Let's find the area of the surface \( z = 19 - 2x - 2y \) above the square \( |x| \leq 4 \), \( |y| \leq 4 \). ### Step 1: Parametric Representation Let’s parametrize the surface: \[ \vec{r}(x, y) = (x, y, 19 - 2x - 2y) \] where \( -4 \leq x \leq 4 \), \( -4 \leq y \leq 4 \). ### Step 2: Surface Area Formula For a surface \( z = f(x, y) \) above a region \( R \): \[ \text{Area} = \iint_R \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2}\, dx\,dy \] ### Step 3: Compute the Partial Derivatives Given \( f(x, y) = 19 - 2x - 2y \): - \( f_x = \frac{\partial f}{\partial x} = -2 \) - \( f_y = \frac{\partial f}{\partial y} = -2 \) ### Step 4: Plug into the Formula \[ \sqrt{1 + f_x^2 + f_y^2} = \sqrt{1 + (-2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \] ### Step 5: Set up the Double Integral The region is \( -4 \leq x \leq 4 \), \( -4 \leq y \leq 4 \): \[ \text{Area} = \iint_R 3\, dx\,dy \] \[ = 3 \iint_R dx\,dy \] \[ = 3 \times \text{area of the square} \] The area of the square is \( (4 - (-4)) \times (4 - (-4)) = 8 \times 8 = 64 \). ### Step 6: Calculate the Final Area \[ \text{Area} = 3 \times 64 = 192 \] --- ## Conclusion **The area of the surface \( z = 19 - 2x - 2y \) above the square \( |x| \leq 4, |y| \leq 4 \) is \(\boxed{192}\) square units.** This is found by calculating the surface area integral with the parametric description and recognizing that the integrand is constant due to the plane's linearity.