1. **Problem Statement:** Find the area of the surface generated by revolving the curve $y=\sqrt{9 - x^2}$ for $-1 \leq x \leq 1$ about the y-axis. 2. **Formula:** The surface area $S$ generated by revolving a curve $y=f(x)$ about the y-axis from $x=a$ to $x=b$ is given by: $$S = \int_a^b 2\pi x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$$ 3. **Find $\frac{dy}{dx}$:** Given $y = \sqrt{9 - x^2} = (9 - x^2)^{1/2}$, $$\frac{dy}{dx} = \frac{1}{2}(9 - x^2)^{-1/2} \cdot (-2x) = -\frac{x}{\sqrt{9 - x^2}}$$ 4. **Compute $1 + \left(\frac{dy}{dx}\right)^2$:** $$1 + \left(-\frac{x}{\sqrt{9 - x^2}}\right)^2 = 1 + \frac{x^2}{9 - x^2} = \frac{9 - x^2 + x^2}{9 - x^2} = \frac{9}{9 - x^2}$$ 5. **Substitute into the integral:** $$S = \int_{-1}^1 2\pi x \sqrt{\frac{9}{9 - x^2}} \, dx = \int_{-1}^1 2\pi x \frac{3}{\sqrt{9 - x^2}} \, dx = 6\pi \int_{-1}^1 \frac{x}{\sqrt{9 - x^2}} \, dx$$ 6. **Evaluate the integral:** Note the integrand $\frac{x}{\sqrt{9 - x^2}}$ is an odd function because $x$ is odd and $\sqrt{9 - x^2}$ is even. Since the limits are symmetric about zero, the integral evaluates to zero: $$\int_{-1}^1 \frac{x}{\sqrt{9 - x^2}} \, dx = 0$$ 7. **Conclusion:** The surface area generated by revolving the curve about the y-axis is: $$S = 6\pi \times 0 = 0$$ This means the surface area contribution from this segment about the y-axis is zero because the curve segment is symmetric and the integrand is odd. **Final answer:** $$\boxed{0}$$