| 模型 | 模型答案 | 正确答案 |
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| 金声玉亮2.0 | To solve the integral \(\int_{\Gamma} \frac{x\,dy - y\,dx}{x^2 + y^2}\), where \(\Gamma\) is any piecewise smooth, closed curve that encloses the origin but does not pass through it, we can use the concept of the Cauchy integral theorem from complex analysis. First, let's rewrite the integral in a form that is more familiar in complex analysis. Consider the complex number \(z = x + iy\). The differential form \(x\,dy - y\,dx\) can be written as the imaginary part of \(dz/z\), since: \[ dz = dx + i\,dy \quad \text{and} \quad \frac{dz}{z} = \frac{dx + i\,dy}{x + iy}. \] Multiplying the numerator and denominator by the conjugate of the denominator \(x - iy\), we get: \[ \frac{dz}{z} = \frac{(dx + i\,dy)(x - iy)}{x^2 + y^2} = \frac{x\,dx + y\,dy + i(x\,dy - y\,dx)}{x^2 + y^2}. \] The imaginary part of this expression is: \[ \text{Im}\left(\frac{dz}{z}\right) = \frac{x\,dy - y\,dx}{x^2 + y^2}. \] Thus, the integral we need to compute is: \[ \int_{\Gamma} \frac{x\,dy - y\,dx}{x^2 + y^2} = \text{Im} \left( \int_{\Gamma} \frac{dz}{z} \right). \] According to the Cauchy integral theorem, for any closed contour \(\Gamma\) that encloses the origin and does not pass through it, the integral of \(1/z\) around \(\Gamma\) is: \[ \int_{\Gamma} \frac{dz}{z} = 2\pi i. \] Taking the imaginary part of this result, we get: \[ \text{Im}(2\pi i) = 2\pi. \] Therefore, the value of the given integral is: \[ \int_{\Gamma} \frac{x\,dy - y\,dx}{x^2 + y^2} = 2\pi. \] |
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| deepseekr1 | To compute the integral \(\int_{\Gamma} \frac{x\,dy - y\,dx}{x^2 + y^2}\), where \(\Gamma\) is a piecewise smooth, closed curve that encloses the origin but does not pass through it, we can use the following approach: 1. **Identify the Integrand**: The given integral is \(\int_{\Gamma} \frac{x\,dy - y\,dx}{x^2 + y^2}\). We recognize this as a line integral involving differential forms. 2. **Green's Theorem Consideration**: Green's theorem relates a line integral around a closed curve to a double integral over the region it encloses. However, since the integrand has a singularity at the origin, Green's theorem cannot be directly applied to the entire region enclosed by \(\Gamma\). 3. **Excluding the Origin**: To avoid the singularity, we consider a small circle \(C_\varepsilon\) of radius \(\varepsilon\) centered at the origin, oriented counterclockwise. The region between \(\Gamma\) and \(C_\varepsilon\) does not include the origin, allowing us to apply Green's theorem here. 4. **Applying Green's Theorem**: By Green's theorem, the integral over \(\Gamma\) minus the integral over \(C_\varepsilon\) (with opposite orientation) equals the double integral over the region between them. However, the integrand's partial derivatives result in zero, leading to the conclusion that the integral over \(\Gamma\) equals the integral over \(C_\varepsilon\). 5. **Parametrizing the Small Circle**: Parametrize \(C_\varepsilon\) as \(x = \varepsilon \cos \theta\), \(y = \varepsilon \sin \theta\) for \(\theta\) from 0 to \(2\pi\). Compute the integrand: \[ \frac{x\,dy - y\,dx}{x^2 + y^2} = \frac{\varepsilon \cos \theta (\varepsilon \cos \theta \, d\theta) - \varepsilon \sin \theta (-\varepsilon \sin \theta \, d\theta)}{\varepsilon^2} = \frac{\varepsilon^2 (\cos^2 \theta + \sin^2 \theta) \, d\theta}{\varepsilon^2} = d\theta \] The integral over \(C_\varepsilon\) is \(\int_0^{2\pi} d\theta = 2\pi\). 6. **Conclusion**: Since the integral over \(\Gamma\) must equal the integral over \(C_\varepsilon\), the value of the integral is \(2\pi\). Thus, the value of the integral is \(\boxed{2\pi}\). |
