1, n)∫18π/x dx (that should be read as "the limit as in approaches infinity of the definite integral from 1 to n of 18 pi over x with respect to x"). The antiderivative of 18π/x is 18π ln (x) and thus the by the fundamental theorem of calculus, (n→∞)lim (1, n)∫18π/x dx= (n→∞)lim (18π ln (n) - 18π ln (1)). Since the limit of ln n as n→∞ is ∞, the surface area is infinite.

Repeating for the volume: the volume is (n→∞)lim (1, n)∫81π/x² dx . Our antiderivative is then -81π/x, and so the final volume is (n→∞)lim -81π/n - (-81π/1). Since the limit of 1/n as n→∞ is zero, the first term in this expression is zero, thus the final answer is - (-81π/1) or simply 81π.

Answer 2

Area : Integral of 9/x from 1 to infinite

the integral is 9 ln (x) where x ranges from 1 to infinite.

so the area is infinite

Volume integral of pi (9/x) ^2 from 1 to infinite, the integral is 81pi (-1/x) where x ranges form 1 to infinite, so the volume is 81pi

Answer 3

I don't get it? at all