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

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

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Answer 1 · 2014-09-16T06:09:54

The answer is $\frac{π}{2} [e^{2} - 1]$ $pi/2[e^2-1]$ .

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:

$V = \int_{a}^{b} A d x$ $V=int_a^b Adx$
$= \int_{a}^{b} π r^{2} d x$ $=int_a^b pi r^2dx$
$= \int_{a}^{b} π {[f (x)]}^{2} d x$ $=int_a^b pi [f(x)]^2dx$

We have the known values:

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

And now we can substitute:

$V = \int_{0}^{1} π {(e^{x})}^{2} d x$ $V=int_0^1 pi (e^x)^2dx$
$= π \int_{0}^{1} e^{2 x} d x$ $=pi int_0^1 e^(2x)dx$
$= π \frac{e^{2 x}}{2} {∣ ∣ ∣}_{0}^{1}$ $=pi (e^(2x))/2|_0^1$
$= \frac{π}{2} [e^{2} - e^{0}]$ $=pi/2[e^2-e^0]$
$= \frac{π}{2} [e^{2} - 1]$ $=pi/2[e^2-1]$

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

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