[!user] User Use Stokes’ Theorem to evaluate $\int_C \mathbf{F} \cdot d \mathbf{r}$. In each case $C$ is oriented counterclockwise as viewed from above, unless otherwise stated : $\mathbf{F}(x, y, z)=x^2 y \mathbf{i}+x^3 \mathbf{j}+e^z \tan ^{-1} z \mathbf{k}$, $C$ is the curve with parametric equations $x=\cos t, y=\sin t$, $z=\sin t, 0 \leqslant t \leqslant 2 \pi$
[!gpt] ChatGPT The integral of the given vector field $\mathbf{F}(x, y, z)=x^2 y \mathbf{i}+x^3 \mathbf{j}+e^z \tan ^{-1} z \mathbf{k}$ along the curve $C$ with parametric equations $x=\cos t, y=\sin t, z=\sin t$ for $0 \leqslant t \leqslant 2 \pi$ appears to be complex to solve directly. This complexity arises from the trigonometric and exponential functions involved in the vector field and the curve.
To simplify this problem, we can use Stokes’ Theorem, which relates a surface integral over a vector field to a line integral around the boundary of the surface. Stokes’ Theorem states that: \(\int_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}\) where $C$ is the boundary curve of the surface $S$, and $\nabla \times \mathbf{F}$ is the curl of the vector field $\mathbf{F}$.
Let’s calculate the curl of $\mathbf{F}$ and then evaluate the surface integral over a surface $S$ bounded by the curve $C$.
Analysis
The curl of the vector field $\mathbf{F}(x, y, z)=x^2 y \mathbf{i}+x^3 \mathbf{j}+e^z \tan ^{-1} z \mathbf{k}$ simplifies to zero. This result implies that the integral of $\mathbf{F}$ along the curve $C$ is also zero, as per Stokes’ Theorem.
In Stokes’ Theorem, the surface integral of the curl of $\mathbf{F}$ over a surface $S$ bounded by the curve $C$ is equal to the line integral of $\mathbf{F}$ along $C$. Since the curl of $\mathbf{F}$ is zero, the line integral of $\mathbf{F}$ along $C$ is also zero.
Therefore, $\int_C \mathbf{F} \cdot d \mathbf{r} = 0$.
손풀이한 내용과 비교해보니, 중간 과정에서 ChatGPT가 curl $\mathbf F$를 잘못 계산한 것 같아 이를 지적합니다. 지적할 때에는 지적사항만 간단히 말해도 됩니다.
[!user] User I think curl F = (0, 0, 2x^2).
[!gpt] ChatGPT
The recalculated curl of the vector field $\mathbf{F}(x, y, z)=x^2 y \mathbf{i}+x^3 \mathbf{j}+e^z \tan ^{-1} z \mathbf{k}$ still simplifies to zero. This means the integral of $\mathbf{F}$ along the curve $C$ should also be zero according to Stokes’ Theorem.
However, if you believe the curl of $\mathbf{F}$ should be $(0, 0, 2x^2)$, let’s manually calculate the curl of $\mathbf{F}$ to verify this.
The curl of a vector field $\mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}$ is given by the determinant of the following matrix: \(\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{pmatrix}\) Let’s calculate this determinant for $\mathbf{F}$. You are correct. The curl of the vector field $\mathbf{F}(x, y, z)=x^2 y \mathbf{i}+x^3 \mathbf{j}+e^z \tan ^{-1} z \mathbf{k}$ is indeed $(0, 0, 2x^2)$. Now, using Stokes’ Theorem, we can evaluate the line integral of $\mathbf{F}$ along the curve $C$ by calculating the surface integral of the curl of $\mathbf{F}$ over a surface $S$ bounded by $C$. Since the curl of $\mathbf{F}$ has only a $k$ component (which is $2x^2$), the surface integral simplifies to: \(\iint_S 2x^2 \, dS\)
Given the parametric equations $x=\cos t, y=\sin t, z=\sin t$ for $0 \leqslant t \leqslant 2 \pi$, we can choose the surface $S$ to be the disk $x^2 + y^2 \leq 1$ in the $xy$-plane. Let’s calculate this integral. The evaluated surface integral of $2x^2$ over the disk $x^2 + y^2 \leq 1$ in the $xy$-plane, using Stokes’ Theorem, is approximately $1.5708$. This value is the result of the line integral of the vector field $\mathbf{F}(x, y, z)=x^2 y \mathbf{i}+x^3 \mathbf{j}+e^z \tan ^{-1} z \mathbf{k}$ along the curve $C$ with parametric equations $x=\cos t, y=\sin t, z=\sin t$ for $0 \leqslant t \leqslant 2 \pi$.
Therefore, $\int_C \mathbf{F} \cdot d \mathbf{r} \approx 1.5708$.
처음에는 지적사항을 반박하는가 싶더니, 정확히 다시 확인해보는 절차를 걸친 뒤, 잘못된 것을 확인하고 사과합니다. 그 이후로는 정확한 풀이과정을 제시하였습니다. 그런데 $\int_C \mathbf{F} \cdot d \mathbf{r}$를 어떻게 계산하는지가 궁금했기 때문에,
[!user] User Explain the last eqaution step by step.
[!gpt] ChatGPT Certainly! Let’s break down the last calculation step by step:
Curl in Polar Coordinates: The curl of the vector field $\mathbf{F}$ was found to be $(0, 0, 2x^2)$. For the $k$-component, $2x^2$, we converted this into polar coordinates because the surface $S$ is a disk in the $xy$-plane. In polar coordinates, $x = r\cos(\theta)$ and $y = r\sin(\theta)$. So, $2x^2$ becomes $2r^2\cos^2(\theta)$.
Choosing the Surface $S$: We chose the surface $S$ to be the disk in the $xy$-plane bounded by the curve $C$. This disk can be represented in polar coordinates as $0 \leq r \leq 1$ and $0 \leq \theta \leq 2\pi$, corresponding to the entire area of a unit circle.
Surface Integral in Polar Coordinates: The surface integral of a scalar function over a surface in polar coordinates is given by $\iint_S f(r, \theta) \, r \, dr \, d\theta$, where $r \, dr \, d\theta$ is the area element in polar coordinates.
Setting Up the Integral: We set up the integral of $2r^2\cos^2(\theta)$ over the unit disk. The integral becomes: \(\int_{0}^{2\pi} \int_{0}^{1} 2r^2\cos^2(\theta) \, r \, dr \, d\theta\)
Evaluating the Integral: The integral is evaluated in two steps: first with respect to $r$ from 0 to 1, and then with respect to $\theta$ from 0 to $2\pi$. This double integral computes the total flux of the curl of $\mathbf{F}$ through the surface $S$.
Result: The result of the integral, approximately $1.5708$, is the value of the line integral of $\mathbf{F}$ along the curve $C$ as per Stokes’ Theorem.
In summary, we converted the problem into polar coordinates for ease of computation, chose an appropriate surface $S$, and then evaluated the surface integral of the curl of $\mathbf{F}$ over $S$, which, according to Stokes’ Theorem, equals the line integral of $\mathbf{F}$ along $C$.