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| gpto1 | To compute the integral: \[ I = \int_{\Gamma} \frac{x\,dy - y\,dx}{x^2 + y^2} \] where \(\Gamma\) is any piecewise smooth, closed curve that encloses the origin but does not pass through it, we can proceed as follows: **Step 1: Express the integrand in terms of \(d\theta\) using polar coordinates.** Let’s parameterize the curve using polar coordinates. Let \( (x, y) = (r \cos\theta, r \sin\theta) \). We compute the differentials: \[ dx = -r \sin\theta\, d\theta + \cos\theta\, dr \\ dy = r \cos\theta\, d\theta + \sin\theta\, dr. \] Since we are integrating over a closed curve that doesn't pass through the origin (\( r \neq 0 \)), we can assume \( r \) is not zero and, for a fixed curve, \( r \) is a function of \(\theta\). Compute \( x\,dy - y\,dx \): \[ x\,dy - y\,dx = [r \cos\theta][r \cos\theta\, d\theta + \sin\theta\, dr] - [r \sin\theta][-r \sin\theta\, d\theta + \cos\theta\, dr] \\ = r^2 \cos^2\theta\, d\theta + r \cos\theta \sin\theta\, dr + r^2 \sin^2\theta\, d\theta - r \sin\theta \cos\theta\, dr \\ = r^2 (\cos^2\theta + \sin^2\theta)\, d\theta \\ = r^2\, d\theta. \] Since \( x^2 + y^2 = r^2 \), the integrand simplifies to: \[ \frac{x\,dy - y\,dx}{x^2 + y^2} = \frac{r^2\, d\theta}{r^2} = d\theta. \] **Step 2: Evaluate the integral.** Therefore, the integral becomes: \[ I = \int_{\Gamma} d\theta = \Delta \theta, \] where \( \Delta \theta \) represents the total change in the angle \( \theta \) as we traverse the curve \(\Gamma\). **Step 3: Relate \( \Delta \theta \) to the winding number.** Since \(\Gamma\) is closed and encloses the origin, \( \Delta \theta \) corresponds to \( 2\pi n \), where \( n \) is the winding number (the number of times the curve winds around the origin). **Conclusion:** \[ I = \int_{\Gamma} \frac{x\,dy - y\,dx}{x^2 + y^2} = 2\pi n, \] where \( n \) is the integer winding number of the curve \(\Gamma\) around the origin. **Answer:** The value of the integral is 2 π times the winding number of Γ; that is, ∫ = 2π × n. |
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| GPT-4o | To solve the given integral \(\int_{\Gamma} \frac{x \, dy - y \, dx}{x^2 + y^2}\), where \(\Gamma\) is any piecewise smooth, closed curve that encloses the origin but does not pass through it, we can make use of complex analysis. Consider the complex variable \( z = x + iy \) and its conjugate \( \bar{z} = x - iy \). The differential forms \( dy \) and \( dx \) can be expressed in terms of \( dz \) and \( d\bar{z} \): \[ dz = dx + i \, dy \quad \text{and} \quad d\bar{z} = dx - i \, dy. \] From these, we can derive \[ dy = \frac{dz - d\bar{z}}{2i} \quad \text{and} \quad dx = \frac{dz + d\bar{z}}{2}. \] The given integral can be rewritten in terms of \( z \) and \( \bar{z} \): \[ \int_{\Gamma} \frac{x \, dy - y \, dx}{x^2 + y^2}. \] Note that \( x = \frac{z + \bar{z}}{2} \) and \( y = \frac{z - \bar{z}}{2i} \). Therefore, \[ x \, dy - y \, dx = \frac{z + \bar{z}}{2} \cdot \frac{dz - d\bar{z}}{2i} - \frac{z - \bar{z}}{2i} \cdot \frac{dz + d\bar{z}}{2}. \] Simplifying this: \[ x \, dy - y \, dx = \frac{(z + \bar{z})(dz - d\bar{z})}{4i} - \frac{(z - \bar{z})(dz + d\bar{z})}{4i}. \] This further simplifies to: \[ x \, dy - y \, dx = \frac{z dz - z d\bar{z} + \bar{z} dz - \bar{z} d\bar{z} - z dz - z d\bar{z} + \bar{z} dz + \bar{z} d\bar{z}}{4i} = \frac{\bar{z} dz - z d\bar{z}}{2i}. \] However, it is simpler to notice that \[ \frac{x \, dy - y \, dx}{x^2 + y^2} = \frac{x \, dy - y \, dx}{z \bar{z}} = \frac{x \, dy - y \, dx}{|z|^2}. \] By recognizing that \(\frac{1}{z \bar{z}}\) can be written as \(\frac{1}{z\bar{z}} = \frac{1}{|z|^2}\), we can transform the integral into a form involving complex variables. Specifically, \[ \int_{\Gamma} \frac{x \, dy - y \, dx}{x^2 + y^2} = \int_{\Gamma} \frac{1}{|z|^2} (x \, dy - y \, dx). \] But another way to see this is by using the polar representation of complex numbers, where \( z = re^{i\theta} \). Noting that \( \frac{x \, dy - y \, dx}{x^2 + y^2} \) is actually the differential form of \( d\theta \) over the curve \(\Gamma\). Hence, \[ \int_{\Gamma} \frac{x \, dy - y \, dx}{x^2 + y^2} = \int_{\Gamma} d\theta. \] Since \(\Gamma\) is a closed curve enclosing the origin, the integral of \( d\theta \) around the entire curve \(\Gamma\) is \( 2\pi \), which is the total change in \(\theta\) over a complete loop around the origin. Thus, the value of the integral is: \[ \boxed{2\pi}. \] |
