Surface area of revolution $x=\ln(y)$ about $x$ axis

My question is regarding finding the surface area of revolution of an equation. The equation is $x=\ln(y)$, with bounds of $0 \le x \le 1$ about the $x$-axis. I know the integral for solving this would be:

$$2\pi \int_1^e y\sqrt{1+\frac{1}{y^2}}\,dy$$ But I've tried integrating by parts and u-substitution but I can't seem to get the answer of:

$$\pi\left[e\sqrt{1+e^2} + \ln\left(e + \sqrt{1+e^2}\right) - \sqrt{2} - \ln\left(\sqrt{2} +1\right)\right]$$

If anyone can help me with this integration I would appreciate it.

• Your answer doesn't have the same number of left and right parenthesis. – Randy Randerson Mar 4 '16 at 6:29
• Sorry about that, I fixed it – John Mar 4 '16 at 6:32
• No worries. Use the math formatting next time. – Randy Randerson Mar 4 '16 at 6:33
• I couldn't figure it out but I will for next time thanks – John Mar 4 '16 at 6:43

Rewrite the argument of the square root function. Then use trigonometric substitution followed by integration by parts.

In more detail:

$$\int_1^e y\sqrt{1 + \frac{1}{y^2}}\,dy = \int_1^e \sqrt{y^2 + 1}\,dy.$$

Using trigonometric substitution with $y = \tan \phi$, we obtain $$\int_1^e \sqrt{y^2 + 1}\,dy = \int_{\frac{\pi}{4}}^{\tan^{-1}(e)} \sec^3 \phi \,d\phi.$$ Integrating by parts with $u = \sec\phi$, $dv = \sec^2\phi\,d\phi$ yields $$\int_{\frac{\pi}{4}}^{\tan^{-1}(e)} \sec^3 \phi \,d\phi = e\sqrt{1 + e^2} - \sqrt{2} - \int_{\frac{\pi}{4}}^{\tan^{-1}(e)} \sec\phi \tan^2\phi\,d\phi.$$ Since $$\int_{\frac{\pi}{4}}^{\tan^{-1}(e)} \sec\phi \tan^2\phi = \int_{\frac{\pi}{4}}^{\tan^{-1}(e)} \sec^3 \phi\,d\phi - \int_{\frac{\pi}{4}}^{\tan^{-1}(e)} \sec\phi \,d\phi,$$ it follows that \begin{align} 2 \int_{\frac{\pi}{4}}^{\tan^{-1}(e)} \sec^3 \phi\,d\phi &= e\sqrt{1 + e^2} - \sqrt{2} + \int_{\frac{\pi}{4}}^{\tan^{-1}(e)} \sec\phi\,d\phi \\ &= e\sqrt{1 + e^2} - \sqrt{2} + \ln\left(e + \sqrt{1 + e^2}\right) - \ln\left(1 + \sqrt{2}\right). \end{align}

Hence the result on multiplying by $\pi$.

• Thank you so much, the only thing I don't follow is how you got the sec^3? I understand why you changed the bounds, but how did you get sec^3? When I tried to work out the problem I got sec^2, I'm not sure where I went wrong – John Mar 4 '16 at 8:47
• You're welcome. The integrand is $\sec\phi$ and $dy = \sec^2 \phi\,d\phi$. – Randy Randerson Mar 4 '16 at 8:49
• It helps me if you hit +1. – Randy Randerson Mar 4 '16 at 8:49
• I would but I don't have a rep of 15 yet so unfortunately I can only choose it as the correct answer – John Mar 4 '16 at 8:55
• No worries John – Randy Randerson Mar 4 '16 at 8:59

HINT

Would not straightforward $y = e^x$ be easier?

$$SA= 2 \pi \int_1^e y/\cos\phi \,dx ;\ \,\,\tan \phi = e^x$$

• It would, but I need to solve it this way – John Mar 4 '16 at 7:57
• OK, for a check in the working both give the same expressions, same result. – Narasimham Mar 4 '16 at 7:59
• I know, but my professor said this might be on the test and I can't figure it out so I thought I'd ask on here – John Mar 4 '16 at 8:00