Surface Integral Evaluation

Investigating the task

The user needs help solving a problem with a specific format: identifying the subject, explaining concepts, using formulas, and providing a step-by-step solution with a final answer in a sentence.

Breaking down the surface integral

I’m splitting $S$ into $S_1$ (paraboloid) and $S_2$ (disk) to simplify the outward-oriented surface integral.

Parametrizing and integrating

I’m working through parametrizing surfaces $S_1$ and $S_2$, then evaluating the integrals by substituting the missing parts in the given expression.

Formulating the solution

I’m developing the final evaluation by addressing S1 and S2. The task involves breaking down the solution into defined steps, covering concepts, formulas, and calculations, leading to a comprehensive final answer.

Understanding the surface

OK, let me see. We're working with a vector field $\mathbf{F} = -2y \mathbf{j} + z \mathbf{k}$. The surface $S$ includes a paraboloid $S_1$ and a disk $S_2$ at $y = 1$, aiming to evaluate the integral $\iint_S \mathbf{F} \cdot d\mathbf{S}$ by splitting it into integrals over $S_1$ and $S_2$.

Identifying parametrizations

OK, let’s break this down: I need to parametrize $S_1$ and $S_2$ to find $d\mathbf{S}_1$ and $d\mathbf{S}_2$, then calculate their dot products and integrals.

Mapping the integration

I’m thinking through the vector field applied to surfaces $S_1$ and $S_2$. First, parametrize and find $d\mathbf{S}$ for each surface segment.

Parametrizing the paraboloid

OK, let me see. I’m parametrizing the paraboloid $S_1$ with cylindrical coordinates. The radial coordinate $s$ and angular coordinate $t$ are chosen for their symmetry.

Mapping parametrizations for surfaces:

I’m noting the parametrizations for $S_1$ and $S_2$, detailing $s$ and $t$ limits for both surfaces. This helps in visualizing the geometric representation.

Mapping out ranges

I'm thinking through the ranges for $S_1$ and $S_2$. For $S_1$, $s$ spans $0$ to $\sqrt{1}$, and for $S_2$, $s$ ranges from $0$ to $1$, both with $t$ from $0$ to $2\pi$.

Determining parameter ranges

I’m adjusting the parameterization of $s$ and $t$ for $S_1$ and $S_2$ where $s$ spans from 0 to 1 and $t$ spans from 0 to $2\pi$, considering $y = s^2$.

Clarifying parametrizations

I’m mapping out the parametrizations of $S_1$ and $S_2$ with their associated limits of integration.

Calculating surface element

I’m figuring out the infinitesimal surface element $d\mathbf{S}_1$ for the size in question. This involves the unit normal vector $\mathbf{n}_1$ and the infinitesimal area element $dA_1$.

Breaking down the surface area

I'm calculating the surface area vector differential by first determining partial derivatives of the parametrized surface, then computing their cross product to find the normal vector, ensuring orientation accuracy.

Confirming orientation

OK, let me see. I’m ensuring the normal vector $\mathbf{N}$ points outward from the paraboloid surface $S_1$, as the outward orientation seems crucial.

Mapping the surface

OK, let's break this down: I’m calculating the normal vector for $S_2$. After determining $\frac{\partial \mathbf{r}_2}{\partial s}$ and $\frac{\partial \mathbf{r}_2}{\partial t}$, I cross these to find $\mathbf{N}$ via the determinant method.

Reversing the normal vector

I’m thinking through the equation for $d\mathbf{S}_2$ by reversing the direction of the normal vector, which initially points in the negative $y$-direction. This adjustment is necessary to ensure the normal vector points outward, aligning with the positive $y$-direction.

Adjusting the normal

I’m piecing together the normal vector for $S_1$ to ensure it points outward, given the paraboloid's orientation and the $y$-component's negativity. Toy with the $d\mathbf{S}_1$ direction to make the $y$-component positive.

Analyzing surface integrals

I’m working through surface integrals by evaluating dot products and setting up integrals. Coincidentally, both surface elements have a positive $y$-component.

Calculating dot product

I’m working through the dot product of the vector field $\mathbf{F}$ and the surface element $d\mathbf{S}_1$. The first term is zero, the second term simplifies to $-2s^3$ using $y = s^2$, and the third term is $-2s^2 \sin t \cdot z$.

Handling the dot product

OK, I’m rechecking the dot product for $S_1$ with $z = s \sin t$. The expression looks off, so I’m carefully re-evaluating to ensure accuracy.

