Let's solve the **first problem (1)** step by step: --- ## 1. Let \( \mathbf{F}(x, y, z) = (y e^x + e^z)\mathbf{i} + e^x \mathbf{j} + x e^z \mathbf{k} \) be a vector field. ### (a) **Find** \( \mathbf{curl}(\mathbf{F}) \) **and** \( \mathbf{div}(\mathbf{F}) \). #### **Step 1: Components of \( \mathbf{F} \)** \[ \mathbf{F}(x, y, z) = \left( F_1, F_2, F_3 \right) = \left( y e^x + e^z, \ e^x, \ x e^z \right) \] #### **Step 2: Divergence** \[ \text{div}(\mathbf{F}) = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \] \[ \frac{\partial F_1}{\partial x} = y e^x,\quad \frac{\partial F_2}{\partial y} = 0,\quad \frac{\partial F_3}{\partial z} = x e^z \] So, \[ \boxed{\text{div}(\mathbf{F}) = y e^x + x e^z} \] #### **Step 3: Curl** \[ \text{curl}(\mathbf{F}) = \nabla \times \mathbf{F} \] \[ = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y e^x + e^z & e^x & x e^z \end{vmatrix} \] Compute each component: - \( \mathbf{i} \) component: \[ \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} = 0 - 0 = 0 \] - \( \mathbf{j} \) component: \[ \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x} = e^z - e^z = 0 \] - \( \mathbf{k} \) component: \[ \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} = e^x - e^x = 0 \] So, \[ \boxed{\text{curl}(\mathbf{F}) = \mathbf{0}} \] --- ### (b) **Is \( \mathbf{F} \) a conservative vector field? If yes, find the potential function.** - Since \( \text{curl}(\mathbf{F}) = 0 \) **everywhere** and the domain is all of \( \mathbb{R}^3 \) (which is simply connected), **\( \mathbf{F} \) is conservative**. #### **Find the potential function \( f(x, y, z) \) such that \( \nabla f = \mathbf{F} \):** \[ \frac{\partial f}{\partial x} = y e^x + e^z \] \[ \frac{\partial f}{\partial y} = e^x \] \[ \frac{\partial f}{\partial z} = x e^z \] **Step 1:** Integrate \( \frac{\partial f}{\partial x} \) with respect to \( x \): \[ f(x, y, z) = \int (y e^x + e^z) dx = y e^x + x e^z + C(y, z) \] **Step 2:** Take \( \frac{\partial}{\partial y} \): \[ \frac{\partial f}{\partial y} = e^x + \frac{\partial C}{\partial y} \] Set equal to \( e^x \implies \frac{\partial C}{\partial y} = 0 \implies C = C(z) \). **Step 3:** Take \( \frac{\partial}{\partial z} \): \[ \frac{\partial f}{\partial z} = x e^z + C'(z) \] Set equal to \( x e^z \implies C'(z) = 0 \implies C(z) = \text{constant} \). **Final Potential Function:** \[ \boxed{f(x, y, z) = y e^x + x e^z + \text{constant}} \] --- ## **Final Answers for (1):** ### (a) - \( \boxed{\text{curl}(\mathbf{F}) = \mathbf{0}} \) - \( \boxed{\text{div}(\mathbf{F}) = y e^x + x e^z} \) ### (b) - \( \mathbf{F} \) **is a conservative vector field.** - **Potential function:** \( \boxed{f(x, y, z) = y e^x + x e^z + C} \) --- If you want **solutions for the other questions**, let me know which one you want to proceed with!

