**Step 1: Express the Surface \( S \) Above the \(xy\)-Plane as \(z = f(x,y)\) and Find Its Domain** **Explanation:** The ellipsoid is given by \(4x^2 + 4y^2 + z^2 = 4\). Rearrange for \(z\): \[ z^2 = 4 - 4x^2 - 4y^2 \] \[ z = \sqrt{4 - 4x^2 - 4y^2} \] The portion above the \(xy\)-plane corresponds to \(z \geq 0\), so only the positive root is needed. For real values, require \(4 - 4x^2 - 4y^2 \geq 0\): \[ x^2 + y^2 \leq 1 \] This describes a disk of radius 1 centered at the origin in the \(xy\)-plane. --- **Step 2: Write the Double Integral for the Surface Area** **Explanation:** The surface area of a graph \(z = f(x, y)\) over domain \(D\) is: \[ A = \iint_D \sqrt{1 + (f_x)^2 + (f_y)^2} \, dx\,dy \] Here, \(f(x, y) = \sqrt{4 - 4x^2 - 4y^2}\). First, compute the partial derivatives: \[ f_x = \frac{1}{2}(4 - 4x^2 - 4y^2)^{-1/2} \cdot (-8x) = \frac{-4x}{\sqrt{4 - 4x^2 - 4y^2}} \] \[ f_y = \frac{-4y}{\sqrt{4 - 4x^2 - 4y^2}} \] Now plug into the formula: \[ 1 + (f_x)^2 + (f_y)^2 = 1 + \frac{16x^2}{4 - 4x^2 - 4y^2} + \frac{16y^2}{4 - 4x^2 - 4y^2} \] \[ = 1 + \frac{16(x^2 + y^2)}{4(1 - x^2 - y^2)} \] \[ = \frac{4(1 - x^2 - y^2) + 16(x^2 + y^2)}{4(1 - x^2 - y^2)} \] \[ = \frac{4 - 4x^2 - 4y^2 + 16x^2 + 16y^2}{4(1 - x^2 - y^2)} \] \[ = \frac{4 + 12x^2 + 12y^2}{4(1 - x^2 - y^2)} \] The surface area integral is: \[ A = \iint_{x^2 + y^2 \leq 1} \sqrt{\frac{4 + 12x^2 + 12y^2}{4(1 - x^2 - y^2)}} \, dx\,dy \] \[ = \iint_{x^2 + y^2 \leq 1} \frac{\sqrt{4 + 12x^2 + 12y^2}}{2\sqrt{1 - x^2 - y^2}} \, dx\,dy \] --- **Final Answer Summary:** The double integral giving the surface area of the part of the ellipsoid \(4x^2 + 4y^2 + z^2 = 4\) lying above the \(xy\)-plane is: \[ \boxed{ A = \iint_{x^2 + y^2 \leq 1} \frac{\sqrt{4 + 12x^2 + 12y^2}}{2\sqrt{1 - x^2 - y^2}} \; dx\,dy } \]

**Step 1: Express the surface \( S \) as \( z = f(x,y) \) and determine its domain** **Explanation:** The given ellipsoid is described by the equation: \[ 4x^2 + 4y^2 + z^2 = 4 \] To analyze the part of the surface above the \( xy \)-plane, solve for \( z \): \[ z^2 = 4 - 4x^2 - 4y^2 \] Since we are considering the surface above the \( xy \)-plane, take the positive root: \[ z = f(x,y) = \sqrt{4 - 4x^2 - 4y^2} \] For \( z \) to be real and non-negative, the expression under the square root must be non-negative: \[ 4 - 4x^2 - 4y^2 \geq 0 \] Divide through by 4: \[ 1 - x^2 - y^2 \geq 0 \] which simplifies to: \[ x^2 + y^2 \leq 1 \] This describes a disk in the \( xy \)-plane with radius 1 centered at the origin. Therefore, the domain \( D \) over which the surface exists above the \( xy \)-plane is: \[ D = \{ (x, y) \mid x^2 + y^2 \leq 1 \} \] --- **Step 2: Write the double integral for the surface area** **Explanation:** The surface area \( A \) of a graph \( z = f(x,y) \) over a domain \( D \) is given by: \[ A = \iint_D \sqrt{1 + (f_x)^2 + (f_y)^2} \, dx\,dy \] where \( f_x \) and \( f_y \) are the partial derivatives of \( f \) with respect to \( x \) and \( y \), respectively. Given: \[ f(x,y) = \sqrt{4 - 4x^2 - 4y^2} \] Calculate the derivatives: \[ f_x = \frac{1}{2}(4 - 4x^2 - 4y^2)^{-1/2} \times (-8x) = \frac{-4x}{\sqrt{4 - 4x^2 - 4y^2}} \] \[ f_y = \frac{1}{2}(4 - 4x^2 - 4y^2)^{-1/2} \times (-8y) = \frac{-4y}{\sqrt{4 - 4x^2 - 4y^2}} \] Now, compute: \[ 1 + (f_x)^2 + (f_y)^2 = 1 + \frac{16x^2}{4 - 4x^2 - 4y^2} + \frac{16y^2}{4 - 4x^2 - 4y^2} \] Combine the fractions: \[ = 1 + \frac{16(x^2 + y^2)}{4 - 4x^2 - 4y^2} \] Express 1 as a fraction with denominator \( 4 - 4x^2 - 4y^2 \): \[ = \frac{4 - 4x^2 - 4y^2}{4 - 4x^2 - 4y^2} + \frac{16(x^2 + y^2)}{4 - 4x^2 - 4y^2} \] \[ = \frac{4 - 4x^2 - 4y^2 + 16x^2 + 16y^2}{4 - 4x^2 - 4y^2} \] Simplify numerator: \[ 4 + (16x^2 - 4x^2) + (16y^2 - 4y^2) = 4 + 12x^2 + 12y^2 \] Thus: \[ 1 + (f_x)^2 + (f_y)^2 = \frac{4 + 12x^2 + 12y^2}{4 - 4x^2 - 4y^2} \] The integrand becomes the square root of this expression: \[ \sqrt{1 + (f_x)^2 + (f_y)^2} = \sqrt{\frac{4 + 12x^2 + 12y^2}{4 - 4x^2 - 4y^2}} = \frac{\sqrt{4 + 12x^2 + 12y^2}}{\sqrt{4 - 4x^2 - 4y^2}} \] Factor numerator and denominator to simplify: \[ \frac{\sqrt{4 + 12x^2 + 12y^2}}{2 \sqrt{1 - x^2 - y^2}} \] **Final expression for the surface area integral:** \[ A = \iint_{x^2 + y^2 \leq 1} \frac{\sqrt{4 + 12x^2 + 12y^2}}{2 \sqrt{1 - x^2 - y^2}} \, dx\,dy \] This integral represents the surface area of the specified part of the ellipsoid above the \( xy \)-plane. --- **Final Answer Summary:** The surface area integral is: \[ \boxed{ A = \iint_{x^2 + y^2 \leq 1} \frac{\sqrt{4 + 12x^2 + 12y^2}}{2 \sqrt{1 - x^2 - y^2}} \; dx\,dy } \] This double integral captures the surface area of the relevant portion of the ellipsoid.