| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve this problem using Green's Theorem, we first need to identify the functions \(P(x,y)\) and \(Q(x,y)\) in the given line integral. In this case, we have: \[ P(x,y) = y^3, \quad Q(x,y) = x^3 \] Green's Theorem states that for a positively oriented, piecewise-smooth, simple closed curve \(C\) in the plane and a function \(F(x,y) = P(x,y)i + Q(x,y)j\) defined on a region \(D\) that contains \(C\), the following is true: \[ \oint_{C} P\,dx + Q\,dy = \iint_{D} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA \] So, we calculate the partial derivatives: \[ \frac{\partial Q}{\partial x} = 3x^2, \quad \frac{\partial P}{\partial y} = 3y^2 \] So, the integrand in the double integral becomes: \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 3x^2 - 3y^2 \] The region \(D\) is the disk of radius 2 centered at the origin, so we can use polar coordinates to evaluate the double integral. In polar coordinates, \(x = r\cos\theta\) and \(y = r\sin\theta\), and the Jacobian of the transformation is \(r\). The double integral becomes: \[ \iint_{D} (3x^2 - 3y^2) dA = \int_{0}^{2\pi} \int_{0}^{2} (3r^2\cos^2\theta - 3r^2\sin^2\theta) r\,dr\,d\theta \] Simplifying the integrand: \[ \int_{0}^{2\pi} \int_{0}^{2} 3r^3(\cos^2\theta - \sin^2\theta)\,dr\,d\theta = \int_{0}^{2\pi} \int_{0}^{2} 3r^3\cos(2\theta)\,dr\,d\theta \] We can separate the integrals: \[ \int_{0}^{2\pi} \cos(2\theta)\,d\theta \int_{0}^{2} 3r^3\,dr \] The first integral is zero because \(\cos(2\theta)\) is an odd function over a full period: \[ \int_{0}^{2\pi} \cos(2\theta)\,d\theta = 0 \] The second integral evaluates to: \[ \int_{0}^{2} 3r^3\,dr = \left. \frac{3}{4}r^4 \right|_{0}^{2} = \frac{3}{4}(2^4) = 12 \] Since the first integral is zero, the entire integral is zero: \[ \oint_{C} y^3\,dx + x^3\,dy = 0 \] So, the solution is: \[ \boxed{0} \] |
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| deepseekr1 | To evaluate the line integral \(\oint_{C} y^3 \, dx + x^3 \, dy\) using Green's Theorem, where \(C\) is the positively oriented circle of radius 2 centered at the origin, we start by applying Green's Theorem: \[ \oint_{C} (P \, dx + Q \, dy) = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \] Here, \(P = y^3\) and \(Q = x^3\). We compute the partial derivatives: \[ \frac{\partial P}{\partial y} = 3y^2 \quad \text{and} \quad \frac{\partial Q}{\partial x} = 3x^2 \] Applying Green's Theorem, the line integral becomes: \[ \iint_{D} (3x^2 - 3y^2) \, dA = 3 \iint_{D} (x^2 - y^2) \, dA \] We convert to polar coordinates where \(x = r \cos \theta\), \(y = r \sin \theta\), and \(dA = r \, dr \, d\theta\). The integrand \(x^2 - y^2\) becomes \(r^2 \cos(2\theta)\). The region \(D\) is a disk of radius 2, so the limits are \(0 \leq r \leq 2\) and \(0 \leq \theta \leq 2\pi\). The integral becomes: \[ 3 \int_{0}^{2\pi} \int_{0}^{2} r^2 \cos(2\theta) \cdot r \, dr \, d\theta = 3 \int_{0}^{2\pi} \cos(2\theta) \, d\theta \int_{0}^{2} r^3 \, dr \] First, we evaluate the radial integral: \[ \int_{0}^{2} r^3 \, dr = \left[ \frac{1}{4} r^4 \right]_{0}^{2} = \frac{1}{4} (16) = 4 \] Next, we evaluate the angular integral: \[ \int_{0}^{2\pi} \cos(2\theta) \, d\theta = \left[ \frac{1}{2} \sin(2\theta) \right]_{0}^{2\pi} = 0 \] Multiplying the results of the integrals and the constant factor 3: \[ 3 \cdot 4 \cdot 0 = 0 \] Thus, the value of the line integral is: \[ \boxed{0} \] |
