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I am just completely lost with this problem...thanks if you can help!

Find the exact area of the surface obtained by rotating the given curve about the x-axis.

x = 6t - 2t^3

y = 6t^2

0 ≤ t ≤ 1

I think you set up the integral of (dy/dt)/(dx/dt) but then I don't know where to go from there. Please help!! thanks a bunch!

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The answer you want is the integral of dA(t) from t = 0 to 1

For each value of t, you have a circular frustum:

http://www.analyzemath.com/Geometry_calculators/su...

whose outer surface (not the surface of the two circles) is dA(t). One radius of the frustum is y(t); the other is y(t) + dy(t). The height of the frustum is dx(t)

You can think of this outer surface as a strip of width sqrt(dx^2 + dy^2) and length (2)(pi)(y). Then dA(t) = width(t) height(t).

Then just integrate.

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