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| llama3.1-70B-Instruct | A classic problem in complex analysis! The integral you're asking about is a well-known example of a contour integral that can be evaluated using the Cauchy Integral Formula. Let's break it down step by step: 1. We can rewrite the integral as: $$\int_{\Gamma} \frac{x\,dy - y\,dx}{x^2 + y^2} = \int_{\Gamma} \frac{x\,dy - y\,dx}{(x + iy)(x - iy)}$$ where we've used the fact that $x^2 + y^2 = (x + iy)(x - iy)$. 2. Now, let's consider the function $f(z) = \frac{1}{z}$, which is analytic everywhere except at $z = 0$. We can write: $$f(z) = \frac{1}{z} = \frac{1}{x + iy}$$ 3. Using the Cauchy Integral Formula, we know that: $$\int_{\Gamma} \frac{f(z)}{z - a} dz = 2\pi i f(a)$$ where $a$ is a point inside the contour $\Gamma$. 4. In our case, we have $a = 0$, so we can write: $$\int_{\Gamma} \frac{1}{z} dz = 2\pi i f(0) = 2\pi i \cdot \frac{1}{0} = 2\pi i$$ 5. Now, we can relate this result to our original integral: $$\int_{\Gamma} \frac{x\,dy - y\,dx}{x^2 + y^2} = \int_{\Gamma} \frac{1}{z} dz = 2\pi i$$ So, the value of the integral is $2\pi i$, which is a purely imaginary number. Note that this result is independent of the specific contour $\Gamma$, as long as it encloses the origin and is piecewise smooth. |
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| llama3.1-405B-Instruct | To compute the integral \int_{\Gamma} \frac{x*dy-y*dx}{x^2+y^2}, where $\Gamma$ is any piecewise smooth, closed curve that encloses the origin but does not pass through it, we can use Green's theorem. Green's theorem states that for a piecewise smooth, closed curve $\Gamma$ that encloses a region $D$, we have: \int_{\Gamma} (Pdx + Qdy) = \iint_{D} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dxdy In this case, we have: P = -y/(x^2+y^2) Q = x/(x^2+y^2) Using Green's theorem, we can rewrite the integral as: \int_{\Gamma} \frac{x*dy-y*dx}{x^2+y^2} = \iint_{D} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dxdy Computing the partial derivatives, we get: \frac{\partial Q}{\partial x} = \frac{y^2-x^2}{(x^2+y^2)^2} \frac{\partial P}{\partial y} = \frac{x^2-y^2}{(x^2+y^2)^2} Substituting these expressions into the integral, we get: \int_{\Gamma} \frac{x*dy-y*dx}{x^2+y^2} = \iint_{D} \left(\frac{y^2-x^2}{(x^2+y^2)^2} - \frac{x^2-y^2}{(x^2+y^2)^2}\right)dxdy Simplifying the integrand, we get: \int_{\Gamma} \frac{x*dy-y*dx}{x^2+y^2} = \iint_{D} \frac{2(y^2-x^2)}{(x^2+y^2)^2}dxdy Now, we can convert the integral to polar coordinates using the transformation: x = rcosθ y = rsinθ The Jacobian of this transformation is: J = \frac{\partial(x,y)}{\partial(r,\theta)} = r Substituting the transformation and the Jacobian into the integral, we get: \int_{\Gamma} \frac{x*dy-y*dx}{x^2+y^2} = \int_{0}^{2\pi} \int_{0}^{\infty} \frac{2(r^2sin^2\theta-r^2cos^2\theta)}{(r^2)^2}rdrd\theta Simplifying the integrand, we get: \int_{\Gamma} \frac{x*dy-y*dx}{x^2+y^2} = \int_{0}^{2\pi} \int_{0}^{\infty} \frac{2sin^2\theta-2cos^2\theta}{r}drd\theta Evaluating the inner integral, we get: \int_{\Gamma} \frac{x*dy-y*dx}{x^2+y^2} = \int_{0}^{2\pi} \left[2sin^2\theta-2cos^2\theta\right]d\theta Evaluating the outer integral, we get: \int_{\Gamma} \frac{x*dy-y*dx}{x^2+y^2} = 2\pi Therefore, the value of the integral is 2π. |
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| DeepSeek-Coder-V2 | To solve the integral \(\int_{\Gamma} \frac{x \, dy - y \, dx}{x^2 + y^2}\), where \(\Gamma\) is any piecewise smooth, closed curve that encloses the origin but does not pass through it, we can use the Residue Theorem from complex analysis. First, let's rewrite the integral in a form that is more amenable to complex analysis. Consider the complex function \(f(z) = \frac{1}{z}\), where \(z = x + iy\). The integral can be interpreted as a contour integral in the complex plane. The given integral can be rewritten using the differential form \(dz = dx + i \, dy\). We can express \(x \, dy - y \, dx\) in terms of \(dz\) and \(\overline{dz}\) (the complex conjugate of \(dz\)): \[ x \, dy - y \, dx = \text{Im}(z \, \overline{dz}) \] Thus, the integral becomes: \[ \int_{\Gamma} \frac{x \, dy - y \, dx}{x^2 + y^2} = \int_{\Gamma} \frac{\text{Im}(z \, \overline{dz})}{z \overline{z}} \] Since \(z \overline{z} = |z|^2 = x^2 + y^2\), the integral simplifies to: \[ \int_{\Gamma} \frac{\text{Im}(z \, \overline{dz})}{z \overline{z}} = \int_{\Gamma} \frac{\text{Im}(z \, \overline{dz})}{|z|^2} \] Now, consider the contour integral of the function \(f(z) = \frac{1}{z}\) around the closed curve \(\Gamma\). By the Residue Theorem, the integral of \(f(z)\) around \(\Gamma\) is \(2\pi i\) times the sum of the residues of \(f(z)\) inside \(\Gamma\). Since \(f(z) = \frac{1}{z}\) has a simple pole at \(z = 0\) (the origin), and \(\Gamma\) encloses the origin, the residue of \(f(z)\) at \(z = 0\) is 1. Therefore, the integral is: \[ \int_{\Gamma} \frac{1}{z} \, dz = 2\pi i \cdot 1 = 2\pi i \] Thus, the value of the original integral is: \[ \int_{\Gamma} \frac{x \, dy - y \, dx}{x^2 + y^2} = 2\pi i \] Since we are dealing with a real-valued integral, we take the imaginary part of the result: \[ \int_{\Gamma} \frac{x \, dy - y \, dx}{x^2 + y^2} = 2\pi \] Therefore, the solution to the integral is: \[ \boxed{2\pi} \] |
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| Qwen2-72B-Instruct | To solve this integral, we can use the concept of Green's Theorem, which relates a line integral around a simple closed curve to a double integral over the plane region bounded by that curve. The theorem states that for a positively oriented, piecewise-smooth, simple closed curve C and a region D contained in the plane, if P and Q have continuous partial derivatives in an open region containing D, then \[ \oint_C P\,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\,dx\,dy. \] In our case, we have \[ P = -\frac{y}{x^2+y^2}, \quad Q = \frac{x}{x^2+y^2}. \] We need to compute the partial derivatives \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}\left(\frac{x}{x^2+y^2}\right) = \frac{(x^2+y^2) - 2x^2}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}, \] \[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}\left(-\frac{y}{x^2+y^2}\right) = -\frac{(x^2+y^2) - 2y^2}{(x^2+y^2)^2} = -\frac{x^2-y^2}{(x^2+y^2)^2}. \] Thus, \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{y^2-x^2}{(x^2+y^2)^2} + \frac{x^2-y^2}{(x^2+y^2)^2} = \frac{2y^2-2x^2}{(x^2+y^2)^2} = \frac{2(y^2-x^2)}{(x^2+y^2)^2}. \] Now, we can apply Green's Theorem to the given line integral: \[ \oint_{\Gamma} \frac{x\,dy - y\,dx}{x^2+y^2} = \iint_D \frac{2(y^2-x^2)}{(x^2+y^2)^2}\,dx\,dy. \] The integrand is an odd function with respect to both x and y, i.e., \[ f(x, y) = \frac{2(y^2-x^2)}{(x^2+y^2)^2}, \quad f(-x, -y) = -f(x, y). \] Since the region D is symmetric about the origin (as it encloses the origin), the double integral over D of any odd function is zero. Therefore, \[ \iint_D \frac{2(y^2-x^2)}{(x^2+y^2)^2}\,dx\,dy = 0. \] However, this conclusion is incorrect for the given integral. The reason is that the integrand has a singularity at the origin (0, 0), which is inside the curve $\Gamma$. Therefore, we cannot apply Green's Theorem directly. Instead, we should consider the line integral as a special case of a more general theorem known as the Cauchy integral theorem from complex analysis. The given integral is a special case of the Cauchy integral formula for a circle centered at the origin with radius R: \[ \oint_{\Gamma} \frac{dz}{z} = 2\pi i, \] where $\Gamma$ is the circle $|z| = R$ and $z = x + iy$. Since the integrand is the complex conjugate of $1/z$, we have \[ \oint_{\Gamma} \frac{x\,dy - y\,dx}{x^2+y^2} = \oint_{\Gamma} \frac{dz}{z} = 2\pi i. \] However, since the integral is real, the imaginary part must be zero. Thus, the value of the integral is the real part of $2\pi i$, which is \[ \oint_{\Gamma} \frac{x\,dy - y\,dx}{x^2+y^2} = 2\pi. \] Therefore, the value of the given integral is $2\pi$. |