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| gpto1 | To evaluate the line integral \(\oint_C y^3\, dx + x^3\, dy\) using Green's Theorem, where \(C\) is the positively oriented circle of radius 2 centered at the origin, we proceed as follows. **Green's Theorem** states that for a positively oriented, simple closed curve \(C\) and a region \(D\) it encloses: \[ \oint_C P\, dx + Q\, dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx\, dy \] Given: - \( P = y^3 \) - \( Q = x^3 \) Compute the partial derivatives: \[ \frac{\partial Q}{\partial x} = \frac{\partial x^3}{\partial x} = 3x^2 \] \[ \frac{\partial P}{\partial y} = \frac{\partial y^3}{\partial y} = 3y^2 \] Thus, the integrand becomes: \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 3x^2 - 3y^2 = 3(x^2 - y^2) \] Now, the double integral over the disk \(D\) (with \(x^2 + y^2 \leq 4\)) is: \[ \iint_D 3(x^2 - y^2) \, dx\, dy \] We convert to polar coordinates where \(x = r\cos\theta\), \(y = r\sin\theta\), and \(dx\, dy = r\, dr\, d\theta\): \[ x^2 - y^2 = r^2\cos^2\theta - r^2\sin^2\theta = r^2(\cos^2\theta - \sin^2\theta) = r^2\cos(2\theta) \] Therefore, the integral becomes: \[ 3 \int_0^{2\pi} \int_0^2 r^2\cos(2\theta) \cdot r\, dr\, d\theta = 3 \int_0^{2\pi} \left( \int_0^2 r^3\, dr \right) \cos(2\theta)\, d\theta \] Compute the inner integral: \[ \int_0^2 r^3\, dr = \left[ \frac{r^4}{4} \right]_0^2 = \frac{16}{4} = 4 \] Substituting back: \[ 3 \times 4 \int_0^{2\pi} \cos(2\theta)\, d\theta = 12 \int_0^{2\pi} \cos(2\theta)\, d\theta \] But: \[ \int_0^{2\pi} \cos(2\theta)\, d\theta = 0 \] Because \(\cos(2\theta)\) is periodic over \([0, 2\pi]\) and its positive and negative areas cancel out. Therefore: \[ 12 \int_0^{2\pi} \cos(2\theta)\, d\theta = 12 \times 0 = 0 \] **Answer:** 0 |
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| GPT-4o | To solve the given problem using Green's Theorem, we will follow these steps: 1. **State Green's Theorem:** Green's Theorem relates a line integral around a simple closed curve \(C\) to a double integral over the plane region \(D\) bounded by \(C\). Specifically, \[ \oint_C P \, dx + Q \, dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA, \] where \(P(x, y)\) and \(Q(x, y)\) are functions of \(x\) and \(y\). 2. **Identify \(P\) and \(Q\):** In the given integral, \(P(x, y) = y^3\) and \(Q(x, y) = x^3\). 3. **Compute the partial derivatives:** \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^3) = 3x^2, \] \[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(y^3) = 3y^2. \] 4. **Substitute the derivatives into Green's Theorem:** \[ \oint_C y^3 \, dx + x^3 \, dy = \iint_D \left( 3x^2 - 3y^2 \right) \, dA. \] 5. **Describe the region \(D\):** The region \(D\) is the disk of radius 2 centered at the origin, given by \(x^2 + y^2 \leq 4\). 6. **Set up the double integral in polar coordinates:** In polar coordinates, \(x = r \cos \theta\) and \(y = r \sin \theta\), and the area element \(dA = r \, dr \, d\theta\). The region \(D\) in polar coordinates is given by \(0 \leq r \leq 2\) and \(0 \leq \theta \leq 2\pi\). Convert the integrand \(3x^2 - 3y^2\) to polar coordinates: \[ 3x^2 - 3y^2 = 3(r^2 \cos^2 \theta) - 3(r^2 \sin^2 \theta) = 3r^2 (\cos^2 \theta - \sin^2 \theta). \] The integral becomes: \[ \iint_D (3x^2 - 3y^2) \, dA = \int_0^{2\pi} \int_0^2 3r^2 (\cos^2 \theta - \sin^2 \theta) \, r \, dr \, d\theta. \] 7. **Simplify and evaluate the integral:** Factor out constants and integrate: \[ \int_0^{2\pi} \int_0^2 3r^3 (\cos^2 \theta - \sin^2 \theta) \, dr \, d\theta. \] \[ = 3 \int_0^{2\pi} (\cos^2 \theta - \sin^2 \theta) \left( \int_0^2 r^3 \, dr \right) d\theta. \] Evaluate the inner integral: \[ \int_0^2 r^3 \, dr = \left[ \frac{r^4}{4} \right]_0^2 = \frac{16}{4} = 4. \] Thus, the integral becomes: \[ 3 \int_0^{2\pi} 4 (\cos^2 \theta - \sin^2 \theta) \, d\theta = 12 \int_0^{2\pi} (\cos^2 \theta - \sin^2 \theta) \, d\theta. \] Since \(\cos^2 \theta - \sin^2 \theta = \cos 2\theta\), the integral further simplifies to: \[ 12 \int_0^{2\pi} \cos 2\theta \, d\theta. \] The integral of \(\cos 2\theta\) over one full period is zero: \[ \int_0^{2\pi} \cos 2\theta \, d\theta = 0. \] 8 |