Calculating the dot product

I’m figuring out the dot product of $\mathbf{F}$ and $d\mathbf{S}_1$. By breaking down each term, I simplify the expression to $-2s^3 (1 + \sin^2 t)$.

Dot product calculation

I’m noting the dot product of the force field $\mathbf{F}$ and the differential surface element $d\mathbf{S}_2$ at $S_2$ by focusing on their $\mathbf{j}$ components. The result is $-2s$.

Calculating the dot product

I’m working through the dot product $\mathbf{F} \cdot d\mathbf{S}_2$, which simplifies to $-2s$. Next, I’ll integrate this over $S_2$.

Breaking down integrals

I started with separating the integral into two $s$-integrals. Evaluating these gave $-\frac{1}{2}$. Combining with $t$-integrals, the final expression was derived.

Calculating surface integral

I’m working through the surface integral by evaluating $I_1$ and $I_2$, integrating over their respective domains. Summing these results, I reach the final surface integral value of $-\frac{7\pi}{2}$.

Revisiting arithmetic precision

Breaking down the integral

I’m examining $\iint_S (-2yj + zk) \cdot dS$. Splitting into $S_1$ and $S_2$, the final value is $-\frac{7\pi}{2}$.

Understanding the problem

I’m mapping out the surface integral $\iint_S \mathbf{F} \cdot d\mathbf{S}$ by parametrizing the paraboloid $y = x^2 + z^2$ and the disk $x^2 + z^2 \leq 1$ at $y = 1$. Calculating $d\mathbf{S}$ for each surface is next.

Noticing the interval

OK, so it looks like $t$ is set to span from $0$ to $2\pi$, covering a full cycle.

Piecing together surface integrals

First, I calculate $\iint_{S_1} \mathbf{F} \cdot d\mathbf{S}_1$ to get $-\frac{3\pi}{2}$. Then, I find $\iint_{S_2} \mathbf{F} \cdot d\mathbf{S}_2$ to be $-2\pi$. Adding these, I arrive at the final integral value of $-\frac{7\pi}{2}$.

Subject: Vector Calculus - Surface Integrals

Step 1: Understanding the Problem and Setting Up the Parametrizations

Given:

• A vector field $\mathbf{F} = -2y\,\mathbf{j} + z\,\mathbf{k}$.
• Surface $S$ consists of:
  • $S_1$: The paraboloid $y = x^2 + z^2$ for $0 \leq y \leq 1$.
  • $S_2$: The disk $x^2 + z^2 \leq 1$ at $y = 1$.
• The surface has outward orientation.

To Find:

• Evaluate the surface integral: $$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_{S_1} \mathbf{F} \cdot d\mathbf{S}_1 + \iint_{S_2} \mathbf{F} \cdot d\mathbf{S}_2$$

Explanation and Concepts Used:

• Parametrization: We'll parametrize each part of the surface using cylindrical coordinates due to symmetry.
• Surface Element $d\mathbf{S}$: Calculated using the cross product of partial derivatives.
• Orientation: Ensure normals point outward for correct orientation.

Parametrizations and Limits:

• For $S_1$: $$\mathbf{r}_1(s, t) = \langle s\cos t,\, s^2,\, s\sin t \rangle$$
  • $s$ ranges from $0$ to $1$.
  • $t$ ranges from $0$ to $2\pi$.
• For $S_2$: $$\mathbf{r}_2(s, t) = \langle s\cos t,\, 1,\, s\sin t \rangle$$
  • $s$ ranges from $0$ to $1$.
  • $t$ ranges from $0$ to $2\pi$.

Step 2: Calculating the Surface Integrals

Computing $d\mathbf{S}_1$ and Evaluating $\iint_{S_1} \mathbf{F} \cdot d\mathbf{S}_1$:

1. Calculate Partial Derivatives: $$\frac{\partial \mathbf{r}_1}{\partial s} = \langle \cos t,\, 2s,\, \sin t \rangle,\quad \frac{\partial \mathbf{r}_1}{\partial t} = \langle -s\sin t,\, 0,\, s\cos t \rangle$$
2. Compute Cross Product for $d\mathbf{S}_1$: $$d\mathbf{S}_1 = \left( \frac{\partial \mathbf{r}_1}{\partial s} \times \frac{\partial \mathbf{r}_1}{\partial t} \right) ds\,dt = \langle -2s^2\cos t,\, s,\, -2s^2\sin t \rangle ds\,dt$$
   • Adjusted for outward orientation.
3. Compute $\mathbf{F} \cdot d\mathbf{S}_1$: $$\mathbf{F} \cdot d\mathbf{S}_1 = \left( -2s^2 \right) s + \left( s\sin t \right) \left( -2s^2\sin t \right) = -2s^3 - 2s^3\sin^2 t$$
4. Evaluate the Integral: $$\iint_{S_1} \mathbf{F} \cdot d\mathbf{S}_1 = \int_{0}^{2\pi} \int_{0}^{1} \left( -2s^3 - 2s^3\sin^2 t \right) ds\,dt = -\frac{3\pi}{2}$$

