Let's solve the problem step by step: ### **Step 1: Write the Problem Clearly** Given: \[ \mathbf{F}(x, y, z) = \left(z \arctan(y^2),\ z^3 \ln(x^2 + 1),\ x^2 z\right) \] Surface \( S \): Part of \( x^2 + y^2 + z = 2 \) **above** \( z = 1 \), oriented **upward**. We are to find the **flux** of \( \mathbf{F} \) **across \( S \)**. --- ### **Step 2: Surface and Its Projection** The surface \( S \) is given by: \[ z = 2 - x^2 - y^2, \quad \text{with } z \geq 1 \] So, \[ 2 - x^2 - y^2 \geq 1 \implies x^2 + y^2 \leq 1 \] Thus, the projection \( D \) on the \( xy \)-plane is the disk \( x^2 + y^2 \leq 1 \). --- ### **Step 3: Find the Upward Unit Normal Vector** For a surface \( z = f(x, y) \), the upward normal is: \[ \mathbf{n} = \frac{(-f_x, -f_y, 1)}{\sqrt{f_x^2 + f_y^2 + 1}} \] But for flux, we use the surface integral: \[ \iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_D \mathbf{F}(x, y, f(x, y)) \cdot \mathbf{n} \, dA \] Or, equivalently, \[ \iint_D \left[ \mathbf{F}(x, y, f(x, y)) \cdot (-f_x, -f_y, 1) \right] dA \] For \( z = 2 - x^2 - y^2 \): \[ f_x = -2x,\quad f_y = -2y \] So, \( (-f_x, -f_y, 1) = (2x, 2y, 1) \). --- ### **Step 4: Substitute \( z \) and Compute the Dot Product** Evaluate \( \mathbf{F}(x, y, z) \) at \( z = 2 - x^2 - y^2 \): - \( z = 2 - x^2 - y^2 \) - \( F_1 = z \arctan(y^2) \) - \( F_2 = z^3 \ln(x^2 + 1) \) - \( F_3 = x^2 z \) So, \[ \mathbf{F}(x, y, z) = \left((2 - x^2 - y^2)\arctan(y^2),\ (2 - x^2 - y^2)^3 \ln(x^2 + 1),\ x^2 (2 - x^2 - y^2)\right) \] Dot product with \( (2x, 2y, 1) \): \[ \begin{align*} &\mathbf{F} \cdot (2x, 2y, 1) \\ =& 2x (2 - x^2 - y^2)\arctan(y^2) + 2y (2 - x^2 - y^2)^3 \ln(x^2 + 1) + x^2(2 - x^2 - y^2) \end{align*} \] --- ### **Step 5: Write the Flux Integral** \[ \iint_D \left[ 2x (2 - x^2 - y^2)\arctan(y^2) + 2y (2 - x^2 - y^2)^3 \ln(x^2 + 1) + x^2(2 - x^2 - y^2) \right] dA \] where \( D: x^2 + y^2 \leq 1 \). --- ### **Step 6: Change to Polar Coordinates** Let \( x = r\cos\theta, y = r\sin\theta \), \( dA = r dr d\theta \), \( r \in [0, 1], \theta \in [0, 2\pi] \). - \( x^2 + y^2 = r^2 \) - \( x = r\cos\theta \) - \( y = r\sin\theta \) So, the terms become: 1. \( 2x (2 - x^2 - y^2) \arctan(y^2) = 2r\cos\theta (2 - r^2)\arctan(r^2\sin^2\theta) \) 2. \( 2y (2 - x^2 - y^2)^3 \ln(x^2 + 1) = 2r\sin\theta (2 - r^2)^3 \ln(r^2\cos^2\theta + 1) \) 3. \( x^2(2 - x^2 - y^2) = r^2\cos^2\theta (2 - r^2) \) --- ### **Step 7: Integrate Each Term** #### **First Two Terms:** Notice that integrating any function that is **odd** in \( x \) or \( y \) over a disk centered at the origin will yield **zero**. - \( x \) and \( y \) are odd functions. - The disk \( D \) is symmetric about the origin. - The first and second terms are odd in \( x \) and \( y \), respectively. So, the integrals of the first two terms over the disk are **zero**. #### **Only Third Term Remains:** \[ \iint_D x^2(2 - x^2 - y^2)\ dA \] In polar coordinates: \[ \int_0^{2\pi} \int_0^1 r^2\cos^2\theta (2 - r^2) r dr d\theta \] --- ### **Step 8: Compute the Remaining Integral** \[ \int_0^{2\pi} \int_0^1 r^2\cos^2\theta (2 - r^2) r dr d\theta = \int_0^{2\pi} \cos^2\theta\ d\theta \int_0^1 r^3 (2 - r^2) dr \] Compute the \( r \) integral: \[ \int_0^1 r^3 (2 - r^2) dr = 2\int_0^1 r^3 dr - \int_0^1 r^5 dr = 2\left[\frac{r^4}{4}\right]_0^1 - \left[\frac{r^6}{6}\right]_0^1 = 2 \left(\frac{1}{4}\right) - \left(\frac{1}{6}\right) = \frac{1}{2} - \frac{1}{6} = \frac{1}{3} \] Now, the \( \theta \) integral: \[ \int_0^{2\pi} \cos^2\theta\ d\theta = \int_0^{2\pi} \frac{1+\cos 2\theta}{2} d\theta = \frac{1}{2}\int_0^{2\pi} 1 d\theta + \frac{1}{2} \int_0^{2\pi} \cos 2\theta d\theta = \frac{1}{2}(2\pi) + 0 = \pi \] --- ### **Step 9: Final Answer** Multiply the two results: \[ \text{Flux} = \pi \times \frac{1}{3} = \boxed{\frac{\pi}{3}} \] --- ## **Final Answer** \[ \boxed{\frac{\pi}{3}} \]