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| llama3.1-70B-Instruct | ## Step 1: Understand the problem and identify the components of Green's Theorem We are given a line integral of the form $\oiint_{s} y^3 dx + x^3dy$ where $C$ is the positively oriented circle of radius 2 centered at the origin. To apply Green's Theorem, we need to identify the components $P$ and $Q$ of the vector field $\vec{F} = P\hat{i} + Q\hat{j}$, which in this case are $P = y^3$ and $Q = x^3$. ## Step 2: Apply Green's Theorem Green's Theorem states that $\oiint_{C} Pdx + Qdy = \iint_{R} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dA$, where $R$ is the region enclosed by the curve $C$. In this case, we have $\frac{\partial Q}{\partial x} = 3x^2$ and $\frac{\partial P}{\partial y} = 3y^2$. ## Step 3: Evaluate the double integral The region $R$ is the circle of radius 2 centered at the origin, which can be described in polar coordinates as $0 \leq r \leq 2$ and $0 \leq \theta \leq 2\pi$. The double integral becomes $\iint_{R} (3x^2 - 3y^2)dA = \int_{0}^{2\pi} \int_{0}^{2} (3r^2\cos^2\theta - 3r^2\sin^2\theta)rdrd\theta$. ## Step 4: Simplify and evaluate the integral Using the trigonometric identity $\cos^2\theta - \sin^2\theta = \cos(2\theta)$, the integral simplifies to $\int_{0}^{2\pi} \int_{0}^{2} 3r^3\cos(2\theta)drd\theta$. Evaluating the integral, we get $\int_{0}^{2\pi} 3\cdot\frac{2^4}{4}\cos(2\theta)d\theta = 12\int_{0}^{2\pi} \cos(2\theta)d\theta = 12\cdot\frac{1}{2}\sin(2\theta)\Big|_{0}^{2\pi} = 0$. The final answer is: $\boxed{0}$ |
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| llama3.1-405B-Instruct | We can use Green's Theorem to evaluate the line integral by converting it to a double integral over the region enclosed by the circle. First, we need to find the region enclosed by the circle. The circle is centered at the origin and has a radius of 2, so the region is the disk of radius 2 centered at the origin. Next, we need to find the functions P and Q such that: $$\oiint_{s} y^3 dx + x^3dy = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx dy$$ Comparing the line integral with the formula, we can see that: $$P = y^3$$ $$Q = x^3$$ Now, we can find the partial derivatives: $$\frac{\partial P}{\partial y} = 3y^2$$ $$\frac{\partial Q}{\partial x} = 3x^2$$ Substituting these expressions into the formula, we get: $$\oiint_{s} y^3 dx + x^3dy = \iint_{D} \left( 3x^2 - 3y^2 \right) dx dy$$ Now, we can evaluate the double integral over the region D. Since the region is a disk of radius 2 centered at the origin, we can use polar coordinates to evaluate the integral. In polar coordinates, the region D is described by: $$0 \leq r \leq 2$$ $$0 \leq \theta \leq 2\pi$$ We can convert the integrand to polar coordinates using the following substitutions: $$x = r \cos \theta$$ $$y = r \sin \theta$$ Substituting these expressions into the integrand, we get: $$3x^2 - 3y^2 = 3r^2 \cos^2 \theta - 3r^2 \sin^2 \theta$$ Now, we can evaluate the double integral: $$\iint_{D} \left( 3x^2 - 3y^2 \right) dx dy = \int_{0}^{2\pi} \int_{0}^{2} \left( 3r^2 \cos^2 \theta - 3r^2 \sin^2 \theta \right) r dr d\theta$$ Evaluating the inner integral, we get: $$\int_{0}^{2} \left( 3r^2 \cos^2 \theta - 3r^2 \sin^2 \theta \right) r dr = \left[ \frac{3r^4}{4} \cos^2 \theta - \frac{3r^4}{4} \sin^2 \theta \right]_{0}^{2}$$ $$= \frac{3(2)^4}{4} \cos^2 \theta - \frac{3(2)^4}{4} \sin^2 \theta$$ $$= 12 \cos^2 \theta - 12 \sin^2 \theta$$ Now, we can evaluate the outer integral: $$\int_{0}^{2\pi} \left( 12 \cos^2 \theta - 12 \sin^2 \theta \right) d\theta = \left[ 12 \frac{\theta}{2} + 12 \frac{\sin 2\theta}{4} \right]_{0}^{2\pi}$$ $$= 12 \pi$$ Therefore, the value of the line integral is: $$\oiint_{s} y^3 dx + x^3dy = 12 \pi$$ |
