| 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 \( Q(x,y) \) from the given line integral. In this case, \( P(x,y) = xy \) and \( Q(x,y) = x^2y^3 \). Green's Theorem states that: $$ \oint_{C} P dx + Q dy = \iint_{D} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA $$ where \( D \) is the region enclosed by the curve \( C \). First, we calculate the partial derivatives: $$ \frac{\partial Q}{\partial x} = 2xy^3 $$ $$ \frac{\partial P}{\partial y} = x $$ Now, we subtract the partial derivative of \( P \) with respect to \( y \) from the partial derivative of \( Q \) with respect to \( x \): $$ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2xy^3 - x = x(2y^3 - 1) $$ The next step is to set up the double integral over the region \( D \): $$ \iint_{D} (2xy^3 - x) dA $$ We can see that the region \( D \) is a triangle with vertices at (0,0), (1,0), and (1,2). We can describe this region using the following limits of integration: $$ 0 \leq x \leq 1 $$ $$ 0 \leq y \leq 2x $$ Thus, the double integral becomes: $$ \int_{0}^{1} \int_{0}^{2x} (2xy^3 - x) dy dx $$ Now, we integrate with respect to \( y \): $$ \int_{0}^{1} \left[ \frac{1}{2}xy^4 - xy \right]_{0}^{2x} dx = \int_{0}^{1} \left( 8x^5 - 2x^2 \right) dx $$ Next, we integrate with respect to \( x \): $$ \left[ \frac{4}{3}x^6 - \frac{2}{3}x^3 \right]_{0}^{1} = \frac{4}{3} - \frac{2}{3} = \frac{2}{3} $$ Therefore, the value of the line integral is \( \boxed{\frac{2}{3}} \). |
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| deepseekr1 | To evaluate the line integral \(\oint_{C} xy \, dx + x^2y^3 \, dy\) using Green's Theorem, we first identify \(P\) and \(Q\) from the given integral. Here, \(P = xy\) and \(Q = x^2y^3\). According to Green's Theorem, we need to compute the partial derivatives \(\frac{\partial Q}{\partial x}\) and \(\frac{\partial P}{\partial y}\): \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^2y^3) = 2xy^3 \] \[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(xy) = x \] The integrand for the double integral is \(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2xy^3 - x\). The region \(D\) is the triangle with vertices \((0,0)\), \((1,0)\), and \((1,2)\). We describe this region as bounded by \(y = 0\), \(x = 1\), and \(y = 2x\). We set up the double integral in the order \(dy \, dx\): 1. For \(x\) ranging from 0 to 1, \(y\) ranges from 0 to \(2x\). 2. The integrand is \(2xy^3 - x\). First, we integrate with respect to \(y\): \[ \int_{0}^{2x} (2xy^3 - x) \, dy = 2x \int_{0}^{2x} y^3 \, dy - x \int_{0}^{2x} \, dy \] \[ = 2x \left[ \frac{y^4}{4} \right]_{0}^{2x} - x \left[ y \right]_{0}^{2x} \] \[ = 2x \left( \frac{(2x)^4}{4} - 0 \right) - x (2x - 0) \] \[ = 2x \left( \frac{16x^4}{4} \right) - 2x^2 \] \[ = 8x^5 - 2x^2 \] Next, we integrate with respect to \(x\): \[ \int_{0}^{1} (8x^5 - 2x^2) \, dx \] \[ = 8 \int_{0}^{1} x^5 \, dx - 2 \int_{0}^{1} x^2 \, dx \] \[ = 8 \left[ \frac{x^6}{6} \right]_{0}^{1} - 2 \left[ \frac{x^3}{3} \right]_{0}^{1} \] \[ = 8 \left( \frac{1}{6} - 0 \right) - 2 \left( \frac{1}{3} - 0 \right) \] \[ = \frac{8}{6} - \frac{2}{3} \] \[ = \frac{4}{3} - \frac{2}{3} \] \[ = \frac{2}{3} \] Thus, the value of the line integral is \(\boxed{\dfrac{2}{3}}\). |
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| gpto1 | To evaluate the line integral \(\oint_{C} xy\, dx + x^2 y^3\, dy\) over the triangle \(C\) with vertices \((0,0)\), \((1,0)\), and \((1,2)\) in positive (counterclockwise) orientation using Green's Theorem, we'll proceed through the following steps: **1. Apply Green's Theorem:** Green's Theorem states that for a positively oriented, simple closed curve \(C\) enclosing a region \(D\): \[ \oint_{C} P\, dx + Q\, dy = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \] Given \(P = xy\) and \(Q = x^2 y^3\), compute the partial derivatives: \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^2 y^3) = 2x y^3 \] \[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(xy) = x \] So, the integrand is: \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x y^3 - x = x(2y^3 - 1) \] **2. Set Up the Double Integral:** We need to integrate over the region \(D\) bounded by the triangle. The sides of the triangle are: - From \((0,0)\) to \((1,0)\): \(y = 0\) - From \((1,0)\) to \((1,2)\): \(x = 1\) - From \((1,2)\) to \((0,0)\): \(y = 2x\) For a vertical slice (integrating with respect to \(x\) first), for each \(y\) from \(0\) to \(2\), \(x\) varies from \(x = y/2\) to \(x = 1\). So the double integral becomes: \[ \iint_{D} x(2y^3 - 1)\, dA = \int_{y=0}^{2} \int_{x=y/2}^{1} x(2y^3 - 1)\, dx\, dy \] **3. Compute the Inner Integral with Respect to \(x\):** Since \(2y^3 - 1\) is constant with respect to \(x\) for fixed \(y\), we integrate \(x\): \[ \int_{x=y/2}^{1} x(2y^3 - 1)\, dx = (2y^3 - 1) \left[ \frac{x^2}{2} \right]_{x=y/2}^{x=1} = (2y^3 - 1) \left( \frac{1}{2} - \frac{(y/2)^2}{2} \right) \] \[ = (2y^3 - 1) \left( \frac{1}{2} - \frac{y^2}{8} \right) = \frac{1}{8} (4 - y^2)(2y^3 - 1) \] **4. Expand and Simplify the Integrand:** \[ \frac{1}{8} (4 - y^2)(2y^3 - 1) = \frac{1}{8} \left( 8y^3 - 4 - 2y^5 + y^2 \right) = \frac{1}{8} \left( -2y^5 + 8y^3 + y^2 - 4 \right) \] **5. Integrate with Respect to \(y\):** Now, integrate term by term over \(y\) from \(0\) to \(2\): \[ \int_{y=0}^{2} \left( -\frac{1}{4} y^5 + y^3 + \frac{1}{8} y^2 - \frac{1}{2} \right) dy = \left[ -\frac{1}{24} y^6 + \frac{1}{4} y^4 + \frac{1}{24} y^3 - \frac{1}{2} y \right]_{0}^{2} \] Compute the antiderivatives: - \(-\frac{1}{4} y^5\) integrates to \(-\frac{1}{24} y^6\) - \(y^3\) integrates to \(\frac{1}{4} y^4\) - \(\frac{1}{8} y^2\) integrates to \(\frac{1}{24} y^3\) - \(-\frac{1}{2}\) integrates to \(-\frac{1}{2} y\) **6. Evaluate the Antiderivatives at the Limits:** Plug in \(y = 2\) and \(y = 0\): At \(y = 2\): \[ -\frac{1}{24} (2)^6 + \frac{1}{4} (2)^4 + \frac{1}{24} (2)^3 - \frac{1}{2} (2) = -\frac{64}{24} + \frac{16}{4} + \frac{8}{24} - 1 = -\frac{8}{3} + 4 + \frac{1}{3} - 1 \] Simplify: \[ \left( -\frac{8}{3} + \frac{1}{3} \right) + (4 - 1) = -\frac{7}{3} + 3 = \frac{2}{3} \] At \(y = 0\), all terms are zero. **7. Conclude the Result:** Therefore, the value of the line integral is \(\frac{2}{3}\). **Answer:** \(\displaystyle \oint_{C} xy\,dx + x^{2}y^{3}\,dy\;=\;\frac{2}{3}\) |
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| GPT-4o | To solve the given problem using Green's Theorem, we start by recalling the statement of Green's Theorem: \[ \oint_{C} P \, dx + Q \, dy = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \] where \( P(x, y) = xy \) and \( Q(x, y) = x^2 y^3 \), and \( C \) is the boundary of the region \( D \). First, we need to compute the partial derivatives of \( Q \) and \( P \): \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (x^2 y^3) = 2xy^3 \] \[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (xy) = x \] Now, we substitute these partial derivatives into Green's Theorem: \[ \oint_{C} xy \, dx + x^2 y^3 \, dy = \iint_{D} \left( 2xy^3 - x \right) \, dA \] Next, we need to describe the region \( D \). The region \( D \) is the triangle with vertices \((0,0)\), \((1,0)\), and \((1,2)\). To set up the double integral, we observe the bounds for \( D \). For \( x \) ranging from 0 to 1, \( y \) ranges from 0 to \( 2x \). Therefore, the double integral becomes: \[ \iint_{D} (2xy^3 - x) \, dA = \int_{0}^{1} \int_{0}^{2x} (2xy^3 - x) \, dy \, dx \] We now integrate with respect to \( y \): \[ \int_{0}^{1} \left[ \int_{0}^{2x} (2xy^3 - x) \, dy \right] \, dx \] First, evaluate the inner integral: \[ \int_{0}^{2x} (2xy^3 - x) \, dy = \int_{0}^{2x} 2xy^3 \, dy - \int_{0}^{2x} x \, dy \] \[ = 2x \int_{0}^{2x} y^3 \, dy - x \int_{0}^{2x} 1 \, dy \] \[ = 2x \left[ \frac{y^4}{4} \right]_{0}^{2x} - x \left[ y \right]_{0}^{2x} \] \[ = 2x \left( \frac{(2x)^4}{4} - 0 \right) - x \left( 2x - 0 \right) \] \[ = 2x \left( \frac{16x^4}{4} \right) - 2x^2 \] \[ = 2x \cdot 4x^4 - 2x^2 \] \[ = 8x^5 - 2x^2 \] Now, integrate with respect to \( x \): \[ \int_{0}^{1} (8x^5 - 2x^2) \, dx \] \[ = \int_{0}^{1} 8x^5 \, dx - \int_{0}^{1} 2x^2 \, dx \] \[ = 8 \left[ \frac{x^6}{6} \right]_{0}^{1} - 2 \left[ \frac{x^3}{3} \right]_{0}^{1} \] \[ = 8 \left( \frac{1}{6} - 0 \right) - 2 \left( \frac{1}{3} - 0 \right) \] \[ = 8 \cdot \frac{1}{6} - 2 \cdot \frac{1}{3} \] \[ = \frac{8}{6} - \frac{2}{3} \] \[ = \frac{4}{3} - \frac{2}{3} \] \[ = \frac{2}{3} \] Therefore, the value of the line integral is: \[ \oint_{C} xy \, dx + x^2 y^3 \, dy = \frac{2}{3} \] |
