Question

c4 - arc length and surface area: problem 6
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set up an integral that represents the area of the surface obtained by rotating the curve ( g(x)=2 x e^{-x} ) about the ( x ) - axis for ( 1 leq x leq 3 ) in the form
int_{a}^{b} f(x) d x.
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Explanation:

Step1: Recall the formula for surface - area

The formula for the surface area \(S\) of a curve \(y = g(x)\) rotated about the \(x\) - axis from \(x=a\) to \(x = b\) is \(S=\int_{a}^{b}2\pi y\sqrt{1+(y')^{2}}dx\).
Given \(y = g(x)=2xe^{-x}\), first find \(y'\).
Using the product rule \((uv)^\prime=u^\prime v + uv^\prime\) where \(u = 2x\) and \(v=e^{-x}\), \(u^\prime=2\) and \(v^\prime=-e^{-x}\).
So \(y^\prime=2e^{-x}-2xe^{-x}=2e^{-x}(1 - x)\).

Step2: Substitute \(y\) and \(y'\) into the surface - area formula

Substitute \(y = 2xe^{-x}\) and \(y^\prime=2e^{-x}(1 - x)\) into \(S=\int_{a}^{b}2\pi y\sqrt{1+(y')^{2}}dx\).
\(1+(y')^{2}=1 + 4e^{-2x}(1 - x)^{2}\).
Then \(f(x)=2\pi(2xe^{-x})\sqrt{1 + 4e^{-2x}(1 - x)^{2}}\).

Answer:

\(a = 1\), \(b = 3\), \(f(x)=4\pi xe^{-x}\sqrt{1+4e^{-2x}(1 - x)^{2}}\)