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| DeepSeek-Coder-V2 | To solve the given problem using Green's Theorem, we need to evaluate the line integral \(\oint_C y^3 dx + x^3 dy\) over the positively oriented circle of radius 2 centered at the origin. Green's Theorem states that for a positively oriented closed curve \(C\) enclosing a region \(D\), \[ \oint_C P dx + Q dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \] In our case, \(P = y^3\) and \(Q = x^3\). We need to compute the partial derivatives: \[ \frac{\partial Q}{\partial x} = \frac{\partial (x^3)}{\partial x} = 3x^2 \] \[ \frac{\partial P}{\partial y} = \frac{\partial (y^3)}{\partial y} = 3y^2 \] Substituting these into Green's Theorem, we get: \[ \oint_C y^3 dx + x^3 dy = \iint_D \left( 3x^2 - 3y^2 \right) dA \] We can factor out the 3: \[ \oint_C y^3 dx + x^3 dy = 3 \iint_D \left( x^2 - y^2 \right) dA \] Next, we need to evaluate the double integral over the region \(D\), which is the disk of radius 2 centered at the origin. We can use polar coordinates to simplify the integration. In polar coordinates, \(x = r \cos \theta\) and \(y = r \sin \theta\), and the Jacobian of the transformation is \(r\). The region \(D\) is described by \(0 \leq r \leq 2\) and \(0 \leq \theta \leq 2\pi\). Substituting \(x\) and \(y\) in terms of \(r\) and \(\theta\): \[ x^2 - y^2 = (r \cos \theta)^2 - (r \sin \theta)^2 = r^2 (\cos^2 \theta - \sin^2 \theta) = r^2 \cos(2\theta) \] Thus, the integral becomes: \[ 3 \iint_D (x^2 - y^2) dA = 3 \int_0^{2\pi} \int_0^2 r^2 \cos(2\theta) r dr d\theta \] Simplifying the integrand: \[ 3 \int_0^{2\pi} \int_0^2 r^3 \cos(2\theta) dr d\theta \] First, we integrate with respect to \(r\): \[ \int_0^2 r^3 dr = \left[ \frac{r^4}{4} \right]_0^2 = \frac{2^4}{4} - \frac{0^4}{4} = \frac{16}{4} = 4 \] So the integral becomes: \[ 3 \int_0^{2\pi} 4 \cos(2\theta) d\theta = 12 \int_0^{2\pi} \cos(2\theta) d\theta \] Next, we integrate with respect to \(\theta\): \[ \int_0^{2\pi} \cos(2\theta) d\theta = \left[ \frac{\sin(2\theta)}{2} \right]_0^{2\pi} = \frac{\sin(4\pi)}{2} - \frac{\sin(0)}{2} = 0 \] Thus, the entire integral evaluates to: \[ 12 \cdot 0 = 0 \] Therefore, the value of the line integral is: \[ \boxed{0} \] |
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| Qwen2-72B-Instruct | To solve this problem using Green's Theorem, we first need to identify the functions $P(x,y)$ and $Q(x,y)$ in the given line integral. In this case, we have: \[ P(x,y) = y^3, \quad Q(x,y) = x^3 \] Green's Theorem states that for a positively oriented, piecewise-smooth, simple closed curve $C$ in the plane and a function $F(x,y) = P(x,y)i + Q(x,y)j$ defined on a region $D$ that contains $C$, the following is true: \[ \oint_{C} P\,dx + Q\,dy = \iint_{D} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA \] So, we calculate the partial derivatives: \[ \frac{\partial Q}{\partial x} = 3x^2, \quad \frac{\partial P}{\partial y} = 3y^2 \] So, the integrand in the double integral becomes: \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 3x^2 - 3y^2 \] The region $D$ is the disk of radius 2 centered at the origin, so we can use polar coordinates to evaluate the double integral. In polar coordinates, $x = r\cos\theta$ and $y = r\sin\theta$, and the Jacobian of the transformation is $r$. The double integral becomes: \[ \iint_{D} (3x^2 - 3y^2) dA = \int_{0}^{2\pi} \int_{0}^{2} (3r^2\cos^2\theta - 3r^2\sin^2\theta) r\,dr\,d\theta \] Simplifying the integrand: \[ \int_{0}^{2\pi} \int_{0}^{2} 3r^3(\cos^2\theta - \sin^2\theta)\,dr\,d\theta = \int_{0}^{2\pi} \int_{0}^{2} 3r^3\cos(2\theta)\,dr\,d\theta \] We can separate the integrals: \[ \int_{0}^{2\pi} \cos(2\theta)\,d\theta \int_{0}^{2} 3r^3\,dr \] The first integral is zero because $\cos(2\theta)$ is an odd function over a full period: \[ \int_{0}^{2\pi} \cos(2\theta)\,d\theta = 0 \] The second integral evaluates to: \[ \int_{0}^{2} 3r^3\,dr = \left. \frac{3}{4}r^4 \right|_{0}^{2} = \frac{3}{4}(2^4) = 12 \] Since the first integral is zero, the entire integral is zero: \[ \oint_{C} y^3\,dx + x^3\,dy = 0 \] So, the solution is: \[ \boxed{0} \] |
