Step 1: Conceptualization: Determine the surface area of a planar segment. The segment's projection onto the xy-plane is delineated by \(x=0, y=0\) and \(x^2+y^2=16\), identifying it as the first-quadrant sector of the circle \(x^2+y^2=16\).
Step 2: Methodologies: The surface area (S) of a surface \(z = f(x, y)\) over a region R in the xy-plane is computed using \( S = \iint_R \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \, dA \). Alternatively, if \( \vec{n} \) is the normal vector to the plane and \( \hat{k} \) is the unit normal to the xy-plane, the surface area is \( S = \frac{A_{xy}}{|\vec{n} \cdot \hat{k}|} \), where \(A_{xy}\) is the projected area.
Step 3: Execution:
Method 1: Integral Calculation
From the plane equation \(x + 2y + 2z = 12\), derive \( z = 6 - \frac{1}{2}x - y \). The partial derivatives are \( \frac{\partial z}{\partial x} = -\frac{1}{2} \) and \( \frac{\partial z}{\partial y} = -1 \). The surface area integrand is \( \sqrt{1 + \left(-\frac{1}{2}\right)^2 + (-1)^2} = \sqrt{1 + \frac{1}{4} + 1} = \frac{3}{2} \). The surface area integral becomes \( S = \iint_R \frac{3}{2} \, dA = \frac{3}{2} \times (\text{Area of R}) \). The region R is a quarter circle of radius 4, so its area is \( \text{Area of R} = \frac{1}{4} (\pi (4)^2) = 4\pi \). Consequently, \( S = \frac{3}{2} \times (4\pi) = 6\pi \).
Method 2: Projection Formula
The normal vector to the plane \(x + 2y + 2z = 12\) is \( \vec{n} = \langle 1, 2, 2 \rangle \). The projected area \(A_{xy}\) is \(4\pi\). The dot product with the xy-plane unit normal \( \hat{k} = \langle 0, 0, 1 \rangle \) is \( |\vec{n} \cdot \hat{k}| = |\langle 1, 2, 2 \rangle \cdot \langle 0, 0, 1 \rangle| = 2 \). The magnitude of the normal vector is \( |\vec{n}| = \sqrt{1^2+2^2+2^2} = 3 \). Using the formula \( S = \frac{|\vec{n}|}{|\vec{n} \cdot \hat{k}|} A_{xy} \), we get \( S = \frac{3}{2} \times (4\pi) = 6\pi \).
Step 4: Conclusion: The calculated surface area is \(6\pi\).