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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 $\oint_{C} Pdx + Qdy$ where $P = xy$ and $Q = x^2y^3$. The curve $C$ is a triangle with vertices (0,0), (1,0), and (1,2) with a positive orientation. To apply Green's Theorem, we need to find $\frac{\partial Q}{\partial x}$ and $\frac{\partial P}{\partial y}$. ## Step 2: Calculate $\frac{\partial Q}{\partial x}$ and $\frac{\partial P}{\partial y}$ Given $Q = x^2y^3$, we find $\frac{\partial Q}{\partial x} = 2xy^3$. Given $P = xy$, we find $\frac{\partial P}{\partial y} = x$. ## Step 3: Apply Green's Theorem Green's Theorem states that $\oint_{C} Pdx + Qdy = \iint_{D} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dA$, where $D$ is the region enclosed by $C$. Substituting the calculated values, we get $\iint_{D} (2xy^3 - x)dA$. ## Step 4: Evaluate the double integral To evaluate $\iint_{D} (2xy^3 - x)dA$, we first need to define the limits of integration. The region $D$ is a triangle with vertices (0,0), (1,0), and (1,2). The limits for $x$ are from 0 to 1, and for $y$, the limits are from 0 to $2x$ (since the line connecting (1,0) and (1,2) can be expressed as $y = 2x$ for the region of interest). ## Step 5: Perform the integration We integrate with respect to $y$ first, then $x$. The integral becomes $\int_{0}^{1} \int_{0}^{2x} (2xy^3 - x)dydx$. Performing the inner integration, we get $\int_{0}^{1} \left[\frac{2xy^4}{4} - xy\right]_{0}^{2x}dx = \int_{0}^{1} (2x^5 - 2x^2)dx$. ## Step 6: Complete the integration Evaluating the integral, we get $\left[\frac{2x^6}{6} - \frac{2x^3}{3}\right]_{0}^{1} = \left(\frac{1}{3} - \frac{2}{3}\right) - 0 = -\frac{1}{3}$. The final answer is: $\boxed{-\frac{1}{3}}$ |
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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 triangle. First, we need to find the region enclosed by the triangle. The vertices of the triangle are (0,0), (1,0), and (1,2), so the region is a right triangle with base 1 and height 2. Next, we need to find the function P(x,y) and Q(x,y) such that: P(x,y) = xy Q(x,y) = x^2y^3 Now, we can apply Green's Theorem, which states that: ∮[P(x,y)dx + Q(x,y)dy] = ∬[∂Q/∂x - ∂P/∂y]dA where the double integral is taken over the region enclosed by the triangle. Evaluating the partial derivatives, we get: ∂Q/∂x = 2xy^3 ∂P/∂y = x So, the integrand becomes: ∂Q/∂x - ∂P/∂y = 2xy^3 - x Now, we can evaluate the double integral: ∬[2xy^3 - x]dA To evaluate this integral, we can use the fact that the region is a right triangle with base 1 and height 2. We can divide the region into two sub-regions: the lower half of the triangle, where y ranges from 0 to 1, and the upper half of the triangle, where y ranges from 1 to 2. For the lower half of the triangle, the integral becomes: ∫[0,1]∫[0,y][2xy^3 - x]dxdy Evaluating the inner integral, we get: ∫[0,1][y^4 - y^2/2]dy Evaluating this integral, we get: 1/5 - 1/6 = 1/30 For the upper half of the triangle, the integral becomes: ∫[1,2]∫[0,1][2xy^3 - x]dxdy Evaluating the inner integral, we get: ∫[1,2][y^4 - y^2/2]dy Evaluating this integral, we get: 32/5 - 7/3 = 23/15 Adding the two results, we get: 1/30 + 23/15 = 49/30 Therefore, the value of the line integral is: ∮[xydx + x^2y^3dy] = 49/30 This is the final answer. |