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| baichuan4 | To solve this problem, we will use Green's Theorem, which relates the line integral around a closed curve $C$ to the double integral over the region $D$ enclosed by $C$. Green's Theorem states that for a two-dimensional region $D$ with a piecewise-smooth boundary $C$, oriented counterclockwise, the line integral of a vector field $\mathbf{F} = P\mathbf{i} + Q\mathbf{j}$ around $C$ is equal to the double integral over $D$ of the partial derivatives of $Q$ with respect to $x$ minus the partial derivatives of $P$ with respect to $y$: $$\oint_C P \,dx + Q \,dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$$ In our case, we have $P(x, y) = y^3$ and $Q(x, y) = x^3$, so we need to calculate the partial derivatives: $$\frac{\partial Q}{\partial x} = 3x^2$$ $$\frac{\partial P}{\partial y} = 3y^2$$ Substituting these into Green's Theorem, we get: $$\oint_C y^3 \,dx + x^3 \,dy = \iint_D (3x^2 - 3y^2) \,dA$$ Now, we need to determine the region $D$ over which we are integrating. Since $C$ is the circle of radius 2 centered at the origin, $D$ is the disk of radius 2 centered at the origin. We can parameterize this disk using polar coordinates: $$x = r\cos(\theta)$$ $$y = r\sin(\theta)$$ where $0 \leq r \leq 2$ and $0 \leq \theta \leq 2\pi$. The Jacobian determinant for the transformation to polar coordinates is $r$, so our integral becomes: $$\int_0^{2\pi} \int_0^2 (3r^2\cos^2(\theta) - 3r^2\sin^2(\theta)) r \,dr \,d\theta$$ Simplifying, we get: $$\int_0^{2\pi} \int_0^2 3r^3(\cos^2(\theta) - \sin^2(\theta)) \,dr \,d\theta$$ Since $\cos^2(\theta) - \sin^2(\theta) = \cos(2\theta)$, we have: $$\int_0^{2\pi} \int_0^2 3r^3 \cos(2\theta) \,dr \,d\theta$$ Now, we can integrate with respect to $r$ first: $$\int_0^{2\pi} \left[\frac{3}{4}r^4\right]_0^2 \cos(2\theta) \,d\theta = \int_0^{2\pi} 12 \cos(2\theta) \,d\theta$$ Finally, we integrate with respect to $\theta$: $$\left[6\sin(2\theta)\right]_0^{2\pi} = 0$$ Therefore, the value of the line integral is 0. |
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| doubao | Let $D$ be the disk enclosed by the circle $C$. Using Green's Theorem, we have: $\oiint_{s} y^3 dx + x^3dy = \iint_{D} (\frac{\partial(x^3)}{\partial x} - \frac{\partial(y^3)}{\partial y}) dA$. $\frac{\partial(x^3)}{\partial x} = 3x^2$ and $\frac{\partial(y^3)}{\partial y} = 3y^2$. So, the integral becomes: $\iint_{D} (3x^2 - 3y^2) dA$. Converting to polar coordinates, $x = r\cos\theta$, $y = r\sin\theta$, and the area element $dA = r dr d\theta$. The limits of integration for $r$ are from 0 to 2 and for $\theta$ are from 0 to $2\pi$. $\int_{0}^{2\pi} \int_{0}^{2} (3r^2\cos^2\theta - 3r^2\sin^2\theta) r dr d\theta$. $= 3\int_{0}^{2\pi} \int_{0}^{2} r^3(\cos^2\theta - \sin^2\theta) dr d\theta$. Using the identity $\cos 2\theta = \cos^2\theta - \sin^2\theta$, we get: $= 3\int_{0}^{2\pi} \int_{0}^{2} r^3\cos 2\theta dr d\theta$. Integrating with respect to $r$ first: $= 3\int_{0}^{2\pi} [\frac{r^4}{4}\cos 2\theta]_{0}^{2} d\theta = 3\int_{0}^{2\pi} 4\cos 2\theta d\theta$. Integrating with respect to $\theta$: $= 3[2\sin 2\theta]_{0}^{2\pi} = 0$. |
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| minimax | Green's Theorem relates a line integral around a simple closed curve $C$ to a double integral over the plane region $D$ bounded by $C$. It states that if $P(x, y)$ and $Q(x, y)$ are continuously differentiable functions on an open region that contains $D$, then: $$ \oint_C P(x, y)dx + Q(x, y)dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA $$ In this problem, we have $P(x, y) = y^3$ and $Q(x, y) = x^3$. We need to compute the partial derivatives of $P$ and $Q$: $$ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(y^3) = 3y^2 $$ $$ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^3) = 3x^2 $$ Now, we apply Green's Theorem: $$ \oint_C y^3 dx + x^3dy = \iint_D (3x^2 - 3y^2) dA $$ Since $C$ is the positively oriented circle of radius 2 centered at the origin, the region $D$ is the disk of radius 2. We can use polar coordinates to evaluate the double integral, where $x = r\cos(\theta)$ and $y = r\sin(\theta)$. The Jacobian determinant for the transformation from Cartesian to polar coordinates is $r$, so the area element $dA$ becomes $r dr d\theta$. The integral becomes: $$ \iint_D (3x^2 - 3y^2) dA = \int_0^{2\pi} \int_0^2 (3r^2\cos^2(\theta) - 3r^2\sin^2(\theta)) r dr d\theta $$ Simplify the integrand: $$ = \int_0^{2\pi} \int_0^2 3r^3(\cos^2(\theta) - \sin^2(\theta)) dr d\theta $$ Using the trigonometric identity $\cos^2(\theta) - \sin^2(\theta) = \cos(2\theta)$, we get: $$ = \int_0^{2\pi} \int_0^2 3r^3\cos(2\theta) dr d\theta $$ Now we can integrate with respect to $r$ first: $$ = \int_0^{2\pi} \cos(2\theta) \left[ \int_0^2 3r^3 dr \right] d\theta $$ $$ = \int_0^{2\pi} \cos(2\theta) \left[ \frac{3}{4}r^4 \right]_0^2 d\theta $$ $$ = \int_0^{2\pi} \cos(2\theta) \left[ \frac{3}{4}(2^4) \right] d\theta $$ $$ = \int_0^{2\pi} \cos(2\theta) \cdot 3 \cdot 4 d\theta $$ $$ = 12 \int_0^{2\pi} \cos(2\theta) d\theta $$ Now integrate with respect to $\theta$: $$ = 12 \left[ \frac{1}{2}\sin(2\theta) \right]_0^{2\pi} $$ $$ = 6 \left[ \sin(4\pi) - \sin(0) \right] $$ Since $\sin(4\pi) = \sin(0) = 0$, the integral evaluates to: $$ = 6 \cdot 0 = 0 $$ Therefore, the line integral of $y^3 dx + x^3dy$ over the circle $C$ is $0$. |
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| kimi | Green's Theorem relates a line integral around a simple closed curve to a double integral over the plane region it encloses. The theorem states that for a vector field $\mathbf{F} = P(x, y)\mathbf{i} + Q(x, y)\mathbf{j}$, the line integral around a curve $C$ is equal to the double integral over the region $D$ enclosed by $C$: $$ \oint_C P\,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA $$ In your problem, the vector field is $\mathbf{F} = y^3\mathbf{i} + x^3\mathbf{j}$, so $P(x, y) = y^3$ and $Q(x, y) = x^3$. We need to compute the partial derivatives of $P$ and $Q$: $$ \frac{\partial Q}{\partial x} = \frac{\partial x^3}{\partial x} = 3x^2 $$ $$ \frac{\partial P}{\partial y} = \frac{\partial y^3}{\partial y} = 3y^2 $$ Now, we apply Green's Theorem: $$ \oint_C y^3\,dx + x^3\,dy = \iint_D \left(\frac{\partial x^3}{\partial x} - \frac{\partial y^3}{\partial y}\right) dA = \iint_D (3x^2 - 3y^2) dA $$ The region $D$ is the interior of the circle of radius 2 centered at the origin. We can use polar coordinates to evaluate the double integral, where $x = r\cos\theta$ and $y = r\sin\theta$, and $dA = r\,dr\,d\theta$. The limits for $r$ will be from 0 to 2 (the radius of the circle), and the limits for $\theta$ will be from 0 to $2\pi$ (the full circle). $$ \iint_D (3x^2 - 3y^2) dA = \int_0^{2\pi} \int_0^2 (3(r\cos\theta)^2 - 3(r\sin\theta)^2) r\,dr\,d\theta $$ Simplify the integrand: $$ = \int_0^{2\pi} \int_0^2 (3r^3\cos^2\theta - 3r^3\sin^2\theta)\,dr\,d\theta $$ Now, integrate with respect to $r$: $$ = \int_0^{2\pi} \left[\frac{3r^4}{4}\cos^2\theta - \frac{3r^4}{4}\sin^2\theta\right]_0^2\,d\theta $$ Evaluate the limits: $$ = \int_0^{2\pi} \left(\frac{3 \cdot 2^4}{4}\cos^2\theta - \frac{3 \cdot 2^4}{4}\sin^2\theta\right)\,d\theta $$ $$ = \int_0^{2\pi} \left(12\cos^2\theta - 12\sin^2\theta\right)\,d\theta $$ Now, integrate with respect to $\theta$: $$ = \left[12\int_0^{2\pi} \cos^2\theta\,d\theta - 12\int_0^{2\pi} \sin^2\theta\,d\theta\right] $$ Using the identity $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$ and $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$, we get: $$ = 12\left[\int_0^{2\pi} \frac{1 + \cos(2\theta)}{2}\,d\theta - \int_0^{2\pi} \frac{1 - \cos(2\theta)}{2}\,d\theta\right] $$ $$ = 12\left[\frac{1}{2}\int_0^{2\pi} 1\,d\theta + \frac |
