Calculate the surface area generated by rotating the arc

Calculate the surface area generated by rotating the arc of $y=\sin x$ about the $x$-axis between $x=0$ and $x=\pi$.

We rotate one full arch of the sine curve about the $x$-axis.

Step 1: Set up

\[y'=\cos x \implies S=2\pi\int_0^\pi\sin x\sqrt{1+\cos^2x}\,dx\]

Step 2: Substitute $u=\cos x$

$du=-\sin x\,dx$; the limits $x=0,\pi$ become $u=1,-1$:

\[S=2\pi\int_{-1}^{1}\sqrt{1+u^2}\,du\]

Step 3: Standard antiderivative

\[\int\sqrt{1+u^2}\,du = \frac{u\sqrt{1+u^2}}2+\frac12\ln\left(u+\sqrt{1+u^2}\right)+C\]

Step 4: Evaluate (using the integrand's evenness)

At $u=1$: $\frac{\sqrt2}2+\frac12\ln(1+\sqrt2)$. At $u=-1$: $-\frac{\sqrt2}2-\frac12\ln(1+\sqrt2)$ (since $\ln(\sqrt2-1)=-\ln(\sqrt2+1)$).

Step 5: Combine

\[S=2\pi\left[\left(\frac{\sqrt2}2+\frac12\ln(1+\sqrt2)\right)-\left(-\frac{\sqrt2}2-\frac12\ln(1+\sqrt2)\right)\right] = 2\pi\left[\sqrt2+\ln(1+\sqrt2)\right]\]

\[\boxed{S=2\pi\left(\sqrt2+\ln(1+\sqrt2)\right)}\]

Step 6: Numerical value

\[S\approx2\pi(1.414+0.881)\approx2\pi(2.295)\approx14.42\]