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I got a curve $y=a \cdot cosh \frac{x}{a}$ where $|x|\le b$. The task is to find area of the surface obtained by rotating curve around x-axis.

Here is my solution. Unfortunately the result is not identical with the result of the textbook. Would you please look at this and tell me where is my mistake? Maybe there is a mistake in my textbook. Thanks

My attempt at a solution:

Formula for area of the surface obtained by rotating curve: $$S=2\pi \int_{a}^{b}f(x)\sqrt{1+(f'(x))^2}\,dx$$

I calculated $(f'(x))$: $$f'(x)=\frac{d}{dx}(a\cdot \cosh\frac{x}{a})=\ldots =\sinh\frac{x}{a}$$ so: $(f'(x))^2=\sinh^2 \frac{x}{a}$

Then I put it into the formula for area of the surface obtained by rotating curve: $$S=2\pi \int_{a}^{b}a\cdot \cosh\frac{x}{a}\sqrt{1+\sinh^2 \frac{x}{a}}\,dx=2\pi a \int_{a}^{b} \cosh\frac{x}{a}\sqrt{1+\sinh^2 \frac{x}{a}}dx$$

I made 1st substitution $u=\frac{x}{a} $ and $du=\frac{1}{a}dx $ then I obtained: $$2\pi a^2 \int_{a}^{b} \cosh(u)\sqrt{1+\sinh^2 (u)}\,du $$ I made 2nd substitution $v=\sinh(u)$ and $dv=\cosh(u)du$ then I got: $$2\pi a^2\int_{\sinh a}^{\sinh b}\sqrt{1+v^2}dv$$

To solve previous integral I need to calculate this integral $\int_{}^{}\frac{1}{\cos^3\varphi }=\int_{}^{}\sec^3\varphi $: $$\int_{}^{}\sec^3\varphi =\frac{1}{2} \left( \sec\varphi \cdot \tan\varphi +\ln\right|sec\varphi +\tan\varphi \left|\right)$$

Then I returned back to $2\pi a^2\int_{\sinh a}^{\sinh b}\sqrt{1+v^2}\,dv$ and I made 3rd substitution: $v=\tan\varphi $ so $dv=\sec^2\varphi d\varphi$ then I got: $$2\pi a^2\int_{\sinh a}^{\sinh b}\sqrt{1+v^2}\,dv=2\pi a^2\int_{...}^{...}\sqrt{\sec^2\varphi }\sec^2\varphi d\varphi=2\pi a^2\int_{...}^{...}\sec^3 \varphi d\varphi$$

Now we can see that we just calculated the value of $\int_{...}^{...}\sec^3 \varphi d\varphi$ so: $$S=2\pi \int_{a}^{b}a\cdot \cosh\frac{x}{a}\sqrt{1+\sinh^2 \frac{x}{a}}\,dx=2\pi a^2\int_{\sinh a}^{\sinh b}\sqrt{1+v^2}\,dv=2\pi a^2\int_{...}^{...}\sec^3 \varphi d\varphi=2\pi a^2\frac{1}{2}\left[ \sec\varphi \cdot \tan\varphi +\ln\left|sec\varphi +\tan\varphi \right|\right]_{...}^{...}$$

Then I made a substitution back from $\varphi $ to $x$ , I obtained: $$S=2\pi a^2\frac{1}{2}\left[ \sec\varphi \cdot \tan\varphi +\ln\left|\sec\varphi +\tan\varphi \right|\right]_{...}^{...}=2\pi a^2\frac{1}{2}\left[ v\sqrt{1+v^2}+\ln\left|\sqrt{1+v^2}+v\right|\right]_{...}^{...}=$$ $$=\pi a^2 \left[ \sinh\frac{x}{a}\sqrt{1+\sinh^2\frac{x}{a}}+\ln \left| \sqrt{1+\sinh^2\frac{x}{a}}+\sinh\frac{x}{a} \right|\right]_{a}^{b}=$$ $$=\pi a^2 \left[ \sinh\frac{x}{a}\cosh\frac{x}{a}+\ln \left| \cosh\frac{x}{a}+\sinh\frac{x}{a} \right|\right]_{a}^{b}=$$ $$=\pi a^2 \left[ \frac{1}{2}\sinh\frac{2x}{a}+\ln \left| e^{\frac{x}{a}}\right|\right]_{a}^{b}=\pi a^2 \left( \frac{1}{2}\sinh\frac{2b}{a}+\ln \left| e^{\frac{b}{a}}\right| - \frac{1}{2}\sinh 2 -\ln \left| e \right| \right)$$

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up vote 0 down vote accepted

Hint: Use whe well-known hyperbolic identity

$$\cosh^2 x-\sinh^2x=1\iff 1+\sinh^2x=\cosh^2x$$

Use the above and you'll save about 80% of the work you did...:)

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Yes I got it. Thanks a lot for this hint. – user84086 Jun 27 '13 at 11:29

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