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| wenxinyiyan | 首先，我们需要找到向量场$F(x, y) = (y^3, x^3)$的散度。散度是一个标量函数，表示向量场在某一点的“流入”和“流出”之差。 散度的计算公式为： $$ \text{div} F = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} $$ 其中，$F_x$ 和 $F_y$ 分别是向量场$F$的$x$和$y$分量。 对于给定的向量场$F(x, y) = (y^3, x^3)$，我们有： $$ \frac{\partial F_x}{\partial x} = \frac{\partial y^3}{\partial x} = 0, \quad \frac{\partial F_y}{\partial y} = \frac{\partial x^3}{\partial y} = 0 $$ 因此，散度$\text{div} F = 0 + 0 = 0$。 接下来，我们使用散度定理（Green's Theorem的一种形式）来计算给定的环路积分。散度定理告诉我们： $$ \oiint_{S} (\text{div} F) dA = \oint_{C} F \cdot ds $$ 其中，$S$是一个封闭曲面，$C$是$S$的边界，$dA$是面积元素，$ds$是弧长元素。 由于我们已经计算出$\text{div} F = 0$，所以左边的积分等于0。因此，右边的环路积分也必须等于0： $$ \oint_{C} F \cdot ds = \oint_{C} y^3 dx + x^3 dy = 0 $$ 所以，给定的环路积分的值为0。 |
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| zhipu | To evaluate the line integral \(\oint_{C} y^3 \, dx + x^3 \, dy\) where \(C\) is the positively oriented circle of radius 2 centered at the origin using Green's Theorem, we follow these steps: ### Step 1: State Green's Theorem Green's Theorem states that for a positively oriented, simple closed curve \(C\) enclosing a region \(D\), \[ \oint_{C} (P \, dx + Q \, dy) = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA, \] where \(P\) and \(Q\) are functions of \(x\) and \(y\) with continuous partial derivatives on an open region that contains \(D\). ### Step 2: Identify \(P\) and \(Q\) In our problem, \(P = y^3\) and \(Q = x^3\). ### Step 3: Compute the Partial Derivatives We need to compute \(\frac{\partial Q}{\partial x}\) and \(\frac{\partial P}{\partial y}\): \[ \frac{\partial Q}{\partial x} = \frac{\partial (x^3)}{\partial x} = 3x^2, \] \[ \frac{\partial P}{\partial y} = \frac{\partial (y^3)}{\partial y} = 3y^2. \] ### Step 4: Apply Green's Theorem Using Green's Theorem, we convert the line integral to a double integral over the region \(D\) enclosed by \(C\): \[ \oint_{C} y^3 \, dx + x^3 \, dy = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA = \iint_{D} (3x^2 - 3y^2) \, dA. \] ### Step 5: Set Up the Double Integral in Polar Coordinates Since \(D\) is a circle of radius 2 centered at the origin, it is convenient to use polar coordinates. In polar coordinates, \[ x = r \cos \theta, \quad y = r \sin \theta, \] and the area element \(dA\) is \(r \, dr \, d\theta\). The region \(D\) is described by \(0 \leq r \leq 2\) and \(0 \leq \theta \leq 2\pi\). ### Step 6: Transform the Integral Transform the integrand \(3x^2 - 3y^2\) into polar coordinates: \[ 3x^2 - 3y^2 = 3(r \cos \theta)^2 - 3(r \sin \theta)^2 = 3r^2 \cos^2 \theta - 3r^2 \sin^2 \theta = 3r^2 (\cos^2 \theta - \sin^2 \theta). \] Using the trigonometric identity \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\), we get: \[ 3r^2 (\cos^2 \theta - \sin^2 \theta) = 3r^2 \cos 2\theta. \] ### Step 7: Evaluate the Double Integral Now, the double integral becomes: \[ \iint_{D} (3x^2 - 3y^2) \, dA = \int_{0}^{2\pi} \int_{0}^{2} 3r^2 \cos 2\theta \, r \, dr \, d\theta. \] Simplify the integrand: \[ \int_{0}^{2\pi} \int_{0}^{2} 3r^3 \cos 2\theta \, dr \, d\theta. \] Separate the integrals: \[ 3 \int_{0}^{2\pi} \cos 2\theta \, d\theta \int_{0}^{2} r^3 \, dr. \] Evaluate the \(r\)-integral: \[ \int_{0}^{2} r^3 \, dr = \left[ \frac{r^4}{4} \right]_{0}^{2} = \frac{2^4}{4} - \frac{0^4}{4} = \frac{16}{4} = 4. \] Evaluate the \(\theta\)-integral: \[ \int_{0}^{2\pi} \cos 2\theta \, d\theta = \left[ \frac{\sin 2\theta}{2} \right]_{0}^{2\pi} = \frac{\sin 4\pi}{2} - \frac{\sin 0}{2} = 0 - 0 = 0. \] ### Step 8: Combine the Results Since the \(\theta\)-integral evaluates to zero, the entire double integral is zero: \[ 3 \int_{0}^{2\pi} \cos 2\theta \, d\theta \int_{0}^{2} r^3 \, dr = 3 \cdot 0 \cdot 4 = 0. \] ### Conclusion Thus, the value of the line integral \(\oint_{C} y^3 \, dx + x^3 \, dy\) is: \[ \boxed{0}. \] |
