What is the surface area of the solid created by revolving $f(x) = xe^(2x) , x in [2,3]$ around the x axis?

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Answer 1 · 2017-01-01T09:07:55

Volume : $(61e^12--23e^8)/32pi$ = 30974.5 cubic units, nearly. The graph is not to scale. I would try to get surface area, soon.

graph{x^2e^(2x) [-5, 5, -10000, 10000]}

The volume = $pi int f^2 dx$ , with x from 2 to 3

$=pi int x^2e^(4x )dx$ , with x from 2 to 3

Now,

$int x^2e^(4x) dx$

$= 1/4int x^2d(e^(4x))$

#=1/4(x^2e^(4x)--int e^(4x)d(x^2))

$=1/4x^2e^4x-1/8 intxd(e^(4x))$

$=1/4x^2e^4x-1/8(xe^(4x)-inte^(4x) dx))$

$=1/4x^2e^4x-1/8xe^(4x)+1/32e^(4x) + C$

Substituting the limits for the difference, the value here is

$(61e^12--23e^8)/32$ .

So, the volume is

$(61e^12--23e^8)/32pi$ = 30974.5 cubic units, nearly.

What is the surface area of the solid created by revolving f(x) = xe^(2x) , x in [2,3]f(x) = xe^(2x) , x in [2,3] around the x axis?