Let's break down the problem **step by step**: --- ## **Step 1: Understanding the Problem** - **Vector field:** \(\mathbf{F} = 4y\mathbf{i} + 4y\mathbf{j} - 4xz\mathbf{k}\) - **Surface \(S\):** \(y = x^2 + z^2\), where \(x^2 + z^2 \leq 9\) - **Orientation:** In the positive \(y\)-direction. We are to compute the **flux** of \(\mathbf{F}\) through \(S\). --- ## **Step 2: Surface Parameterization** Let’s parameterize the surface: - \(x = x\) - \(z = z\) - \(y = x^2 + z^2\) Let \((x, z)\) vary over the disk \(x^2 + z^2 \leq 9\). --- ## **Step 3: Surface Element and Normal Vector** The surface is given by \(y = f(x, z)\). For a surface \(y = f(x, z)\), the **upward (positive \(y\)) normal vector** is: \[ \vec{n} = \frac{(-f_x, 1, -f_z)}{\sqrt{f_x^2 + 1 + f_z^2}} \] But for flux, we use the unnormalized vector for the surface element: \[ d\vec{S} = (-f_x, 1, -f_z) \, dx\,dz \] - Here, \(f(x, z) = x^2 + z^2\), so \(f_x = 2x, f_z = 2z\). So, \[ d\vec{S} = (-2x, 1, -2z)\,dx\,dz \] --- ## **Step 4: Flux Integral** The flux through the surface is: \[ \iint_S \mathbf{F} \cdot d\vec{S} \] Substitute \(\mathbf{F}(x, y, z)\) where \(y = x^2 + z^2\): \[ \mathbf{F} = 4y\,\mathbf{i} + 4y\,\mathbf{j} - 4xz\,\mathbf{k} = (4(x^2 + z^2), 4(x^2 + z^2), -4xz) \] Dot product with \(d\vec{S}\): \[ \mathbf{F} \cdot d\vec{S} = \begin{align*} & [4(x^2 + z^2), 4(x^2 + z^2), -4xz] \cdot [-2x, 1, -2z]\,dx\,dz \\ = & [4(x^2 + z^2)(-2x) + 4(x^2 + z^2)(1) + (-4xz)(-2z)]\,dx\,dz \\ = & [-8x(x^2 + z^2) + 4(x^2 + z^2) + 8xz^2]\,dx\,dz \end{align*} \] Expand \( -8x(x^2 + z^2) = -8x^3 - 8xz^2 \): \[ = [-8x^3 - 8xz^2 + 4x^2 + 4z^2 + 8xz^2]\,dx\,dz \] \[ = [-8x^3 + 4x^2 + 4z^2]\,dx\,dz \] --- ## **Step 5: Set Up the Double Integral** Integrate over the disk \(x^2 + z^2 \leq 9\): \[ \iint_{x^2 + z^2 \leq 9} (-8x^3 + 4x^2 + 4z^2)\,dx\,dz \] Switch to **polar coordinates**: - \(x = r\cos\theta\) - \(z = r\sin\theta\) - \(r^2 = x^2 + z^2\) - \(dx\,dz = r\,dr\,d\theta\) So the integrand becomes: - \(x^3 = r^3\cos^3\theta\) - \(x^2 = r^2\cos^2\theta\) - \(z^2 = r^2\sin^2\theta\) \[ -8x^3 + 4x^2 + 4z^2 = -8r^3\cos^3\theta + 4r^2\cos^2\theta + 4r^2\sin^2\theta \] \[ = -8r^3\cos^3\theta + 4r^2(\cos^2\theta + \sin^2\theta) \] \[ = -8r^3\cos^3\theta + 4r^2 \] --- ## **Step 6: Evaluate the Double Integral** Limits: - \(r: 0 \to 3\) - \(\theta: 0 \to 2\pi\) So, \[ \int_0^{2\pi} \int_0^3 \left[-8r^3\cos^3\theta + 4r^2\right] r\,dr\,d\theta \] \[ = \int_0^{2\pi} \int_0^3 \left[-8r^4\cos^3\theta + 4r^3\right] dr\,d\theta \] Let’s **split the integral**: ### **First Term:** \[ I_1 = \int_0^{2\pi} \int_0^3 -8r^4\cos^3\theta\, dr\, d\theta = -8 \int_0^{2\pi} \cos^3\theta\, d\theta \int_0^3 r^4 dr \] - \(\int_0^3 r^4 dr = \frac{r^5}{5} \Big|_0^3 = \frac{243}{5}\) - \(\int_0^{2\pi} \cos^3\theta\, d\theta = 0\) (since \(\cos^3\theta\) is an odd function over \(0\) to \(2\pi\)) So, \(I_1 = 0\). --- ### **Second Term:** \[ I_2 = \int_0^{2\pi} \int_0^3 4r^3\, dr\, d\theta = 4\int_0^{2\pi} d\theta \int_0^3 r^3 dr \] - \(\int_0^3 r^3 dr = \frac{r^4}{4}\Big|_0^3 = \frac{81}{1} = 81\) - \(\int_0^{2\pi} d\theta = 2\pi\) So, \[ I_2 = 4 \times 2\pi \times 81 = 8\pi \times 81 = 648\pi \] --- ## **Step 7: Final Answer** \[ \boxed{648\pi} \] --- ### **Final Answer:** \[ \text{flux} = \boxed{648\pi} \]

The flux through the surface is computed by converting the surface integral into a double integral over the projection in the \(x\)-\(z\) plane, using the surface parameterization and the surface element \(d\vec{S}\). Symmetry considerations show that integrals involving odd functions over symmetric bounds vanish, simplifying the calculation significantly.

The flux of the vector field \(\mathbf{F} = 4y\,\mathbf{i} + 4y\,\mathbf{j} - 4xz\,\mathbf{k}\) through the surface \(y = x^2 + z^2\) (with \(x^2 + z^2 \leq 9\)) oriented in the positive \(y\)-direction is **\(\boxed{648\pi}\)**.