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Find the area of the surface generated by revolving the given curve about the x-axis.

x=t

y=t^3 /3 for t=0 to t=3

This parametric eqn is the same as the eqn y = (x^3)/3. Since we are finding the surface of revolution, S, about the x axis, the function y = f(x) gives the radius of the circle around the x axis, at a particular x value. The revolution abround the x axis is thus a circle of circumference 2πy (y = (x^3)/3).

At each point x along the curve there corresponds an infinitesimal length, dl, centered at x, that gets revolved around the x axis, i.e, it gets multiplied by 2πy. It thus sweeps out a circular ribbon of width dl in a complete circle. We then need to integrate the slices along the x axis from 0 to 3, i.e., calculate ∫2πy･dl from 0 to 3

The length dl is given by the so-called arclength of the curve y = f(x) from x to x+dx. Just like a right triangle, this length is dl = sqrt( dx^2 + dy^2). This can also be written as

dl = sqrt(1+(dy/dx)^2)･dx.

Calculating df(x)/dx and plugging it, f(x) and dl into the integral ∫2πy･dl gives

S = ∫ (2π/3) (x^3) ((1+x^2)^1/2) dx integrated from 0 to 3.

This is the formula you need to calculate to solve for S given the problem description. Please let me know if you understand (and buy) the above derivation. If you need help evaluating the integral, let me know. Good luck.

Randy

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