What is the surface area produced by rotating $f (x) = \frac{1}{x + 1}, x \in [0, 3]$ $f(x)=1/(x+1), x in [0,3]$ around the x-axis?

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Answer 1 · 2016-06-22T08:11:45

$\frac{3 π}{4}$ $(3 pi)/4$

Desmos

Revolvong a small elemental area about the x axis creates volume of revolution $Delta V = pi y^2 Delta x$

The volume therefore is

$V = pi int_0^3 y^2 \ dx = pi int_0^3 \ 1/(x+1)^2 \ dx$
$= pi [- 1/(x+1)]_0^3 = pi [ 1/(x+1)]_3^0$
$= pi (1 - 1/4) = (3 pi)/4$

What is the surface area produced by rotating f(x)=1/(x+1), x in [0,3]f(x)=1x+1,x∈[0,3]f(x)=1/(x+1), x in [0,3] around the x-axis?