문제를 주면서, 풀이할 때 그린 정리를 사용해달라고 명시해봅시다.
[!user] User Evaluate the line integral $\int_C\left(\sqrt{x^2+1}+y^2\right) d x+\left(e^{\sin y}+3 x^2-y^3\right) d y$, where $C$ is the boundary of the region outside of $r=1$ and inside of $r=2 \cos \theta$ with positively oriented. (Use green theorem)
[!gpt] ChatGPT To evaluate the given line integral using Green’s Theorem, we first need to understand the theorem and the region over which we are integrating.
Green’s Theorem relates a line integral around a simple, closed, positively oriented curve $C$ in the plane to a double integral over the region $D$ bounded by $C$. The theorem states:
\[\oint_C (P \, dx + Q \, dy) = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dx \, dy\]where $P$ and $Q$ are functions of $x$ and $y$.
In your case, $P = \sqrt{x^2+1} + y^2$ and $Q = e^{\sin y} + 3x^2 - y^3$.
Calculate the partial derivatives:
- $\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (e^{\sin y} + 3x^2 - y^3) = 6x$
- $\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (\sqrt{x^2+1} + y^2) = 2y$
Set up the double integral:
The double integral becomes:
\[\iint_D (6x - 2y) \, dx \, dy\]Describe the region $D$:
The region $D$ is bounded by the curves $r = 1$ and $r = 2\cos\theta$ in polar coordinates. We need to find the range of $\theta$ where these two curves intersect.
Convert the integral to polar coordinates:
In polar coordinates, $dx \, dy$ becomes $r \, dr \, d\theta$, $x = r\cos\theta$, and $y = r\sin\theta$. The integral in polar coordinates is:
\[\iint_D (6r\cos\theta - 2r\sin\theta) \, r \, dr \, d\theta\]Determine the limits of integration:
The limits for $r$ are from 1 to $2\cos\theta$. The limits for $\theta$ are determined by the intersection points of $r = 1$ and $r = 2\cos\theta$.
Evaluate the integral:
Finally, evaluate the double integral with the determined limits.
Let’s start by finding the intersection points of $r = 1$ and $r = 2\cos\theta$ to determine the limits for $\theta$.
[!tip] Using Wolframalpha
The intersection points of $r = 1$ and $r = 2\cos\theta$ occur at $\theta = -\frac{\pi}{3}$ and $\theta = \frac{\pi}{3}$. Since we are dealing with a positively oriented curve, we will integrate over these limits in the counterclockwise direction. Therefore, the limits for $\theta$ are from $-\frac{\pi}{3}$ to $\frac{\pi}{3}$.
Now, let’s set up and evaluate the double integral in polar coordinates:
Integral setup:
\[\iint_D (6r\cos\theta - 2r\sin\theta) \, r \, dr \, d\theta\]Simplifies to:
\[\iint_D (6r^2\cos\theta - 2r^2\sin\theta) \, dr \, d\theta\]Limits of integration:
- $r$ varies from 1 to $2\cos\theta$.
- $\theta$ varies from $-\frac{\pi}{3}$ to $\frac{\pi}{3}$.
Evaluate the integral:
\[\int_{-\pi/3}^{\pi/3} \int_{1}^{2\cos\theta} (6r^2\cos\theta - 2r^2\sin\theta) \, dr \, d\theta\]Let’s proceed with the calculation.
[!tip] Using Wolframalpha
The evaluated double integral is:
\[2\sqrt{3}(-1 + 8\cos^3(\theta))\]This result represents the value of the original line integral
\[\int_C\left(\sqrt{x^2+1}+y^2\right) d x+\left(e^{\sin y}+3 x^2-y^3\right) d y\]where $C$ is the boundary of the region outside of $r=1$ and inside of $r=2 \cos \theta$, oriented positively.
풀이 과정이 대부분 옳지만 마지막 계산이 많이 잘못된 것 같습니다. 이를 지적하며 다시 풀도록 요청해봅시다.
[!user] User I think final calculation is quiet wrong.
[!gpt] ChatGPT
[!tip] Using Wolframalpha
ChatGPT
You are correct, and I appreciate your patience. After re-evaluating the double integral, the correct result is:
\[\frac{3 \sqrt{3}}{2} + 4\pi\]This is the value of the original line integral
\[\int_C\left(\sqrt{x^2+1}+y^2\right) d x+\left(e^{\sin y}+3 x^2-y^3\right) d y\]where $C$ is the boundary of the region outside of $r=1$ and inside of $r=2 \cos \theta$, oriented positively.
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