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| baichuan4 | To solve the given problem, we will use the Green's theorem, which relates the line integral around a simple closed curve to the double integral over the region enclosed by the curve. The theorem states that for a two-dimensional region $D$ bounded by a simple closed curve $\Gamma$, oriented counterclockwise, we have: $$\oint_\Gamma (P \, dx + Q \, dy) = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx \, dy$$ In our case, the line integral is given by: $$\int_\Gamma \frac{x \, dy - y \, dx}{x^2 + y^2}$$ Comparing this with the general form of the line integral in Green's theorem, we can identify $P = -\frac{y}{x^2 + y^2}$ and $Q = \frac{x}{x^2 + y^2}$. Now, we need to calculate the partial derivatives: $$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} \left( \frac{x}{x^2 + y^2} \right) = \frac{y^2 - x^2}{(x^2 + y^2)^2}$$ $$\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} \left( -\frac{y}{x^2 + y^2} \right) = -\frac{x^2 - y^2}{(x^2 + y^2)^2}$$ Substituting these derivatives into the right-hand side of Green's theorem, we get: $$\iint_D \left( \frac{y^2 - x^2}{(x^2 + y^2)^2} - (-\frac{x^2 - y^2}{(x^2 + y^2)^2}) \right) dx \, dy = \iint_D \frac{2(y^2 - x^2)}{(x^2 + y^2)^2} dx \, dy$$ However, since the integrand is an odd function (i.e., it changes sign when $x$ and $y$ are interchanged), and the region $D$ is symmetric about the $y$-axis, the integral over the entire region will be zero. Therefore, we conclude that: $$\int_\Gamma \frac{x \, dy - y \, dx}{x^2 + y^2} = 0$$ This result holds for any piecewise smooth, closed curve $\Gamma$ that encloses the origin but does not pass through it. |
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| doubao | Let's use Green's theorem. Consider the vector field $\vec{F} = \frac{-y}{x^2+y^2} \vec{i} + \frac{x}{x^2+y^2} \vec{j}$. The curl of $\vec{F}$ is: $\text{curl}(\vec{F}) = \frac{\partial}{\partial x}(\frac{x}{x^2+y^2}) - \frac{\partial}{\partial y}(\frac{-y}{x^2+y^2}) = 0$. But since the curve $\Gamma$ encloses the origin where the vector field is not defined continuously, we can't directly apply Green's theorem. We introduce a small circle $C_{\epsilon}$ of radius $\epsilon$ centered at the origin. Now consider the region $R$ bounded by $\Gamma$ and $C_{\epsilon}$. Applying Green's theorem to $R$ gives: $\int_{\Gamma \cup C_{\epsilon}} \frac{x dy - y dx}{x^2+y^2} = \iint_{R} 0 \, dA = 0$. So $\int_{\Gamma} \frac{x dy - y dx}{x^2+y^2} = -\int_{C_{\epsilon}} \frac{x dy - y dx}{x^2+y^2}$. On $C_{\epsilon}$, we can parametrize as $x = \epsilon \cos \theta$, $y = \epsilon \sin \theta$. Then $dx = -\epsilon \sin \theta d \theta$, $dy = \epsilon \cos \theta d \theta$. Evaluating the integral over $C_{\epsilon}$ gives: $\int_{C_{\epsilon}} \frac{x dy - y dx}{x^2+y^2} = \int_{0}^{2 \pi} \frac{\epsilon \cos \theta \cdot \epsilon \cos \theta d \theta - \epsilon \sin \theta \cdot (-\epsilon \sin \theta) d \theta}{\epsilon^2} = 2 \pi$. So $\int_{\Gamma} \frac{x dy - y dx}{x^2+y^2} = -2 \pi$. |
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| minimax | The integral given is of the form: \[ \int_{\Gamma} \frac{x\,dy - y\,dx}{x^2 + y^2} \] where \(\Gamma\) is a piecewise smooth, closed curve that encloses the origin. This integral can be solved using Green's theorem, which relates a line integral around a simple closed curve C to a double integral over the plane region D bounded by C. Green's theorem states that if P and Q have continuous partial derivatives on an open region containing D and C, then: \[ \int_C (P\,dx + Q\,dy) = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA \] In our case, we can identify \(P(x,y) = -\frac{y}{x^2 + y^2}\) and \(Q(x,y) = \frac{x}{x^2 + y^2}\). Let's compute the partial derivatives: \[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}\left(-\frac{y}{x^2 + y^2}\right) = \frac{-(x^2 + y^2) - y(2y)}{(x^2 + y^2)^2} = \frac{-x^2 - y^2 - 2y^2}{(x^2 + y^2)^2} = \frac{-x^2 - 3y^2}{(x^2 + y^2)^2} \] \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}\left(\frac{x}{x^2 + y^2}\right) = \frac{(x^2 + y^2) - x(2x)}{(x^2 + y^2)^2} = \frac{x^2 + y^2 - 2x^2}{(x^2 + y^2)^2} = \frac{-x^2 + y^2}{(x^2 + y^2)^2} \] Now, we apply Green's theorem: \[ \int_{\Gamma} \left(\frac{x\,dy - y\,dx}{x^2 + y^2}\right) = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA \] \[ = \iint_D \left(\frac{-x^2 + y^2}{(x^2 + y^2)^2} - \frac{-x^2 - 3y^2}{(x^2 + y^2)^2}\right) dA \] \[ = \iint_D \left(\frac{y^2 + 3y^2}{(x^2 + y^2)^2}\right) dA \] \[ = \iint_D \frac{4y^2}{(x^2 + y^2)^2} dA \] Since the curve \(\Gamma\) encloses the origin, the region D does not include the origin. The integrand \(\frac{4y^2}{(x^2 + y^2)^2}\) is an even function with respect to both x and y (since \(y^2\) appears in the numerator and \(x^2 + y^2\) in the denominator), and the region D is symmetric about both the x-axis and the y-axis. Therefore, the contributions from regions above and below the x-axis, and to the left and right of the y-axis, will cancel each other out. Thus, the integral over the entire region D is zero: \[ \int_{\Gamma} \left(\frac{x\,dy - y\,dx}{x^2 + y^2}\right) = 0 \] This is the solution to the given problem using Green's theorem. |
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| kimi | To solve the integral \(\int_{\Gamma} \frac{x\,dy - y\,dx}{x^2 + y^2}\), we can use the properties of complex analysis. Let's consider the function \(f(z) = \frac{z}{z^2}\), where \(z = x + iy\) is a complex number and \(i\) is the imaginary unit. The integrand can be written as the differential form \(f'(z)\,dz\), where \(f'(z)\) is the derivative of \(f(z)\) with respect to \(z\). First, let's find the derivative of \(f(z)\): \[ f(z) = \frac{z}{z^2} = \frac{x + iy}{(x + iy)^2} = \frac{x + iy}{x^2 - y^2 + 2ixy} \] \[ f'(z) = \frac{d}{dz} \left( \frac{z}{z^2} \right) \] Using the quotient rule for derivatives, we get: \[ f'(z) = \frac{(z^2)' \cdot z - z' \cdot (z^2)}{(z^2)^2} = \frac{2z \cdot z - 1 \cdot z^2}{z^4} = \frac{2z^2 - z^2}{z^4} = \frac{z}{z^4} \] \[ f'(z) = \frac{1}{z^3} \] Now, let's rewrite the integral in terms of \(z\) and \(dz\): \[ \int_{\Gamma} \frac{x\,dy - y\,dx}{x^2 + y^2} = \int_{\Gamma} f'(z)\,dz \] According to Cauchy's integral theorem, for a function that is analytic (holomorphic) in a simply connected domain, the integral of the function over any closed curve in that domain is zero. However, \(f(z) = \frac{z}{z^2}\) has a singularity at \(z = 0\), which is enclosed by the curve \(\Gamma\). Therefore, we cannot directly apply Cauchy's integral theorem. Instead, we can use the argument principle or the residue theorem. The function \(f(z)\) has a simple pole at \(z = 0\), and the residue at this pole is the coefficient of \(\frac{1}{z}\) in the Laurent series expansion of \(f(z)\) around \(z = 0\). The Laurent series expansion of \(f(z)\) is: \[ f(z) = \frac{1}{z} - 1 \] The residue at \(z = 0\) is the coefficient of \(\frac{1}{z}\), which is \(1\). The residue theorem states that for a function with isolated singularities inside a closed curve \(\Gamma\), the integral of the function over \(\Gamma\) is \(2\pi i\) times the sum of the residues of the function at the singularities inside \(\Gamma\). In this case, since there is only one singularity at \(z = 0\) and its residue is \(1\), we have: \[ \int_{\Gamma} f'(z)\,dz = 2\pi i \cdot 1 = 2\pi i \] Therefore, the solution to the integral is: \[ \int_{\Gamma} \frac{x\,dy - y\,dx}{x^2 + y^2} = 2\pi i \] |
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| wenxinyiyan | Let $\Gamma$ be a circle of radius $r$ centered at the origin. We can parametrize $\Gamma$ by $x = r\cos(t)$ and $y = r\sin(t)$ for $t \in [0, 2\pi]$. Then $dx = -r\sin(t)dt$ and $dy = r\cos(t)dt$. We have $$\int_{\Gamma} \frac{x*dy - y*dx}{x^2 + y^2} = \int_0^{2\pi} \frac{r\cos(t)*r\cos(t)dt - r\sin(t)*(-r\sin(t)dt)}{r^2\cos^2(t) + r^2\sin^2(t)}.$$This simplifies to $$\int_0^{2\pi} \frac{r^2\cos^2(t) + r^2\sin^2(t)}{r^2\cos^2(t) + r^2\sin^2(t)} dt = \int_0^{2\pi} dt = 2\pi.$$Now let $\Gamma$ be any piecewise smooth, closed curve that encloses the origin but does not pass through it. By the Cauchy-Goursat theorem, the integral $$\int_{\Gamma} \frac{x*dy - y*dx}{x^2 + y^2}$$is equal to the integral over a circle of radius $r$ centered at the origin, which we have shown to be $2\pi$. |
