How do you find the surface area of the solid obtained by rotating about the $x$ -axis the region bounded by $y=e^x$ on the interval $0<=x<=1$ ?

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Answer 1 · 2014-09-16T06:09:54

The answer is $pi/2[e^2-1]$ .

Since you are only given a single function and we are rotating about the axis of the parameter, this requires the disk method. The disk method is:

$V=int_a^b Adx$
$=int_a^b pi r^2dx$
$=int_a^b pi [f(x)]^2dx$

We have the known values:

$f(x)=e^x$
$a=0$
$b=1$

And now we can substitute:

$V=int_0^1 pi (e^x)^2dx$
$=pi int_0^1 e^(2x)dx$
$=pi (e^(2x))/2|_0^1$
$=pi/2[e^2-e^0]$
$=pi/2[e^2-1]$

How do you find the surface area of the solid obtained by rotating about the xx-axis the region bounded by y=e^xy=e^x on the interval 0<=x<=10<=x<=1 ?