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