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| DeepSeek-Coder-V2 | To solve the given problem using Green's Theorem, we need to follow these steps: 1. **Understand Green's Theorem**: Green's Theorem states that for a positively oriented, piecewise smooth simple closed curve \( C \) and a region \( D \) it encloses, the line integral of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} \) over \( C \) can be transformed into a double integral over \( D \): \[ \oint_{C} P \, dx + Q \, dy = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \] 2. **Identify the vector field and the region**: - The given vector field is \( \mathbf{F} = xy \mathbf{i} + x^2 y^3 \mathbf{j} \), so \( P = xy \) and \( Q = x^2 y^3 \). - The region \( D \) is the triangle with vertices (0,0), (1,0), and (1,2). 3. **Compute the partial derivatives**: \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (x^2 y^3) = 2x y^3 \] \[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (xy) = x \] 4. **Form the integrand for the double integral**: \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x y^3 - x = x(2y^3 - 1) \] 5. **Set up the double integral over the region \( D \)**: - The region \( D \) can be described as \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 2x \). \[ \iint_{D} x(2y^3 - 1) \, dA = \int_{0}^{1} \int_{0}^{2x} x(2y^3 - 1) \, dy \, dx \] 6. **Evaluate the inner integral**: \[ \int_{0}^{2x} x(2y^3 - 1) \, dy = x \left( \int_{0}^{2x} 2y^3 \, dy - \int_{0}^{2x} 1 \, dy \right) \] \[ = x \left( 2 \int_{0}^{2x} y^3 \, dy - [y]_{0}^{2x} \right) \] \[ = x \left( 2 \left[ \frac{y^4}{4} \right]_{0}^{2x} - 2x \right) \] \[ = x \left( 2 \cdot \frac{(2x)^4}{4} - 2x \right) \] \[ = x \left( 2 \cdot 4x^4 - 2x \right) \] \[ = x \left( 8x^4 - 2x \right) \] \[ = 8x^5 - 2x^2 \] 7. **Evaluate the outer integral**: \[ \int_{0}^{1} (8x^5 - 2x^2) \, dx \] \[ = \int_{0}^{1} 8x^5 \, dx - \int_{0}^{1} 2x^2 \, dx \] \[ = 8 \left[ \frac{x^6}{6} \right]_{0}^{1} - 2 \left[ \frac{x^3}{3} \right]_{0}^{1} \] \[ = 8 \cdot \frac{1}{6} - 2 \cdot \frac{1}{3} \] \[ = \frac{8}{6} - \frac{2}{3} \] \[ = \frac{4}{3} - \frac{2}{3} \] \[ = \frac{2}{3} \] Thus, the value of the line integral is: \[ \boxed{\frac{2}{3}} \] |
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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 $Q(x,y)$ from the given line integral. In this case, $P(x,y) = xy$ and $Q(x,y) = x^2y^3$. Green's Theorem states that: $$ \oint_{C} P dx + Q dy = \iint_{D} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA $$ where $D$ is the region enclosed by the curve $C$. First, we calculate the partial derivatives: $$ \frac{\partial Q}{\partial x} = 2xy^3 $$ $$ \frac{\partial P}{\partial y} = x $$ Now, we subtract the partial derivative of $P$ with respect to $y$ from the partial derivative of $Q$ with respect to $x$: $$ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2xy^3 - x = x(2y^3 - 1) $$ The next step is to set up the double integral over the region $D$: $$ \iint_{D} (2xy^3 - x) dA $$ We can see that the region $D$ is a triangle with vertices at (0,0), (1,0), and (1,2). We can describe this region using the following limits of integration: $$ 0 \leq x \leq 1 $$ $$ 0 \leq y \leq 2x $$ Thus, the double integral becomes: $$ \int_{0}^{1} \int_{0}^{2x} (2xy^3 - x) dy dx $$ Now, we integrate with respect to $y$: $$ \int_{0}^{1} \left[ \frac{1}{2}xy^4 - xy \right]_{0}^{2x} dx = \int_{0}^{1} \left( 8x^5 - 2x^2 \right) dx $$ Next, we integrate with respect to $x$: $$ \left[ \frac{4}{3}x^6 - \frac{2}{3}x^3 \right]_{0}^{1} = \frac{4}{3} - \frac{2}{3} = \frac{2}{3} $$ Therefore, the value of the line integral is $\boxed{\frac{2}{3}}$. |