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| gpto1 | To evaluate the line integral \(\oint_C y^3\, dx + x^3\, dy\) using Green's Theorem, where \(C\) is the positively oriented circle of radius 2 centered at the origin, we proceed as follows. **Green's Theorem** states that for a positively oriented, simple closed curve \(C\) and a region \(D\) it encloses: \[ \oint_C P\, dx + Q\, dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx\, dy \] Given: - \( P = y^3 \) - \( Q = x^3 \) Compute the partial derivatives: \[ \frac{\partial Q}{\partial x} = \frac{\partial x^3}{\partial x} = 3x^2 \] \[ \frac{\partial P}{\partial y} = \frac{\partial y^3}{\partial y} = 3y^2 \] Thus, the integrand becomes: \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 3x^2 - 3y^2 = 3(x^2 - y^2) \] Now, the double integral over the disk \(D\) (with \(x^2 + y^2 \leq 4\)) is: \[ \iint_D 3(x^2 - y^2) \, dx\, dy \] We convert to polar coordinates where \(x = r\cos\theta\), \(y = r\sin\theta\), and \(dx\, dy = r\, dr\, d\theta\): \[ x^2 - y^2 = r^2\cos^2\theta - r^2\sin^2\theta = r^2(\cos^2\theta - \sin^2\theta) = r^2\cos(2\theta) \] Therefore, the integral becomes: \[ 3 \int_0^{2\pi} \int_0^2 r^2\cos(2\theta) \cdot r\, dr\, d\theta = 3 \int_0^{2\pi} \left( \int_0^2 r^3\, dr \right) \cos(2\theta)\, d\theta \] Compute the inner integral: \[ \int_0^2 r^3\, dr = \left[ \frac{r^4}{4} \right]_0^2 = \frac{16}{4} = 4 \] Substituting back: \[ 3 \times 4 \int_0^{2\pi} \cos(2\theta)\, d\theta = 12 \int_0^{2\pi} \cos(2\theta)\, d\theta \] But: \[ \int_0^{2\pi} \cos(2\theta)\, d\theta = 0 \] Because \(\cos(2\theta)\) is periodic over \([0, 2\pi]\) and its positive and negative areas cancel out. Therefore: \[ 12 \int_0^{2\pi} \cos(2\theta)\, d\theta = 12 \times 0 = 0 \] **Answer:** 0 |
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| deepseekv3 | To evaluate the line integral \(\oiint_{S} y^3 \, dx + x^3 \, dy\) using Green's Theorem, we follow these steps: ### Green's Theorem Green's Theorem states that for a positively oriented, piecewise-smooth, simple closed curve \(C\) in the plane, and a region \(D\) bounded by \(C\): \[ \oint_{C} P \, dx + Q \, dy = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \] Here, \(P = y^3\) and \(Q = x^3\). We need to compute: \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \] ### Step 1: Compute \(\frac{\partial Q}{\partial x}\) and \(\frac{\partial P}{\partial y}\) \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (x^3) = 3x^2 \] \[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (y^3) = 3y^2 \] ### Step 2: Apply Green's Theorem Substitute the partial derivatives into Green's Theorem: \[ \oint_{C} y^3 \, dx + x^3 \, dy = \iint_{D} (3x^2 - 3y^2) \, dA \] Factor out the 3: \[ \iint_{D} 3(x^2 - y^2) \, dA \] ### Step 3: Change to Polar Coordinates Since the region \(D\) is a circle of radius 2 centered at the origin, it is convenient to use polar coordinates. Recall that in polar coordinates: \[ x = r \cos \theta, \quad y = r \sin \theta, \quad dA = r \, dr \, d\theta \] Substitute \(x\) and \(y\) into the integrand: \[ x^2 - y^2 = (r \cos \theta)^2 - (r \sin \theta)^2 = r^2 (\cos^2 \theta - \sin^2 \theta) = r^2 \cos(2\theta) \] Thus, the double integral becomes: \[ \iint_{D} 3(x^2 - y^2) \, dA = \int_{0}^{2\pi} \int_{0}^{2} 3r^2 \cos(2\theta) \cdot r \, dr \, d\theta \] Simplify the integrand: \[ 3r^3 \cos(2\theta) \] ### Step 4: Integrate with Respect to \(r\) and \(\theta\) First, integrate with respect to \(r\): \[ \int_{0}^{2} 3r^3 \, dr = 3 \left[ \frac{r^4}{4} \right]_{0}^{2} = 3 \left( \frac{16}{4} - 0 \right) = 12 \] Next, integrate with respect to \(\theta\): \[ \int_{0}^{2\pi} 12 \cos(2\theta) \, d\theta \] The integral of \(\cos(2\theta)\) over a full period \([0, 2\pi]\) is zero: \[ \int_{0}^{2\pi} \cos(2\theta) \, d\theta = 0 \] Thus, the final result is: \[ \oint_{C} y^3 \, dx + x^3 \, dy = 0 \] ### Final Answer \[ \boxed{0} \] |
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