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| zhipu | To solve the integral \(\int_{\Gamma} \frac{x \, dy - y \, dx}{x^2 + y^2}\), where \(\Gamma\) is any piecewise smooth, closed curve that encloses the origin but does not pass through it, we can use the concept of complex analysis, specifically the residue theorem. ### Step-by-Step Solution: 1. **Convert to Complex Form:** Let \(z = x + iy\), where \(x\) and \(y\) are real numbers, and \(i\) is the imaginary unit. Then, \(dz = dx + i \, dy\), and its conjugate \(\bar{z} = x - iy\) gives \(d\bar{z} = dx - i \, dy\). 2. **Express the Integral in Terms of \(z\) and \(\bar{z}\):** Notice that: \[ x \, dy - y \, dx = \text{Im}(z \, dz) \] and \[ x^2 + y^2 = |z|^2 = z \bar{z}. \] Therefore, the integral can be rewritten as: \[ \int_{\Gamma} \frac{x \, dy - y \, dx}{x^2 + y^2} = \int_{\Gamma} \frac{\text{Im}(z \, dz)}{z \bar{z}}. \] 3. **Simplify Using Complex Analysis:** The integral can be further simplified by noting that: \[ \frac{\text{Im}(z \, dz)}{z \bar{z}} = \frac{1}{2i} \frac{z \, dz}{z \bar{z}} - \frac{1}{2i} \frac{\bar{z} \, dz}{z \bar{z}} = \frac{1}{2i} \left( \frac{dz}{\bar{z}} - \frac{d\bar{z}}{z} \right). \] Since \(\bar{z} = \frac{|z|^2}{z}\), we have: \[ \frac{dz}{\bar{z}} = \frac{z \, dz}{|z|^2}. \] Thus, the integral becomes: \[ \int_{\Gamma} \frac{x \, dy - y \, dx}{x^2 + y^2} = \frac{1}{2i} \int_{\Gamma} \left( \frac{dz}{\bar{z}} - \frac{d\bar{z}}{z} \right). \] 4. **Residue Theorem Application:** The integrand \(\frac{1}{z}\) has a simple pole at \(z = 0\). According to the residue theorem, for a closed curve \(\Gamma\) enclosing the origin: \[ \oint_{\Gamma} \frac{dz}{z} = 2\pi i. \] Similarly, \[ \oint_{\Gamma} \frac{d\bar{z}}{\bar{z}} = 2\pi i. \] 5. **Combine Results:** Therefore, the integral becomes: \[ \int_{\Gamma} \frac{x \, dy - y \, dx}{x^2 + y^2} = \frac{1}{2i} \left( 2\pi i - 2\pi i \right) = \frac{1}{2i} \cdot 0 = 0. \] ### Correction and Final Answer: The above steps seem to have a mistake in the simplification. Let's correct it by directly using the known result for this type of integral. The integral \(\int_{\Gamma} \frac{x \, dy - y \, dx}{x^2 + y^2}\) is a standard form in complex analysis and is known to evaluate to \(2\pi\) times the winding number around the origin. Since \(\Gamma\) encloses the origin once (winding number = 1): \[ \int_{\Gamma} \frac{x \, dy - y \, dx}{x^2 + y^2} = 2\pi. \] ### Final Solution: \[ \boxed{2\pi} \] This result is derived from the fact that the integral represents the circulation of the vector field \(\frac{(x, -y)}{x^2 + y^2}\) around the origin, which is equivalent to the circulation of the field \(\frac{(y, -x)}{x^2 + y^2}\), both of which yield \(2\pi\) for a curve enclosing the origin once. |
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| gpto1 | To compute the integral: \[ I = \int_{\Gamma} \frac{x\,dy - y\,dx}{x^2 + y^2} \] where \(\Gamma\) is any piecewise smooth, closed curve that encloses the origin but does not pass through it, we can proceed as follows: **Step 1: Express the integrand in terms of \(d\theta\) using polar coordinates.** Let’s parameterize the curve using polar coordinates. Let \( (x, y) = (r \cos\theta, r \sin\theta) \). We compute the differentials: \[ dx = -r \sin\theta\, d\theta + \cos\theta\, dr \\ dy = r \cos\theta\, d\theta + \sin\theta\, dr. \] Since we are integrating over a closed curve that doesn't pass through the origin (\( r \neq 0 \)), we can assume \( r \) is not zero and, for a fixed curve, \( r \) is a function of \(\theta\). Compute \( x\,dy - y\,dx \): \[ x\,dy - y\,dx = [r \cos\theta][r \cos\theta\, d\theta + \sin\theta\, dr] - [r \sin\theta][-r \sin\theta\, d\theta + \cos\theta\, dr] \\ = r^2 \cos^2\theta\, d\theta + r \cos\theta \sin\theta\, dr + r^2 \sin^2\theta\, d\theta - r \sin\theta \cos\theta\, dr \\ = r^2 (\cos^2\theta + \sin^2\theta)\, d\theta \\ = r^2\, d\theta. \] Since \( x^2 + y^2 = r^2 \), the integrand simplifies to: \[ \frac{x\,dy - y\,dx}{x^2 + y^2} = \frac{r^2\, d\theta}{r^2} = d\theta. \] **Step 2: Evaluate the integral.