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| baichuan4 | To evaluate the line integral $\oint_{C} xy \, dx + x^2y^3 \, dy$ around the triangle with vertices (0,0), (1,0), and (1,2) using Green's Theorem, we first need to find the double integral of the partial derivatives of the integrand over the area of the triangle. Green's Theorem states that for a two-dimensional region $D$ bounded by a simple, closed, piecewise smooth curve $C$, oriented counterclockwise, the line integral of a vector field $\mathbf{F}$ around $C$ equals the double integral over $D$ of the curl of $\mathbf{F}$. Mathematically, if $\mathbf{F} = P \mathbf{i} + Q \mathbf{j}$ (where $\mathbf{i}$ and $\mathbf{j}$ are the unit vectors in the $x$ and $y$ directions, respectively), then: $$\oint_{C} P \, dx + Q \, dy = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA$$ In our case, $\mathbf{F} = xy \mathbf{i} + x^2y^3 \mathbf{j}$. So, $P = xy$ and $Q = x^2y^3$. We need to calculate: $$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^2y^3) = 2xy^3$$ $$\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(xy) = x$$ Substituting these into the formula, we get: $$\oint_{C} xy \, dx + x^2y^3 \, dy = \iint_{D} (2xy^3 - x) \, dA$$ Now, we need to set up the double integral over the triangle with vertices (0,0), (1,0), and (1,2). The easiest way to set up this integral is in Cartesian coordinates, where we integrate over the region $0 \leq x \leq 1$ and $0 \leq y \leq 2x$ (since the triangle has a slope of 2). The double integral becomes: $$\int_{0}^{1} \int_{0}^{2x} (2xy^3 - x) \, dy \, dx$$ We can now integrate with respect to $y$ first: $$\int_{0}^{1} \left[ \frac{1}{2}2xy^4 - xy \right]_{0}^{2x} \, dx = \int_{0}^{1} (2x(2x)^4 - 2x^2) \, dx$$ $$= \int_{0}^{1} (32x^5 - 2x^2) \, dx$$ Integrating with respect to $x$ gives us: $$\left[ \frac{32}{6}x^6 - \frac{2}{3}x^3 \right]_{0}^{1} = \frac{32}{6} - \frac{2}{3}$$ Calculating the final result: $$\frac{32}{6} - \frac{2}{3} = \frac{16}{3} - \frac{2}{3} = \frac{14}{3}$$ So, the value of the line integral $\oint_{C} xy \, dx + x^2y^3 \, dy$ around the given triangle is $\frac{14}{3}$. |
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| doubao | Let $D$ be the region enclosed by the triangle $C$. The equation of the line joining (0,0) and (1,2) is $y=2x$. Green's Theorem states that $\oint_{C} Pdx + Qdy = \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dxdy$. Here $P=xy$ and $Q=x^2y^3$. So $\frac{\partial Q}{\partial x}=2xy^3$ and $\frac{\partial P}{\partial y}=x$. Then the integral becomes: \[ \begin{align*} &\iint_{D} (2xy^3 - x)dxdy\\ =&\int_{0}^{1}\int_{0}^{2x} (2xy^3 - x)dydx\\ =&\int_{0}^{1} [\frac{1}{2}xy^4 - xy]_{0}^{2x}dx\\ =&\int_{0}^{1} (\frac{1}{2}x(2x)^4 - x(2x))dx\\ =&\int_{0}^{1} (8x^5 - 2x^2)dx\\ =&[\frac{4}{3}x^6 - \frac{2}{3}x^3]_{0}^{1}\\ =&(\frac{4}{3} - \frac{2}{3})\\ =&\frac{2}{3} \end{align*} \] |
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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 functions with continuous partial derivatives on an open region containing $D$, 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 our case, $P(x, y) = xy$ and $Q(x, y) = x^2y^3$. We need to compute the partial derivatives: $$ \frac{\partial P}{\partial y} = x $$ $$ \frac{\partial Q}{\partial x} = 2xy^3 $$ Now, we apply Green's Theorem: $$ \oint_{C} xy dx + x^2y^3 dy = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA = \iint_{D} (2xy^3 - x) dA $$ The region $D$ is a triangle with vertices at $(0,0)$, $(1,0)$, and $(1,2)$. We can describe this region using two different approaches: either by integrating over $y$ first and then over $x$, or vice versa. Since the region is bounded by the lines $x=0$, $y=0$, and $y=2(1-x)$, it might be easier to integrate over $y$ first and then over $x$. The limits for $y$ go from $0$ to $2(1-x)$, and the limits for $x$ go from $0$ to $1$. So the integral becomes: $$ \int_{0}^{1} \int_{0}^{2(1-x)} (2xy^3 - x) dy dx $$ Let's compute the integral: $$ \int_{0}^{1} \left[ \int_{0}^{2(1-x)} (2xy^3 - x) dy \right] dx = \int_{0}^{1} \left[ \frac{1}{2}xy^4 - xy \right]_{0}^{2(1-x)} dx $$ $$ = \int_{0}^{1} \left[ \frac{1}{2}x(2(1-x))^4 - x(2(1-x)) \right] dx $$ $$ = \int_{0}^{1} \left[ \frac{1}{2}x(16(1-x)^4) - 2x(1-x) \right] dx $$ $$ = \int_{0}^{1} \left[ 8x(1-x)^4 - 2x + 2x^2 \right] dx $$ $$ = \int_{0}^{1} \left[ 8x(1-4x+6x^2-4x^3+x^4) - 2x + 2x^2 \right] dx $$ $$ = \int_{0}^{1} \left[ 8x - 32x^2 + 48x^3 - 32x^4 + 8x^5 - 2x + 2x^2 \right] dx $$ $$ = \int_{0}^{1} \left[ 8x^5 - 30x^2 + 48x^3 - 30x \right] dx $$ Now we integrate term by term: $$ \left[ \frac{8}{6}x^6 - \frac{30}{3}x^3 + \frac{48}{4}x^4 - \frac{30}{2}x^2 \right]_{0}^{1} $$ $$ = \left[ \frac{4}{3}x^6 - 10x^3 + 12x^4 - 15x^2 \right]_{0}^{1} $$ $$ = \left[ \frac{4}{3} - 10 + 12 - 15 \right] - \left[ 0 - 0 + 0 - 0 \right] $$ $$ = \frac{4}{3} - 10 + 12 - 15 $$ $$ = \frac{4}{3} - 13 $$ $$ = \frac{4 - 39}{3} $$ $$ = \frac{-35}{3} $$ So the value of the line integral $\oint_{C} xy dx + x^2y^3 dy$ over the given triangle is $-\frac{35}{3}$. |