Certainly! Let's proceed step-by-step with all the remaining problems. --- ## **Problem 2: Circulation around the boundary using Green's Theorem** Given: \[ \mathbf{F}(x, y) = (y + e^{x^2}) \mathbf{i} + (2z - e^{x}) \mathbf{j} \] **Note:** Based on the problem statement, the region is bounded above by \( y = 3 - 2x^2 \) and below by \( y = 2x^2 + 1 \). ### Step 1: Understand the boundary The boundary \( C \) is the closed curve formed by these two curves. ### Step 2: Green's Theorem Green's Theorem relates the line integral around the closed curve \( C \) to a double integral over the region \( R \): \[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx dy \] Where: \[ \mathbf{F} = P \mathbf{i} + Q \mathbf{j} \] Identify: \[ P = y + e^{x^2}, \quad Q = 2z - e^{x} \] But since the integral is in \( x, y \), and \( z \) appears as a constant (assuming z is fixed), the relevant part is: \[ P = y + e^{x^2} \] \[ Q = - e^{x} \] ### Step 3: Compute derivatives \[ \frac{\partial Q}{\partial x} = - e^{x} \] \[ \frac{\partial P}{\partial y} = 1 \] So, \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = - e^{x} - 1 \] ### Step 4: Set up the double integral over \( R \) \[ \iint_R (- e^{x} - 1) \, dx dy \] where \( R \) is bounded between \( y = 2x^2 + 1 \) (below) and \( y = 3 - 2x^2 \) (above). ### Step 5: Limits of integration - For \( x \), find where the curves intersect: \[ 2x^2 + 1 = 3 - 2x^2 \] \[ 4x^2 = 2 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}} \] - For each \( x \), \( y \) goes from \( y = 2x^2 + 1 \) to \( y = 3 - 2x^2 \). ### Step 6: Write the integral \[ \text{Work} = \int_{x = -\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \int_{y=2x^2 + 1}^{3 - 2x^2} (- e^{x} - 1) dy dx \] ### Step 7: Integrate with respect to \( y \) \[ \int_{2x^2 + 1}^{3 - 2x^2} (- e^{x} - 1) dy = (- e^{x} - 1) \left[ y \right]_{2x^2 + 1}^{3 - 2x^2} \] \[ = (- e^{x} - 1) \left( (3 - 2x^2) - (2x^2 + 1) \right) = (- e^{x} - 1) (2 - 4x^2) \] ### Step 8: Final integral \[ \boxed{ \text{Work} = \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} (- e^{x} - 1) (2 - 4x^2) dx } \] This integral can be computed further by splitting into parts, but the main setup is complete. --- ## **Problem 3: Series convergence analysis** Let's analyze each series: ### (a) \(\sum_{n=0}^\infty \frac{2^{n+1}}{n!}\) - Recognize as related to the exponential series: \[ \sum_{n=0}^\infty \frac{a^n}{n!} = e^a \] - Here: \[ \sum_{n=0}^\infty \frac{2^{n+1}}{n!} = 2 \sum_{n=0}^\infty \frac{2^{n}}{n!} = 2 e^{2} \] - **Converges absolutely** (since exponential series converge for all real \( a \)). ### **Answer: Absolutely convergent.** --- ### (b) \(\sum_{n=1}^\infty \frac{n^2 + 1}{n^3}\) - Simplify: \[ \sum_{n=1}^\infty \left( \frac{n^2}{n^3} + \frac{1}{n^3} \right) = \sum_{n=1}^\infty \left( \frac{1}{n} + \frac{1}{n^3} \right) \] - The \( \frac{1}{n} \) term diverges (harmonic series). - Since the series contains a divergent part, **the series diverges**. ### **Answer: Divergent.** --- ### (c) \(\sum_{n=1}^\infty \frac{\tan^{-1}(n)}{n^2}\) - \( \tan^{-1}(n) \to \frac{\pi}{2} \) as \( n \to \infty \). - For large \( n \), the general term behaves like \( \frac{\pi/2}{n^2} \), which converges (since \( \sum 1/n^2 \) converges). - The series converges **absolutely** (since the terms are dominated by \( 1/n^2 \)). ### **Answer: Absolutely convergent.