Computing $d\mathbf{S}_2$ and Evaluating $\iint_{S_2} \mathbf{F} \cdot d\mathbf{S}_2$:

1. Calculate Partial Derivatives: $$\frac{\partial \mathbf{r}_2}{\partial s} = \langle \cos t,\, 0,\, \sin t \rangle,\quad \frac{\partial \mathbf{r}_2}{\partial t} = \langle -s\sin t,\, 0,\, s\cos t \rangle$$
2. Compute Cross Product for $d\mathbf{S}_2$: $$d\mathbf{S}_2 = \left( \frac{\partial \mathbf{r}_2}{\partial s} \times \frac{\partial \mathbf{r}_2}{\partial t} \right) ds\,dt = \langle 0,\, s,\, 0 \rangle ds\,dt$$
   • Adjusted for outward orientation.
3. Compute $\mathbf{F} \cdot d\mathbf{S}_2$: $$\mathbf{F} \cdot d\mathbf{S}_2 = (-2 \times 1) \times s = -2s$$
4. Evaluate the Integral: $$\iint_{S_2} \mathbf{F} \cdot d\mathbf{S}_2 = \int_{0}^{2\pi} \int_{0}^{1} (-2s) ds\,dt = -2\pi$$

Final Answer:

The surface integral evaluates to:

So, $\iint_S (-2y\,\mathbf{j} + z\,\mathbf{k}) \cdot d\mathbf{S} = -\frac{3\pi}{2} + (-2\pi)$.

Answer: The surface integral evaluates to $-\dfrac{7\pi}{2}$.

Clarifying the approach

The assistant is tasked with breaking down a solution into detailed steps, ensuring clarity and following OpenAI's guidelines.

Ensuring compliance and clear instructions

Analyzing directional derivatives

I’m working through the objective of calculating the directional derivative of the temperature function along specific velocity vectors. This involves expressing the temperature as a function of time and determining how it changes with the vectors.

Evaluating directional derivative

I’m thinking through the directional derivative of $f$ along $\vec{e}_2$ and comparing it to the partial derivative $D_2 f(a)$. Recalculating with $[Df(a)] \vec{v}$ confirms accuracy.

Breaking down the solution

I’m focusing on presenting the next steps in the solution, ensuring clarity and progression. Each step hones in on a specific calculation or concept.

Navigating the problem

I’m analyzing the temperature of a moving car using $g'(0)$ and integrating the chain rule in calculus to express $g'(t)$ and subsequently determine $g'(0)$.

Step 3️⃣: I expanded and simplified the function $g(t) = 22 + 13t + 24t^2 + 9t^3$ to express temperature as a polynomial in $t$.

Expressing car's path

I mapped out the car's trajectory as $\mathbf{r}(t) = \begin{pmatrix} 2 - t \\ 1 - t \end{pmatrix}$, leading to $x(t) = 2 - t$ and $y(t) = 1 - t$.

Breaking down temperature dynamics

I expanded and simplified the temperature function $h(t)$. Then, finding $h'(t)$ and evaluating it at $t = 0$ showed a rate of $-5^\circ \text{C per unit time}$.

Azimuthal angle derivation

OK, let me break this down: I calculated $\frac{d\phi}{d\rho}$ using partial derivatives and the chain rule, which gives $\frac{2}{\rho}$.

Revisiting my approach

I updated equation (2) and tried combining it with equation (1). It turns out the simplest method is to discard equation (3) and solve using the original two equations.

Breaking down the steps

Firstly, I multiplied equation (2') by 3. Subtracting equation (1) from it, I simplified to find $f_x(a) = 1$. Then, through direct computation, I confirmed $f_y(a) = 4$.

Recalculating derivatives

I’m recalculating directional derivatives using the gradient vector $\nabla f(a)$ and verifying them against previous results, ensuring consistency in parts (a) and (b).