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| kimi | Green's Theorem states that for a vector field $\mathbf{F} = P\mathbf{i} + Q\mathbf{j}$ and a positively oriented, simple, closed curve $C$, the line integral around $C$ 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 this case, we have $P(x, y) = xy$ and $Q(x, y) = x^2y^3$. Applying Green's Theorem, we need to compute the partial derivatives: $$\frac{\partial Q}{\partial x} = \frac{\partial (x^2y^3)}{\partial x} = 2xy^3$$ $$\frac{\partial P}{\partial y} = \frac{\partial (xy)}{\partial y} = x$$ Now, we calculate the difference: $$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2xy^3 - x$$ The region $D$ enclosed by the triangle $C$ is a right triangle with base and height given by the x and y coordinates of the vertices. The base of the triangle is along the x-axis from (0,0) to (1,0), and the height is along the line from (1,0) to (1,2). So, the region $D$ is defined by $0 \leq x \leq 1$ and $0 \leq y \leq 2 - x$. Now we can set up the double integral: $$\iint_{D} (2xy^3 - x) dA = \int_{0}^{1} \int_{0}^{2-x} (2xy^3 - x) dy dx$$ Let's compute the inner integral with respect to $y$: $$\int_{0}^{2-x} (2xy^3 - x) dy = \left[ x y^4 - \frac{x y^2}{2} \right]_{0}^{2-x} = x (2-x)^4 - \frac{x (2-x)^2}{2}$$ Now, we compute the outer integral with respect to $x$: $$\int_{0}^{1} \left[ x (2-x)^4 - \frac{x (2-x)^2}{2} \right] dx$$ This integral can be computed by expanding the terms and integrating term by term. However, this is a bit tedious to do by hand, so let's just outline the steps: 1. Expand $(2-x)^4$ and $(2-x)^2$ using the binomial theorem. 2. Multiply each term by $x$ and integrate with respect to $x$ from 0 to 1. 3. Combine like terms and evaluate the integral. After computing the integral, we will get the value of the line integral around the curve $C$. Let's compute it: $$\int_{0}^{1} \left[ x (16 - 32x + 24x^2 - 8x^3 + x^4) - \frac{x (4 - 4x + x^2)}{2} \right] dx$$ $$= \int_{0}^{1} \left[ 16x - 32x^2 + 24x^3 - 8x^4 + x^5 - 2x^2 + x^3 \right] dx$$ $$= \int_{0}^{1} \left[ 16x - 34x^2 + 25x^3 - 8x^4 + x^5 \right] dx$$ Now, integrate term by term: $$\left[ 8x^2 - \frac{34x^3}{3} + \frac{25x^4}{4} - \frac{8x^5}{5} + \frac{x^6}{6} \right]_{0}^{1}$$ Evaluate the antiderivative at the bounds: $$\left( 8(1)^2 - \frac{34(1)^3}{3} + \frac{25(1)^4}{4} - \frac{8(1)^5}{5} + \frac{(1)^6}{6} \ |
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| wenxinyiyan | Let $P = xy$ and $Q = x^2y^3$. Then, we have $\frac{\partial P}{\partial y} = x$ and $\frac{\partial Q}{\partial x} = 2xy^3$. Now, we use Green's Theorem to evaluate the given line integral: $\oint_{C} xy dx + x^2y^3dy = \int\int_{R} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA$ $= \int\int_{R} (2xy^3 - x) dA$ where $R$ is the region enclosed by the triangle with vertices (0,0), (1,0), and (1,2). To evaluate this double integral, we need to find the bounds for $x$ and $y$. The region $R$ is bounded by the lines $y = 0$, $x = 1$, and $y = 2x$. Therefore, we have $0 \leq x \leq 1$ and $0 \leq y \leq 2x$. Now, we can evaluate the double integral: $\int\int_{R} (2xy^3 - x) dA = \int_{0}^{1} \int_{0}^{2x} (2xy^3 - x) dy dx$ $= \int_{0}^{1} [\frac{1}{2}xy^4 - xy]_{0}^{2x} dx$ $= \int_{0}^{1} [\frac{1}{2}x(2x)^4 - x(2x)] dx$ $= \int_{0}^{1} [\frac{1}{2}x(16x^4) - 2x^2] dx$ $= \int_{0}^{1} [8x^5 - 2x^2] dx$ $= [\frac{4}{3}x^6 - \frac{2}{3}x^3]_{0}^{1}$ $= \frac{4}{3} - \frac{2}{3}$ $= \frac{2}{3}$ Therefore, the value of the given line integral is $\frac{2}{3}$. |