** --- ## **Problem 4: Power series representation of \( \int 2z \arctan(z^2) dz \)** ### Step 1: Recall the power series for \( \arctan(w) \): \[ \arctan(w) = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} w^{2n+1} \quad \text{for } |w| \le 1 \] ### Step 2: Substitute \( w = z^2 \): \[ \arctan(z^2) = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} z^{4n+2} \] ### Step 3: Multiply by \( 2z \): \[ 2z \arctan(z^2) = 2z \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} z^{4n+2} = \sum_{n=0}^\infty \frac{2 (-1)^n}{2n+1} z^{4n+3} \] ### **Answer:** \[ \boxed{ \int 2z \arctan(z^2) dz = \int \sum_{n=0}^\infty \frac{2 (-1)^n}{2n+1} z^{4n+3} dz } \] Integrate term-by-term: \[ = \sum_{n=0}^\infty \frac{2 (-1)^n}{2n+1} \cdot \frac{z^{4n+4}}{4n+4} + C \] Simplify: \[ = \sum_{n=0}^\infty \frac{2 (-1)^n}{(2n+1)(4n+4)} z^{4n+4} + C \] --- ## **Problem 5: Interval and radius of convergence of \( \sum_{n=1}^\infty n (z + 3)^n \)** This is a power series centered at \( z = -3 \). ### Step 1: Recognize the series \[ \sum_{n=1}^\infty a_n (z + 3)^n \quad \text{with} \quad a_n = n \] ### Step 2: Use the root or ratio test for convergence - The ratio test: \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{n+1}{n} = 1 \] - For the series \( \sum n r^n \), the radius of convergence \( R \) is 1. ### Step 3: Conclusion - The power series converges for \( |z + 3| < 1 \). - The interval of convergence is \( (-4, -2) \). - To check convergence at endpoints: - At \( z = -4 \): \( (z+3) = -1 \). Series becomes \( \sum n (-1)^n \), which diverges. - At \( z = -2 \): \( (z+3) = 1 \). Series becomes \( \sum n \), which diverges. ### **Final answer:** - **Radius of convergence:** \( R = 1 \) - **Interval of convergence:** \( (-4, -2) \) --- # **Summary of solutions:** | Problem | Final Result | |---|---| | 1 | \( \text{curl}(\mathbf{F})=0 \), \( \text{div}(\mathbf{F})= y e^x + x e^z \), potential function \( f = y e^x + x e^z \) | | 2 | \( \displaystyle \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} (- e^{x} - 1)(2 - 4x^2) dx \) (numeric evaluation possible) | | 3a | Absolutely convergent | | 3b | Divergent | | 3c | Absolutely convergent | | 4 | \( \int \sum_{n=0}^\infty \frac{2 (-1)^n}{(2n+1)(4n+4)} z^{4n+4} + C \) | | 5 | Radius \( R = 1 \), Interval \( (-4, -2) \) | --- Let me know if you'd like detailed steps for any specific part or further assistance!

Certainly! Here are the final answers for all the problems: --- ### **Problem 1: Vector Field \( \mathbf{F} \)** - **Curl:** \(\boxed{\mathbf{0}}\) - **Divergence:** \(\boxed{y e^x + x e^z}\) - **Potential function:** \(\boxed{f(x,y,z) = y e^x + x e^z + C}\) --- ### **Problem 2: Circulation (using Green's Theorem)** - **Integral expression:** \[ \boxed{ \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} (- e^{x} - 1)(2 - 4x^2) \, dx } \] (Computable numerically or further simplified as needed) --- ### **Problem 3: Series Convergence** - **(a)** Absolutely convergent - **(b)** Divergent - **(c)** Absolutely convergent --- ### **Problem 4: Power Series for \(\int 2z \arctan(z^2) dz\)** \[ \boxed{ \int 2z \arctan(z^2) dz = \sum_{n=0}^\infty \frac{2 (-1)^n}{(2n+1)(4n+4)} z^{4n+4} + C } \] --- ### **Problem 5: Interval and Radius of Convergence** - **Radius of convergence:** \(\boxed{1}\) - **Interval of convergence:** \(\boxed{(-4, -2)}\) --- Let me know if you need further details or explanations!