Let's carefully solve the problem step by step, including all detailed calculations. --- ### **Given Data:** - Vector field: \[ \mathbf{F}(x, y, z) = \left(z \arctan(y^2),\ z^3 \ln(x^2 + 1),\ x^2 z\right) \] - Surface \( S \): part of the paraboloid \[ z = 2 - x^2 - y^2 \] lying **above** the plane \( z = 1 \), with upward orientation. --- ### **Step 1: Describe the Surface and Its Projection** The surface \( z = 2 - x^2 - y^2 \) is a downward-opening paraboloid. - The portion above \( z=1 \): \[ 2 - x^2 - y^2 \geq 1 \Rightarrow x^2 + y^2 \leq 1 \] So, the projection \( D \) in the \( xy \)-plane is the disk: \[ x^2 + y^2 \leq 1 \] --- ### **Step 2: Surface Element and Normal Vector** For a surface \( z = f(x, y) = 2 - x^2 - y^2 \), - The upward normal vector is given by: \[ \mathbf{n} = \frac{(-f_x, -f_y, 1)}{\sqrt{f_x^2 + f_y^2 + 1}} \] - The differential surface element: \[ d\mathbf{S} = \mathbf{n} \, dA = \frac{(-f_x, -f_y, 1)}{\sqrt{f_x^2 + f_y^2 + 1}} \, dA \] However, for flux calculation, it's often easier to use the surface integral in the form: \[ \iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_D \mathbf{F}(x, y, f(x,y)) \cdot \mathbf{n} \, dA \] Or, directly: \[ \iint_D \mathbf{F}(x, y, f(x,y)) \cdot (-f_x, -f_y, 1) \, dA \] since the normal points upward. --- ### **Step 3: Compute derivatives \(f_x\) and \(f_y\)** \[ f_x = -2x, \quad f_y = -2y \] Thus, the upward normal vector components (without normalization) are: \[ (-f_x, -f_y, 1) = (2x, 2y, 1) \] --- ### **Step 4: Evaluate \( \mathbf{F} \) at \( z = f(x,y) \)** \[ z = 2 - x^2 - y^2 \] Compute each component: - \(F_1 = z \arctan(y^2) = (2 - x^2 - y^2) \arctan(y^2)\) - \(F_2 = z^3 \ln(x^2 + 1) = (2 - x^2 - y^2)^3 \ln(x^2 + 1)\) - \(F_3 = x^2 z = x^2 (2 - x^2 - y^2)\) --- ### **Step 5: Set up the flux integral** \[ \text{Flux} = \iint_D \left[ \mathbf{F} \cdot (2x, 2y, 1) \right] dA \] Explicitly: \[ \text{Flux} = \iint_D \left[ F_1 \cdot 2x + F_2 \cdot 2y + F_3 \cdot 1 \right] dA \] Substitute: \[ \begin{aligned} \text{Flux} &= \iint_D \left[ (2 - x^2 - y^2) \arctan(y^2) \cdot 2x \right. \\ &\quad + \left. (2 - x^2 - y^2)^3 \ln(x^2 + 1) \cdot 2y + x^2 (2 - x^2 - y^2) \right] dA \end{aligned} \] --- ### **Step 6: Symmetry considerations** - The terms involving \( x \arctan(y^2) \) and \( y \ln(x^2 + 1) \) are **odd** functions in \( x \) or \( y \) over a symmetric disk centered at the origin. - When integrating over the disk \( x^2 + y^2 \leq 1 \), integrals of odd functions across symmetric limits are zero. **Therefore:** \[ \iint_D 2x (2 - x^2 - y^2) \arctan(y^2) dA = 0 \] and \[ \iint_D 2y (2 - x^2 - y^2)^3 \ln(x^2 + 1) dA = 0 \] The remaining term is: \[ \iint_D x^2 (2 - x^2 - y^2) dA \] --- ### **Step 7: Convert remaining integral to polar coordinates** In polar coordinates: \[ x = r \cos \theta,\quad y = r \sin \theta,\quad dA = r dr d\theta \] Also: \[ x^2 = r^2 \cos^2 \theta,\quad x^2 + y^2 = r^2 \] The integral becomes: \[ \text{Flux} = \int_0^{2\pi} \int_0^1 r^2 \cos^2 \theta \cdot (2 - r^2) \cdot r dr d\theta \] which simplifies to: \[ \int_0^{2\pi} \cos^2 \theta \, d\theta \times \int_0^1 r^3 (2 - r^2) dr \] --- ### **Step 8: Compute angular integral** \[ \int_0^{2\pi} \cos^2 \theta d\theta = \pi \] (using the standard integral \( \int_0^{2\pi} \cos^2 \theta d\theta = \pi \)) --- ### **Step 9: Compute radial integral** \[ \int_0^1 r^3 (2 - r^2) dr = 2 \int_0^1 r^3 dr - \int_0^1 r^5 dr \] Calculate each: \[ \int_0^1 r^3 dr = \left[\frac{r^4}{4}\right]_0^1 = \frac{1}{4} \] \[ \int_0^1 r^5 dr = \left[\frac{r^6}{6}\right]_0^1 = \frac{1}{6} \] So: \[ 2 \times \frac{1}{4} - \frac{1}{6} = \frac{1}{2} - \frac{1}{6} = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \] --- ### **Step 10: Final result** Multiply angular and radial integrals: \[ \text{Flux} = \pi \times \frac{1}{3} = \boxed{\frac{\pi}{3}} \] --- ### **### **Final Answer:** \[ \boxed{\frac{\pi}{3}} \] This is the flux of the vector field \( \mathbf{F} \) across the specified surface \( S \).