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| zhipu | Green's Theorem relates a line integral around a simple closed curve \( C \) to a double integral over the plane region \( D \) bounded by \( C \). The theorem states: \[ \oint_{C} P \, dx + Q \, dy = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \] For the given problem, \( P = xy \) and \( Q = x^2y^3 \). We need to compute the partial derivatives \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \): \[ \frac{\partial Q}{\partial x} = \frac{\partial (x^2y^3)}{\partial x} = 2xy^3 \] \[ \frac{\partial P}{\partial y} = \frac{\partial (xy)}{\partial y} = x \] Now, we can apply Green's Theorem: \[ \oint_{C} xy \, dx + x^2y^3 \, dy = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA = \iint_{D} \left( 2xy^3 - x \right) \, dA \] Next, we need to determine the region \( D \) over which we will integrate. The triangle with vertices \((0,0)\), \((1,0)\), and \((1,2)\) can be described by: \[ 0 \leq x \leq 1 \quad \text{and} \quad 0 \leq y \leq 2x \] Now, we set up the double integral: \[ \iint_{D} (2xy^3 - x) \, dA = \int_{0}^{1} \int_{0}^{2x} (2xy^3 - x) \, dy \, dx \] First, integrate with respect to \( y \): \[ \int_{0}^{2x} (2xy^3 - x) \, dy = \int_{0}^{2x} 2xy^3 \, dy - \int_{0}^{2x} x \, dy \] \[ = 2x \int_{0}^{2x} y^3 \, dy - x \int_{0}^{2x} \, dy \] \[ = 2x \left[ \frac{y^4}{4} \right]_{0}^{2x} - x \left[ y \right]_{0}^{2x} \] \[ = 2x \left( \frac{(2x)^4}{4} - 0 \right) - x \left( 2x - 0 \right) \] \[ = 2x \left( \frac{16x^4}{4} \right) - 2x^2 \] \[ = 2x \left( 4x^4 \right) - 2x^2 \] \[ = 8x^5 - 2x^2 \] Now, integrate with respect to \( x \): \[ \int_{0}^{1} (8x^5 - 2x^2) \, dx = \int_{0}^{1} 8x^5 \, dx - \int_{0}^{1} 2x^2 \, dx \] \[ = 8 \int_{0}^{1} x^5 \, dx - 2 \int_{0}^{1} x^2 \, dx \] \[ = 8 \left[ \frac{x^6}{6} \right]_{0}^{1} - 2 \left[ \frac{x^3}{3} \right]_{0}^{1} \] \[ = 8 \left( \frac{1}{6} - 0 \right) - 2 \left( \frac{1}{3} - 0 \right) \] \[ = 8 \cdot \frac{1}{6} - 2 \cdot \frac{1}{3} \] \[ = \frac{8}{6} - \frac{2}{3} \] \[ = \frac{4}{3} - \frac{2}{3} \] \[ = \frac{2}{3} \] Therefore, the value of the line integral \(\oint_{C} xy \, dx + x^2y^3 \, dy\) is: \[ \boxed{\frac{2}{3}} \] |
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| gpto1 | To evaluate the line integral \(\oint_{C} xy\, dx + x^2 y^3\, dy\) over the triangle \(C\) with vertices \((0,0)\), \((1,0)\), and \((1,2)\) in positive (counterclockwise) orientation using Green's Theorem, we'll proceed through the following steps: **1. Apply Green's Theorem:** Green's Theorem states that for a positively oriented, simple closed curve \(C\) enclosing a region \(D\): \[ \oint_{C} P\, dx + Q\, dy = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \] Given \(P = xy\) and \(Q = x^2 y^3\), compute the partial derivatives: \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^2 y^3) = 2x y^3 \] \[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(xy) = x \] So, the integrand is: \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x y^3 - x = x(2y^3 - 1) \] **2. Set Up the Double Integral:** We need to integrate over the region \(D\) bounded by the triangle. The sides of the triangle are: - From \((0,0)\) to \((1,0)\): \(y = 0\) - From \((1,0)\) to \((1,2)\): \(x = 1\) - From \((1,2)\) to \((0,0)\): \(y = 2x\) For a vertical slice (integrating with respect to \(x\) first), for each \(y\) from \(0\) to \(2\), \(x\) varies from \(x = y/2\) to \(x = 1\). So the double integral becomes: \[ \iint_{D} x(2y^3 - 1)\, dA = \int_{y=0}^{2} \int_{x=y/2}^{1} x(2y^3 - 1)\, dx\, dy \] **3. Compute the Inner Integral with Respect to \(x\):** Since \(2y^3 - 1\) is constant with respect to \(x\) for fixed \(y\), we integrate \(x\): \[ \int_{x=y/2}^{1} x(2y^3 - 1)\, dx = (2y^3 - 1) \left[ \frac{x^2}{2} \right]_{x=y/2}^{x=1} = (2y^3 - 1) \left( \frac{1}{2} - \frac{(y/2)^2}{2} \right) \] \[ = (2y^3 - 1) \left( \frac{1}{2} - \frac{y^2}{8} \right) = \frac{1}{8} (4 - y^2)(2y^3 - 1) \] **4. Expand and Simplify