1. **Problem Statement:** We want to find the surface area generated by revolving the curve $y = \cos x$ from $x=0$ to $x=\frac{\pi}{2}$ about the x-axis. 2. **Formula for Surface Area of Revolution:** When a curve $y=f(x)$ is revolved about the x-axis, the surface area $S$ is given by: $$S = 2\pi \int_a^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$$ where $a=0$ and $b=\frac{\pi}{2}$. 3. **Calculate the derivative:** Given $y = \cos x$, we have: $$\frac{dy}{dx} = -\sin x$$ 4. **Substitute into the formula:** $$S = 2\pi \int_0^{\frac{\pi}{2}} \cos x \sqrt{1 + (-\sin x)^2} \, dx = 2\pi \int_0^{\frac{\pi}{2}} \cos x \sqrt{1 + \sin^2 x} \, dx$$ 5. **Simplify the integrand:** The integrand is $\cos x \sqrt{1 + \sin^2 x}$. 6. **Use substitution:** Let $u = \sin x$, then $du = \cos x \, dx$. 7. **Change limits:** When $x=0$, $u=\sin 0=0$; when $x=\frac{\pi}{2}$, $u=\sin \frac{\pi}{2}=1$. 8. **Rewrite the integral:** $$S = 2\pi \int_0^1 \sqrt{1 + u^2} \, du$$ 9. **Integrate:** The integral of $\sqrt{1+u^2}$ is: $$\int \sqrt{1+u^2} \, du = \frac{u}{2} \sqrt{1+u^2} + \frac{1}{2} \ln \left| u + \sqrt{1+u^2} \right| + C$$ 10. **Evaluate definite integral:** $$\int_0^1 \sqrt{1+u^2} \, du = \left[ \frac{u}{2} \sqrt{1+u^2} + \frac{1}{2} \ln \left( u + \sqrt{1+u^2} \right) \right]_0^1$$ Calculate at $u=1$: $$\frac{1}{2} \sqrt{2} + \frac{1}{2} \ln (1 + \sqrt{2})$$ At $u=0$: $$0 + \frac{1}{2} \ln(0 + 1) = 0$$ 11. **Final surface area:** $$S = 2\pi \left( \frac{\sqrt{2}}{2} + \frac{1}{2} \ln (1 + \sqrt{2}) \right) = \pi \sqrt{2} + \pi \ln (1 + \sqrt{2})$$ **Answer:** $$\boxed{S = \pi \sqrt{2} + \pi \ln (1 + \sqrt{2})}$$ This is the surface area of the solid formed by revolving $y=\cos x$ from $0$ to $\frac{\pi}{2}$ about the x-axis.