Verifying directional derivatives

I’m confirming the directional derivatives calculated using the gradient, ensuring they align with earlier results. This confirms both the linearity and correctness of the findings.

Subject: Multivariable Calculus - Directional Derivatives and Differentiability

Step 1: Problem (a) - Calculating the Rate of Change of Temperature Along $\vec{v}_1$

Given:

• Temperature function: $f(x, y) = 20 + x y^2$
• Point of interest: $a = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$
• Velocity vector: $\vec{v}_1 = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$
• At time: $t = 0$, the car is at point $a$

To Find:

• Function $g(t)$: Temperature outside the car as a function of time
• Derivative $g'(0)$: Rate at which the temperature is changing at $t = 0$

Explanation and Concepts Used:

• Parametric Path: The position of the car over time is given by $\mathbf{r}(t) = a + \vec{v}_1 t$.
• Composition of Functions: The temperature as a function of time is $g(t) = f(\mathbf{r}(t))$.
• Single-variable Calculus: Use derivative $g'(t)$ to find the rate of change at $t = 0$.

Step-by-Step Solution:

1. Determine the Path of the Car:

   $$\mathbf{r}(t) = \begin{pmatrix} 2 \\ 1 \end{pmatrix} + t \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = \begin{pmatrix} 2 + t \\ 1 + 3t \end{pmatrix}$$

2. Express Temperature as a Function of Time:

   $$g(t) = f(x(t), y(t)) = 20 + (2 + t)(1 + 3t)^2$$

3. Simplify $g(t)$:

   • Expand $(1 + 3t)^2 = 1 + 6t + 9t^2$
   • Multiply: $$(2 + t)(1 + 6t + 9t^2) = (2)(1 + 6t + 9t^2) + t(1 + 6t + 9t^2)$$ $$= (2 + 12t + 18t^2) + (1t + 6t^2 + 9t^3) = 2 + 13t + 24t^2 + 9t^3$$
   • Add the constant 20: $$g(t) = 20 + 2 + 13t + 24t^2 + 9t^3 = 22 + 13t + 24t^2 + 9t^3$$

4. Compute $g'(t)$:

   $$g'(t) = \frac{d}{dt}[22 + 13t + 24t^2 + 9t^3] = 13 + 48t + 27t^2$$

5. Evaluate $g'(0)$:

   $$g'(0) = 13 + 48(0) + 27(0)^2 = 13$$

Answer to Part (a): The rate at which the temperature is changing at $t = 0$ is $g'(0) = 13^\circ \text{C per unit time}$.

Step 2: Problem (b) - Calculating the Rate of Change of Temperature Along $\vec{v}_2$

Given:

• Velocity vector: $\vec{v}_2 = \begin{pmatrix} -1 \\ -1 \end{pmatrix}$

To Find:

• Function $h(t)$: Temperature outside the car as a function of time
• Derivative $h'(0)$: Rate at which the temperature is changing at $t = 0$

Step-by-Step Solution:

1. Determine the Path of the Car:

   $$\mathbf{r}(t) = \begin{pmatrix} 2 \\ 1 \end{pmatrix} + t \begin{pmatrix} -1 \\ -1 \end{pmatrix} = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = \begin{pmatrix} 2 - t \\ 1 - t \end{pmatrix}$$

2. Express Temperature as a Function of Time:

   $$h(t) = f(x(t), y(t)) = 20 + (2 - t)(1 - t)^2$$

3. Simplify $h(t)$:

   • Expand $(1 - t)^2 = 1 - 2t + t^2$
   • Multiply: $$(2 - t)(1 - 2t + t^2) = 2(1 - 2t + t^2) - t(1 - 2t + t^2)$$ $$= (2 - 4t + 2t^2) - (t - 2t^2 + t^3) = 2 - 5t + 4t^2 - t^3$$
   • Add the constant 20: $$h(t) = 20 + 2 - 5t + 4t^2 - t^3 = 22 - 5t + 4t^2 - t^3$$

4. Compute $h'(t)$:

   $$h'(t) = \frac{d}{dt}[22 - 5t + 4t^2 - t^3] = -5 + 8t - 3t^2$$

5. Evaluate $h'(0)$:

   $$h'(0) = -5 + 8(0) - 3(0)^2 = -5$$

Answer to Part (b): The rate at which the temperature is changing at $t = 0$ is $h'(0) = -5^\circ \text{C per unit time}$.

