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An at first easy looking question has been giving me problems.

Given is the function $f(x)=e^x$ on the interval $[0,1]$, asked are the areas of its surfaces of revolution about the $x$-axis and $y$-axis.


We have $A=2\pi\int\limits_0^1f(x)\sqrt{1+(f'(x))^2}dx=2\pi\int\limits_0^1e^x\sqrt{1+e^{2x}}dx=2\pi\int\limits_1^e\sqrt{1+u^2}du$

Here we can substitute $u=\textrm{tan}(\theta)$ and we get $A=2\pi\int\limits_{\textrm{atan}(1)}^{\textrm{atan}(e)}\frac{1}{\textrm{cos}^3(\theta)}d\theta$. I don't really know how to approach this integral and would appreciate help. Wolfram Alpha suggests using a reduction formula but we haven't learnt this formula and I don't think we're meant to use it.


For this one we have $A=2\pi\int\limits_0^1x\sqrt{1+(f'(x))^2}dx=2\pi\int\limits_0^1x\sqrt{1+e^{2x}}dx$. I have tried many things and not gotten anywhere. Furthermore Wolfram Alpha suggests a complicated function as the indefinite integral of the integrand which leads me to think I have done something else wrong as well.

Any help is appreciated

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x-axis: use $u=\sinh(\theta)$ instead. y-axis: use $u=e^x$ and then $u=\sinh(\theta)$, as before. – Dominik Dec 4 '12 at 22:00
up vote 3 down vote accepted

$$\int \frac {1}{\cos^3x}dx=\int \frac {\cos x}{\cos^4x}dx=\int \frac {\cos x}{(1-\sin^2x)^2}dx$$

Now make the substitution $u=\sin x$ to get: $\displaystyle\int \frac{1}{(1-u^2)^2}du$. Finally, use partial fractions to find the last integral.

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You can use \sin and \cos (and similarly with the other trig functions) to get $\sin$ and $\cos$. – Brian M. Scott Dec 4 '12 at 22:04
Thank you, this works. Could you also help with the second part of the problem? – user50407 Dec 4 '12 at 22:21
I think the second integral is non elementary – Amr Dec 4 '12 at 23:06
O my bad, I just saw the exercise said you could leave the integral without calculating it. Thanks for helping. – user50407 Dec 4 '12 at 23:42

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