If the infinite curve y=e^(-3x)...

Question

If the infinite curve y=e^(-3x), x>=0, is rotated about the x-axis, find the area of the resulting surface.

Answer 1

Guide on the formula S = 2π∫(y)ds or 2π∫(e^-3x)ds or 2π∫(e^-3x)√[1 + {d(e^-3x)/dx}²]dx.

Rewrite this last as 2π∫(e^-3x)√[1 + {-3e^-3x}²]dx or 2π∫(e^-3x)√[1 + {9e^-6x}]dx.

Establish:

u = e^-3x

du = -3e^-3xdx or -3udx

dx = -du/3u

Then (2π)∫(e^-3x)√[1 + {-3e^-3x}²]dx becomes (2π)∫u√[1 + {du/dx}²](-du/3u) or 

(-2π/3)∫√[1 + {-3u}²]du.

Next, from(-2π/3)∫√[1 + {-3u}²]du, go to (-2π/3)∫√[1 + 9u²]du.

Draw a right triangle with horizontal leg of 1, vertical leg of 3u, and hypotenuse of √[1 + 9u²].

The angle opposite the vertical leg is θ=arctan {3u/1} or arctan 3u.

That is, tan θ = 3u and u is (1/3)tan θ.

Furthermore, du equals (1/3)sec²θdθ.

Now set down (-2π/3)∫√[1 + 9u²]du which translates to

(-2π/3)∫√[1 + 9(1/9)tan²θ]{3^-1sec²θ)}dθ or (-2π/3)∫√[1 + tan²θ]{3^-1sec²θ)}dθ.

(-2π/3)∫√[1 + tan²θ]{3^-1sec²θ)}dθ can be developed to

(-2π/9)∫sec³θdθ which integrates to:

(-π/9)[sec θtan θ + ln |sec θ + tan θ|_{(from x=0 to x=∞}] where θ = arctan (3e^-3x).

Find the limit of this last expression as x goes to positive infinity. (Approximate ∞ by 1E9 or one billion.)

A Casio Programmable Calculator computes this last expression to 0 − -3.946286982

or 3.946286982 square units.