Surface area of revolution $y=\sin{x}$ for $x\in[0,\pi]$ , I always get half of the correct answer

Question

So we know the formula for the surface area of revolution: $$\left|S\right|=2\pi\displaystyle\int\limits_a^b\left|f(x)\right|\sqrt{1+\left(f'(x)\right)^2} \space dx$$ For this question: $$\left|S\right|=2\pi\displaystyle\int\limits_0^{\pi}\left(\sin{x}\right)\sqrt{1+\cos^2{x}} \space dx$$ So I decided to use substitution method for $\underline{\cos{x}=a}$: $$-2\pi\displaystyle\int\sqrt{1+a^2}\space da$$ After this part, it gets more complicated. Now I know about hyperbolic trigonometric functions, however, I must solve this without using them. Hence I searched and found this and this videos where the integration is solved the supposed way, the answer being $\dfrac{1}{2}\sqrt{1+x^2}\cdot x+\dfrac{1}{2}\ln{\left|\sqrt{1+x^2}+x\right|}+C$, which for me, is: $$-\dfrac{\pi}{2}\left[\sqrt{1+a^2}\cdot a+\ln{\left|\sqrt{1+a^2}+a\right|}\right]+C=-\dfrac{\pi}{2}\left[\sqrt{1+\cos^2{x}}\cdot\cos{x}+\ln{\left|\sqrt{1+\cos^2{x}}+\cos{x}\right|}\right]\Bigg|_0^{\pi}$$ When we calculate, we get: $-\dfrac{\pi}{2}\left[-2\sqrt{2}+\ln{\left(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\right)}\right]\approx7.21$

The problem is, when I calculate this problem on WolframAlpha, it gives me about $14.42$, which is twice my answer. I am unsure where did I make the mistake.

Answer 1

It seems like you are multiplying $2\pi$ by $\frac{1}{2}$ and getting $\frac{\pi}{2}$ instead of just $\pi.$

Answer 2

Based on your working, you are rotating the curve around x-axis. You should explicitly mention that. Coming to the working,

$S = 2\pi\displaystyle\int\limits_{0}^{\pi} \sin x \sqrt{1+\cos^2x} \space dx ​= 2\pi\int\limits_{-1}^{1} \sqrt{1+t^2} \space dt$

$ = 2 \pi \left[\left(\dfrac{1}{\sqrt2} + \dfrac{1}{2}\ln (\sqrt2 + 1)\right) - \left(- \dfrac{1}{\sqrt2} + \dfrac{1}{2}\ln (\sqrt2 - 1)\right)\right]$

$\approx 14.42$

Or you can write it as,

$S = 2\pi\displaystyle\int\limits_{0}^{\pi} \sin x \sqrt{1+\cos^2x} \space dx ​= 4\pi\displaystyle\int\limits_{0}^{\pi/2} \sin x \sqrt{1+\cos^2x} \space dx$

$ = 4\pi\displaystyle\int\limits_{0}^{1} \sqrt{1+t^2} \space dt \approx 14.42$