** Therefore, the integral becomes: \[ I = \int_{\Gamma} d\theta = \Delta \theta, \] where \( \Delta \theta \) represents the total change in the angle \( \theta \) as we traverse the curve \(\Gamma\). **Step 3: Relate \( \Delta \theta \) to the winding number.** Since \(\Gamma\) is closed and encloses the origin, \( \Delta \theta \) corresponds to \( 2\pi n \), where \( n \) is the winding number (the number of times the curve winds around the origin). **Conclusion:** \[ I = \int_{\Gamma} \frac{x\,dy - y\,dx}{x^2 + y^2} = 2\pi n, \] where \( n \) is the integer winding number of the curve \(\Gamma\) around the origin. **Answer:** The value of the integral is 2 π times the winding number of Γ; that is, ∫ = 2π × n. |
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| deepseekv3 | Alright, I have this integral to compute: \[ \int_{\Gamma} \frac{x \, dy - y \, dx}{x^2 + y^2} \] where \(\Gamma\) is any piecewise smooth, closed curve that encloses the origin but does not pass through it. First, I need to understand what this integral represents. It looks like a line integral around a closed curve \(\Gamma\). The integrand \(\frac{x \, dy - y \, dx}{x^2 + y^2}\) seems familiar, but I'm not entirely sure where to start. **Step 1: Recognize the form of the integrand** The expression \(\frac{x \, dy - y \, dx}{x^2 + y^2}\) reminds me of the differential form associated with the angle in polar coordinates. In polar coordinates, \(x = r \cos \theta\) and \(y = r \sin \theta\), so: \[ \frac{x \, dy - y \, dx}{x^2 + y^2} = \frac{r \cos \theta \, d(r \sin \theta) - r \sin \theta \, d(r \cos \theta)}{r^2} \] Let's compute \(d(r \sin \theta)\) and \(d(r \cos \theta)\): \[ d(r \sin \theta) = \sin \theta \, dr + r \cos \theta \, d\theta \] \[ d(r \cos \theta) = \cos \theta \, dr - r \sin \theta \, d\theta \] Substituting these back into the integrand: \[ \frac{r \cos \theta (\sin \theta \, dr + r \cos \theta \, d\theta) - r \sin \theta (\cos \theta \, dr - r \sin \theta \, d\theta)}{r^2} \] Simplifying the numerator: \[ r \cos \theta \sin \theta \, dr + r^2 \cos^2 \theta \, d\theta - r \sin \theta \cos \theta \, dr + r^2 \sin^2 \theta \, d\theta \] The terms \(r \cos \theta \sin \theta \, dr\) and \(-r \sin \theta \cos \theta \, dr\) cancel each other out, leaving: \[ r^2 \cos^2 \theta \, d\theta + r^2 \sin^2 \theta \, d\theta = r^2 (\cos^2 \theta + \sin^2 \theta) \, d\theta = r^2 \, d\theta \] So, the integrand simplifies to: \[ \frac{r^2 \, d\theta}{r^2} = d\theta \] **Step 2: Simplify the integral** Now, the integral becomes: \[ \int_{\Gamma} d\theta \] This is simply the total change in the angle \(\theta\) as we go around the curve \(\Gamma\). Since \(\Gamma\) is a closed curve that encloses the origin, the angle \(\theta\) will increase by \(2\pi\) as we make one full loop around the origin. **Step 3: Compute the integral** Therefore, the value of the integral is: \[ \int_{\Gamma} d\theta = 2\pi \] **Step 4: Verify the result** To ensure this is correct, let's consider a specific simple closed curve that encloses the origin, such as the unit circle \(C\) defined by \(x = \cos t\), \(y = \sin t\) for \(t \in [0, 2\pi]\). Compute the integral over \(C\): \[ \int_{C} \frac{x \, dy - y \, dx}{x^2 + y^2} \] Substitute \(x = \cos t\), \(y = \sin t\): \[ dx = -\sin t \, dt, \quad dy = \cos t \, dt \] So, \[ x \, dy - y \, dx = \cos t (\cos t \, dt) - \sin t (-\sin t \, dt) = \cos^2 t \, dt + \sin^2 t \, dt = (\cos^2 t + \sin^2 t) \, dt = dt \] And, \[ x^2 + y^2 = \cos^2 t + \sin^2 t = 1 \] Thus, the integral becomes: \[ \int_{0}^{2\pi} \frac{dt}{1} = \int_{0}^{2\pi} dt = 2\pi \] This matches our earlier result, so I can be confident that the integral is indeed \(2\pi\). **Final Answer:** \[ \boxed{2\pi} \] |
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