Question

find the surface area obtained by rotating the curve y=e^x on [0,1] around the x-axis

Answer 1

the surface area formula is AREA = integral 2pie*f(x)*sqrt(1+[f '(x)]^2)dx

Answer 2

f'(x) = e^x
radius is e^x +1
\[\rightarrow 2\pi \int\limits_{0}^{1}(e^{x}+1)\sqrt{1+e^{2x}} dx\]

Answer 3

how do you integrate that?

Answer 4

sorry about that, also i was wrong about the radius, its just e^x
\[2\pi \int\limits_{0}^{1}e^{x}\sqrt{1+e^{2x}} dx\]
substitute
u = e^x
du = e^x dx
\[2\pi \int\limits\limits_{0}^{1}\sqrt{1+u^{2}} du\]
substitute
u = tan@
du = sec^2@ d@
\[2\pi \int\limits\limits_{0}^{1}\sec^{3} \theta d \theta\]
use integration by parts
u = sec dv = sec^2
du = sectan v = tan
\[\rightarrow \int\limits_{}^{}\sec^{3}\theta = \sec \theta \tan \theta-\int\limits_{}^{}\sec \theta \tan^{2}\theta\]
using identity tan^2 = 1-sec^2

\[\rightarrow \int\limits_{}^{}\sec^{3}\theta = \sec \theta \tan \theta+\int\limits_{}^{}\sec \theta-\int\limits_{}^{}\sec^{3} \theta \]
\[\rightarrow \int\limits_{}^{}\sec^{3}\theta =\frac{1}{2} \sec \theta \tan \theta +\frac{1}{2}\int\limits_{}^{}\sec \theta \]
\[\rightarrow \int\limits\limits_{}^{}\sec^{3}\theta =\frac{1}{2} \sec \theta \tan \theta +\frac{1}{2}\ln (\sec \theta+\tan \theta)\]
Now notice
tan@ = u = e^x
sec@ = sqrt(1+tan^2) = sqrt(1+e^2x)
Substitute back and evaluate
\[2\pi \int\limits_{0}^{1}e^{x}\sqrt{1+e^{2x}} = \pi[e^{x}\sqrt{1+e^{2x}}+\ln (e^{x}+\sqrt{1+e^{2x}})] from 0->1\]
\[=\pi[e\sqrt{1+e^{2}}+\ln (e+\sqrt{1+e^{2}})] - \pi[\sqrt{2}+\ln (1+\sqrt{2})]\]
\[\approx 22.943\]

Answer 5

http://www.wolframalpha.com/input/?i=integrte+2*pi*e^x*sqrt%281%2Be^%282x%29%29+dx+from+0+to+1