the Integrand:** \[ \frac{1}{8} (4 - y^2)(2y^3 - 1) = \frac{1}{8} \left( 8y^3 - 4 - 2y^5 + y^2 \right) = \frac{1}{8} \left( -2y^5 + 8y^3 + y^2 - 4 \right) \] **5. Integrate with Respect to \(y\):** Now, integrate term by term over \(y\) from \(0\) to \(2\): \[ \int_{y=0}^{2} \left( -\frac{1}{4} y^5 + y^3 + \frac{1}{8} y^2 - \frac{1}{2} \right) dy = \left[ -\frac{1}{24} y^6 + \frac{1}{4} y^4 + \frac{1}{24} y^3 - \frac{1}{2} y \right]_{0}^{2} \] Compute the antiderivatives: - \(-\frac{1}{4} y^5\) integrates to \(-\frac{1}{24} y^6\) - \(y^3\) integrates to \(\frac{1}{4} y^4\) - \(\frac{1}{8} y^2\) integrates to \(\frac{1}{24} y^3\) - \(-\frac{1}{2}\) integrates to \(-\frac{1}{2} y\) **6. Evaluate the Antiderivatives at the Limits:** Plug in \(y = 2\) and \(y = 0\): At \(y = 2\): \[ -\frac{1}{24} (2)^6 + \frac{1}{4} (2)^4 + \frac{1}{24} (2)^3 - \frac{1}{2} (2) = -\frac{64}{24} + \frac{16}{4} + \frac{8}{24} - 1 = -\frac{8}{3} + 4 + \frac{1}{3} - 1 \] Simplify: \[ \left( -\frac{8}{3} + \frac{1}{3} \right) + (4 - 1) = -\frac{7}{3} + 3 = \frac{2}{3} \] At \(y = 0\), all terms are zero. **7. Conclude the Result:** Therefore, the value of the line integral is \(\frac{2}{3}\). **Answer:** \(\displaystyle \oint_{C} xy\,dx + x^{2}y^{3}\,dy\;=\;\frac{2}{3}\) |
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| deepseekv3 | To evaluate the line integral \(\oint_{C} xy \, dx + x^2y^3 \, dy\) using Green's Theorem, we follow these steps: ### Step 1: Understand Green's Theorem Green's Theorem relates a line integral around a simple, closed curve \(C\) to a double integral over the region \(D\) enclosed by \(C\). It states: \[ \oint_{C} P \, dx + Q \, dy = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \] Here, \(P(x, y) = xy\) and \(Q(x, y) = x^2y^3\). ### Step 2: Compute the Partial Derivatives Compute \(\frac{\partial Q}{\partial x}\) and \(\frac{\partial P}{\partial y}\): \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (x^2y^3) = 2xy^3 \] \[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (xy) = x \] ### Step 3: Apply Green's Theorem Substitute these into Green's Theorem: \[ \oint_{C} xy \, dx + x^2y^3 \, dy = \iint_{D} \left( 2xy^3 - x \right) dA \] ### Step 4: Set Up the Double Integral The region \(D\) is a triangle with vertices \((0,0)\), \((1,0)\), and \((1,2)\). To set up the double integral, we need to describe \(D\) in terms of \(x\) and \(y\). - The base of the triangle is along the \(x\)-axis from \(x = 0\) to \(x = 1\). - The top of the triangle is along the line from \((1,0)\) to \((1,2)\), so \(y\) ranges from \(0\) to \(2x\) for each \(x\) in \([0,1]\). Thus, the double integral becomes: \[ \iint_{D} \left( 2xy^3 - x \right) dA = \int_{0}^{1} \int_{0}^{2x} \left( 2xy^3 - x \right) dy \, dx \] ### Step 5: Evaluate the Inner Integral First, evaluate the inner integral with respect to \(y\): \[ \int_{0}^{2x} \left( 2xy^3 - x \right) dy = 2x \int_{0}^{2x} y^3 \, dy - x \int_{0}^{2x} 1 \, dy \] Compute each part: \[ 2x \int_{0}^{2x} y^3 \, dy = 2x \left[ \frac{y^4}{4} \right]_{0}^{2x} = 2x \left( \frac{(2x)^4}{4} - 0 \right) = 2x \left( \frac{16x^4}{4} \right) = 8x^5 \] \[ -x \int_{0}^{2x} 1 \, dy = -x [y]_{0}^{2x} = -x (2x - 0) = -2x^2 \] Combine the results: \[ \int_{0}^{2x} \left( 2xy^3 - x \right) dy = 8x^5 - 2x^2 \] ### Step 6: Evaluate the Outer Integral Now, evaluate the outer integral with respect to \(x\): \[ \int_{0}^{1} \left( 8x^5 - 2x^2 \right) dx = 8 \int_{0}^{1} x^5 \, dx - 2 \int_{0}^{1} x^2 \, dx \] Compute each part: \[ 8 \int_{0}^{1} x^5 \, dx = 8 \left[ \frac{x^6}{6} \right]_{0}^{1} = 8 \left( \frac{1}{6} - 0 \right) = \frac{8}{6} = \frac{4}{3} \] \[ -2 \int_{0}^{1} x^2 \, dx = -2 \left[ \frac{x^3}{3} \right]_{0}^{1} = -2 \left( \frac{1}{3} - 0 \right) = -\frac{2}{3} \] Combine the results: \[ \int_{0}^{1} \left( 8x^5 - 2x^2 \right) dx = \frac{4}{3} - \frac{2}{3} = \frac{2}{3} \] ### Final Answer The value of the line integral is: \[ \boxed{\frac{2}{3}} \] |
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