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Find the area of the surface $S(u,v)=(2u\sin v, 2u\cos v, e^u+e^{-u})$, $0

I'm having problems to solve this exercise.

I have that the surfaces area is given by $\displaystyle{\int |r_u\times r_v|dA}$ where $r_u$ and $r_u$ are vectors whose componentes are the partial derivatives of each component of $S$ with respect to $u,v$.

$r_u=(2\sin v, 2u\cos v, e^u-e^{-u})$. $r_v=(2u\cos v, -2u\sin v, 0)$.

Then

$r_u \times r_v = (2u(e^u-e^{-u})\sin v,2u(e^u-e^{-u})\cos v, -4u\sin^2 v -4u\cos^2 v)$

The computation of $|r_u\times r_v|$ yields $\sqrt{4u^2(e^u-e^{-u})^2\sin^2 v+ 4u^2(e^u-e^{-u})^2\cos^2 v+16u^2\sin^4 v+32u^2\sin^2 v\cos^2 v+16u^2\cos^4 v}$

which I simplified to $2u\sqrt{(e^u-e^{-u})^2+4}$.

But now I can't calculate the surface area:

$$\int_0^{\pi/2}\int_0^1 2u\sqrt{(e^u-e^{-u})^2+4} du dv = \pi \int_0^1 2u\sqrt{(e^u-e^{-u})^2+4}$$

I don't know how to solve the last integral. I tried the change $u=\log x$ but I couldn't find a solution to that integral either.

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    Do you know any hyperbolic trig? I would take a look at $\sinh$ for this. – 2017-02-02
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    Unless I made a mistake, your substitution of $u = \log x$ works. After simplification, you should be left with the integrand (omitting the scalar multiple): $\log x + \frac{\log x}{x^2}$ which you can integrate (separately) by parts. – 2017-02-02
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    @J.Loreaux I don't get quite the same thing, With $u=\log x$ it should be $du=1/x$ $dx$ then $u\sqrt{(e^{u}-e^{-u})^2+4}$ $du$ becomes $\log x\sqrt{(x-(-x)^2+4}(1/x)dx=\frac{\log x}{x}\sqrt{4x^2+4}dx=2\frac{\log x}{x}\sqrt{x^2+1}dx$. – 2017-02-02
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    Your problem lies in the simplification of $e^{-u}$. Note that: $$e^{-u} = e^{- \log x} = e^{\log (1/x)} = \frac{1}{x}.$$ – 2017-02-02
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    Oh! I see! Thank you very much @J.Loreaux – 2017-02-02

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