Step 3: Problem (c) - Determining the Directional Derivative Along $\vec{e}_2$

To Find:

• Directional derivative along $\vec{e}_2$: Use linearity to determine $D_{\vec{e}_2} f(a)$
• Confirm: Verify that $D_{\vec{e}_2} f(a) = D_2 f(a)$

Explanation and Concepts Used:

• Linearity of Directional Derivatives: $D_{\vec{v}} f(a) = f_x(a) v_1 + f_y(a) v_2$
• Partial Derivatives: $D_1 f(a) = f_x(a)$, $D_2 f(a) = f_y(a)$

Step-by-Step Solution:

1. Set Up Equations from Previous Results:

   From Part (a):

   $$D_{\vec{v}_1} f(a) = 13 = f_x(a)(1) + f_y(a)(3)$$

   From Part (b):

   $$D_{\vec{v}_2} f(a) = -5 = f_x(a)(-1) + f_y(a)(-1)$$

2. Solve for Partial Derivatives:

   • Add the equations: $$13 + (-5) = [f_x(a)(1) + f_y(a)(3)] + [f_x(a)(-1) + f_y(a)(-1)]$$ $$8 = f_x(a)(1 - 1) + f_y(a)(3 - 1) = f_y(a)(2)$$
   • Solve for $f_y(a)$: $$f_y(a) = \frac{8}{2} = 4$$
   • Substitute $f_y(a)$ back into one of the equations: $$-5 = -f_x(a) - 4$$ $$-5 = -f_x(a) - 4 \implies f_x(a) = 1$$

3. Determine $D_{\vec{e}_2} f(a)$:

   • Since $f_y(a) = 4$, the directional derivative along $\vec{e}_2$ is $D_{\vec{e}_2} f(a) = 4$.

4. Confirm with Partial Derivative:

   • Compute $f_y(x, y) = \frac{\partial f}{\partial y} = 2 x y$
   • At $a = (2, 1)$: $$f_y(a) = 2 \cdot 2 \cdot 1 = 4$$
   • This confirms $D_2 f(a) = 4$.

Answer to Part (c): The directional derivative along $\vec{e}_2$ is $D_{\vec{e}_2} f(a) = 4$, confirming that $D_2 f(a) = 4$.

Step 4: Problem (d) - Verifying Directional Derivatives Using the Gradient

To Find:

• Recalculate directional derivatives: Use $[Df(a)] \vec{v} = \nabla f(a) \cdot \vec{v}$
• Verify: Ensure results match those from Parts (a) and (b)

Explanation and Concepts Used:

• Gradient Vector: $\nabla f(a) = \begin{pmatrix} f_x(a) \\ f_y(a) \end{pmatrix}$
• Directional Derivative Formula: $D_{\vec{v}} f(a) = \nabla f(a) \cdot \vec{v}$

Step-by-Step Solution:

1. Use Calculated Partial Derivatives:

   • $f_x(a) = 1$
   • $f_y(a) = 4$
   • $\nabla f(a) = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$

2. Calculate $D_{\vec{v}_1} f(a)$:

   $$D_{\vec{v}_1} f(a) = \nabla f(a) \cdot \vec{v}_1 = (1)(1) + (4)(3) = 1 + 12 = 13$$

3. Calculate $D_{\vec{v}_2} f(a)$:

   $$D_{\vec{v}_2} f(a) = \nabla f(a) \cdot \vec{v}_2 = (1)(-1) + (4)(-1) = -1 - 4 = -5$$

4. Verify Results:

   • Both directional derivatives match the results from Parts (a) and (b).

Answer to Part (d): The directional derivatives using the gradient are $D_{\vec{v}_1} f(a) = 13$ and $D_{\vec{v}_2} f(a) = -5$, confirming our earlier calculations.

Final Answer:

By applying single-variable calculus and properties of differentiable functions, we determined:

• (a) The rate at which the temperature changes along $\vec{v}_1$ is $g'(0) = 13^\circ \text{C per unit time}$.
• (b) The rate at which the temperature changes along $\vec{v}_2$ is $h'(0) = -5^\circ \text{C per unit time}$.
• (c) The directional derivative along $\vec{e}_2$ is $D_{\vec{e}_2} f(a) = 4$, matching the partial derivative $f_y(a)$.
• (d) The gradient method verifies our previous results, confirming the calculations.

Therefore, we have successfully calculated the rates of temperature change along different paths and verified the linearity of directional derivatives for the given differentiable function.