1. **State the problem:** We need to compute the surface integral of the function $$G(x,y,z) = y \sqrt{\frac{2}{x^2} + 4}$$ over the surface defined by the parabolic cylinder $$x^2 + 4z = 4$$ bounded by the planes $$y=0$$, $$y=1$$, and $$z=0$$. 2. **Understand the surface:** The surface is part of the parabolic cylinder where $$z \geq 0$$, $$0 \leq y \leq 1$$, and $$x$$ varies such that $$x^2 + 4z = 4$$. Since $$z=\frac{4 - x^2}{4}$$, the domain for $$x$$ is $$-2 \leq x \leq 2$$. 3. **Parametrize the surface:** We can parametrize the surface using parameters $$x$$ and $$y$$: $$$\vec{r}(x,y) = \langle x, y, \frac{4 - x^2}{4} \rangle$$$ with $$x \in [-2,2]$$ and $$y \in [0,1]$$. 4. **Compute the surface element:** The surface element $$dS$$ is given by: $$$dS = \left| \vec{r}_x \times \vec{r}_y \right| dx dy$$$ where $$$\vec{r}_x = \frac{\partial \vec{r}}{\partial x} = \langle 1, 0, -\frac{x}{2} \rangle,$$$ $$$\vec{r}_y = \frac{\partial \vec{r}}{\partial y} = \langle 0, 1, 0 \rangle.$$$ 5. **Calculate the cross product:** $$$\vec{r}_x \times \vec{r}_y = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & -\frac{x}{2} \\ 0 & 1 & 0 \end{vmatrix} = \mathbf{i} \left(0 - 0\right) - \mathbf{j} \left(0 - 0\right) + \mathbf{k} \left(1 \cdot 1 - 0 \cdot 0\right) = \langle 0, \frac{x}{2}, 1 \rangle$$ Correction: The determinant calculation above is incorrect; let's compute carefully: $$$\vec{r}_x \times \vec{r}_y = \langle (0)(0) - (-\frac{x}{2})(1), -\left(1 \cdot 0 - (-\frac{x}{2}) \cdot 0\right), 1 \cdot 1 - 0 \cdot 0 \rangle = \langle \frac{x}{2}, 0, 1 \rangle.$$$ 6. **Magnitude of the cross product:** $$$\left| \vec{r}_x \times \vec{r}_y \right| = \sqrt{\left(\frac{x}{2}\right)^2 + 0^2 + 1^2} = \sqrt{\frac{x^2}{4} + 1} = \sqrt{1 + \frac{x^2}{4}}.$$$ 7. **Express the integrand on the surface:** Since $$G(x,y,z) = y \sqrt{\frac{2}{x^2} + 4}$$, and $$y$$ is a parameter, we substitute directly. 8. **Set up the surface integral:** $$$\iint_S G(x,y,z) \, dS = \int_0^1 \int_{-2}^2 y \sqrt{\frac{2}{x^2} + 4} \cdot \sqrt{1 + \frac{x^2}{4}} \, dx \, dy.$$$ 9. **Simplify the integrand:** $$$\sqrt{\frac{2}{x^2} + 4} = \sqrt{\frac{2 + 4x^2}{x^2}} = \frac{\sqrt{2 + 4x^2}}{|x|}.$$$ So the integrand becomes: $$$y \cdot \frac{\sqrt{2 + 4x^2}}{|x|} \cdot \sqrt{1 + \frac{x^2}{4}} = y \cdot \frac{\sqrt{2 + 4x^2}}{|x|} \cdot \sqrt{\frac{4 + x^2}{4}} = y \cdot \frac{\sqrt{2 + 4x^2} \cdot \sqrt{4 + x^2}}{|x| \cdot 2}.$$$ 10. **Rewrite the integral:** $$$\int_0^1 y \, dy \int_{-2}^2 \frac{\sqrt{(2 + 4x^2)(4 + x^2)}}{2 |x|} \, dx.$$$ 11. **Integrate with respect to y:** $$$\int_0^1 y \, dy = \frac{1}{2}.$$$ 12. **Focus on the x-integral:** $$$I = \int_{-2}^2 \frac{\sqrt{(2 + 4x^2)(4 + x^2)}}{2 |x|} \, dx.$$$ 13. **Use symmetry:** The integrand is even because $$|x|$$ and $$x^2$$ terms are even functions, so $$$I = 2 \int_0^2 \frac{\sqrt{(2 + 4x^2)(4 + x^2)}}{2 x} \, dx = \int_0^2 \frac{\sqrt{(2 + 4x^2)(4 + x^2)}}{x} \, dx.$$$ 14. **Simplify inside the square root:** $$$(2 + 4x^2)(4 + x^2) = 8 + 2x^2 + 16x^2 + 4x^4 = 8 + 18x^2 + 4x^4.$$$ 15. **Integral becomes:** $$$I = \int_0^2 \frac{\sqrt{4x^4 + 18x^2 + 8}}{x} \, dx.$$$ 16. **Substitution:** Let $$u = x^2$$, then $$du = 2x dx$$, so $$dx = \frac{du}{2x}$$. Rewrite the integral: $$$I = \int_0^{4} \frac{\sqrt{4u^2 + 18u + 8}}{\sqrt{u}} \cdot \frac{du}{2 \sqrt{u}} = \frac{1}{2} \int_0^4 \frac{\sqrt{4u^2 + 18u + 8}}{u} \, du.$$$ 17. **Final integral to evaluate:** $$$I = \frac{1}{2} \int_0^4 \frac{\sqrt{4u^2 + 18u + 8}}{u} \, du.$$$ This integral is nontrivial and may require advanced techniques or numerical methods. --- **Alternative approach:** 1. Use the surface integral formula for a surface given by $$z = f(x,y)$$: $$$\iint_S G(x,y,z) \, dS = \iint_D G(x,y,f(x,y)) \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dx dy,$$$ where $$D$$ is the projection of the surface onto the $$xy$$-plane. 2. Here, $$f(x,y) = \frac{4 - x^2}{4}$$, so $$$\frac{\partial f}{\partial x} = -\frac{x}{2}, \quad \frac{\partial f}{\partial y} = 0.$$$ 3. The integrand factor is $$$\sqrt{1 + \left(-\frac{x}{2}\right)^2 + 0} = \sqrt{1 + \frac{x^2}{4}}.$$$ 4. The domain $$D$$ is $$x \in [-2,2]$$, $$y \in [0,1]$$. 5. The integral becomes $$$\int_0^1 \int_{-2}^2 y \sqrt{\frac{2}{x^2} + 4} \cdot \sqrt{1 + \frac{x^2}{4}} \, dx dy,$$$ which matches the previous setup. **Summary:** The problem reduces to evaluating $$$\frac{1}{2} \int_0^4 \frac{\sqrt{4u^2 + 18u + 8}}{u} \, du,$$$ which can be computed numerically or with special functions. **Final answer:** $$$\iint_S G(x,y,z) \, dS = \frac{1}{2} \times I = \frac{1}{2} \times \frac{1}{2} \int_0^4 \frac{\sqrt{4u^2 + 18u + 8}}{u} \, du = \frac{1}{4} \int_0^4 \frac{\sqrt{4u^2 + 18u + 8}}{u} \, du.$$$ This integral can be approximated numerically for a precise value.