| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the given problem using Stoke's Theorem, we start by applying the theorem, which states that the line integral of a vector field \(\vec{F}\) around a closed curve \(C\) is equal to the surface integral of the curl of \(\vec{F}\) over the surface \(S\) bounded by \(C\): \[ \int_C \vec{F} \cdot d \vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d \vec{S} \] First, we need to compute the curl of \(\vec{F}\): \[ \vec{F} = z^2 \vec{i} + y^2 \vec{j} + x \vec{k} \] \[ \nabla \times \vec{F} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^2 & y^2 & x \end{vmatrix} \] \[ = \left( \frac{\partial x}{\partial y} - \frac{\partial y^2}{\partial z} \right) \vec{i} + \left( \frac{\partial z^2}{\partial x} - \frac{\partial x}{\partial z} \right) \vec{j} + \left( \frac{\partial y^2}{\partial x} - \frac{\partial z^2}{\partial y} \right) \vec{k} \] \[ = (0 - 0) \vec{i} + (2z - 0) \vec{j} + (2y - 2z) \vec{k} \] \[ = 2z \vec{j} + (2y - 2z) \vec{k} \] Next, we parameterize the surface \(S\) bounded by the triangle \(C\) with vertices (1,0,0), (0,1,0), and (0,0,1). We can use the parameterization: \[ x = 1 - u - v, \quad y = u, \quad z = v \] where \(0 \leq u \leq 1\) and \(0 \leq v \leq 1 - u\). The normal vector \(d \vec{S}\) is given by the cross product of the partial derivatives of the parameterization: \[ \frac{\partial}{\partial u} (1 - u - v, u, v) = (-1, 1, 0) \] \[ \frac{\partial}{\partial v} (1 - u - v, u, v) = (-1, 0, 1) \] \[ d \vec{S} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix} = (1, 1, 1) \] Now, we compute the surface integral: \[ \iint_S (\nabla \times \vec{F}) \cdot d \vec{S} = \int_0^1 \int_0^{1-u} (2v \vec{j} + (2u - 2v) \vec{k}) \cdot (1, 1, 1) dv du \] \[ = \int_0^1 \int_0^{1-u} (2v + 2u - 2v) dv du = \int_0^1 \int_0^{1-u} 2u dv du \] \[ = \int_0^1 2u (1 - u) du = \int_0^1 (2u - 2u^2) du = \left[ u^2 - \frac{2}{3}u^3 \right]_0^1 = \frac{1}{3} \] Thus, the value of the line integral \(\int_C \vec{F} \cdot d \vec{r}\) is \(\boxed{\frac{1}{3}}\). |
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| deepseekr1 | To evaluate the line integral \(\int_C \vec{F} \cdot d \vec{r}\) using Stokes' Theorem, we first compute the curl of the vector field \(\vec{F} = z^2 \vec{i} + y^2 \vec{j} + x \vec{k}\). The curl of \(\vec{F}\) is calculated as follows: \[ \nabla \times \vec{F} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partialx} & \frac{\partial}{\partialy} & \frac{\partial}{\partialz} \\ z^2 & y^2 & x \end{vmatrix} = \left( \frac{\partial x}{\partial y} - \frac{\partial y^2}{\partial z} \right) \vec{i} - \left( \frac{\partial x}{\partial x} - \frac{\partial z^2}{\partial z} \right) \vec{j} + \left( \frac{\partial y^2}{\partial x} - \frac{\partial z^2}{\partial y} \right) \vec{k} \] Simplifying each component: - The \(\vec{i}\) component is \(0 - 0 = 0\). - The \(\vec{j}\) component is \(- \left( 1 - 2z \right)\). - The \(\vec{k}\) component is \(0 - 0 = 0\). Thus, \(\nabla \times \vec{F} = (0, -1 + 2z, 0)\). Next, we parametrize the surface \(S\) (the triangle with vertices \((1,0,0)\), \((0,1,0)\), and \((0,0,1)\)) using \(x\) and \(y\) as parameters. The equation of the plane is \(x + y + z = 1\), so \(z = 1 - x - y\). The parametrization is: \[ \vec{r}(x, y) = x \vec{i} + y \vec{j} + (1 - x - y) \vec{k} \] We compute the partial derivatives: \[ \vec{r}_x = (1, 0, -1), \quad \vec{r}_y = (0, 1, -1) \] The cross product \(\vec{r}_x \times \vec{r}_y\) is: \[ \vec{r}_x \times \vec{r}_y = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & -1 \\ 0 & 1 & -1 \end{vmatrix} = \vec{i}(1) - \vec{j}(-1) + \vec{k}(1) = (1, 1, 1) \] Substituting \(z = 1 - x - y\) into \(\nabla \times \vec{F}\), we get: \[ \nabla \times \vec{F} = (0, 1 - 2x - 2y, 0) \] The surface integral becomes: \[ \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \iint_D (0, 1 - 2x - 2y, 0) \cdot (1, 1, 1) \, dx \, dy = \iint_D (1 - 2x - 2y) \, dx \, dy \] The domain \(D\) is the triangle in the \(xy\)-plane with vertices \((0,0)\), \((1,0)\), and \((0,1)\). Setting up the integral: \[ \int_{0}^{1} \int_{0}^{1-x} (1 - 2x - 2y) \, dy \, dx \] Evaluating the inner integral: \[ \int_{0}^{1-x} (1 - 2x - 2y) \, dy = \left[ y - 2xy - y^2 \right]_{0}^{1-x} = (1 - x) - 2x(1 - x) - (1 - x)^2 = -x + x^2 \] Evaluating the outer integral: \[ \int_{0}^{1} (-x + x^2) \, dx = \left[ -\frac{x^2}{2} + \frac{x^3}{3} \right]_{0}^{1} = -\frac{1}{2} + \frac{1}{3} = -\frac{1}{6} \] Thus, the value of the line integral is: \[ \boxed{-\dfrac{1}{6}} \] |
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| gpto1 | **Solution:** We are to evaluate the line integral \(\int_C \vec{F} \cdot d\vec{r}\) where \(\vec{F} = z^2 \vec{i} + y^2 \vec{j} + x \vec{k}\) and \(C\) is the triangle with vertices \((1,0,0)\), \((0,1,0)\), and \((0,0,1)\) oriented counterclockwise. We will use **Stokes' Theorem**, which relates a line integral around a closed curve \(C\) to a surface integral over a surface \(S\) bounded by \(C\): \[ \int_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot \vec{n}\, dS \] where \(\vec{n}\) is the unit normal vector to \(S\) consistent with the orientation of \(C\). **Step 1: Compute the Curl of \(\vec{F}\)** Given \(\vec{F} = z^2 \vec{i} + y^2 \vec{j} + x \vec{k}\), we compute \(\nabla \times \vec{F}\): \[ \nabla \times \vec{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) \vec{i} + \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \vec{j} + \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \vec{k} \] Computing the partial derivatives: - \(\dfrac{\partial F_z}{\partial y} = \dfrac{\partial x}{\partial y} = 0\) - \(\dfrac{\partial F_y}{\partial z} = \dfrac{\partial (y^2)}{\partial z} = 0\) Thus, the \(i\)-component is \(0 - 0 = 0\). - \(\dfrac{\partial F_x}{\partial z} = \dfrac{\partial (z^2)}{\partial z} = 2z\) - \(\dfrac{\partial F_z}{\partial x} = \dfrac{\partial x}{\partial x} = 1\) Thus, the \(j\)-component is \(2z - 1\). - \(\dfrac{\partial F_y}{\partial x} = \dfrac{\partial (y^2)}{\partial x} = 0\) - \(\dfrac{\partial F_x}{\partial y} = \dfrac{\partial (z^2)}{\partial y} = 0\) Thus, the \(k\)-component is \(0 - 0 = 0\). Therefore, the curl is: \[ \nabla \times \vec{F} = (0)\vec{i} + (2z - 1)\vec{j} + (0)\vec{k} \] **Step 2: Parametrize the Surface \(S\) and Compute \(\vec{n} \, dS\)** The surface \(S\) is the triangle bounded by \(C\). The vertices are \((1,0,0)\), \((0,1,0)\), and \((0,0,1)\), which lie on the plane \(x + y + z = 1\). We can parametrize the surface \(S\) using parameters \(u\) and \(v\): \[ \begin{cases} x = u \\ y = v \\ z = 1 - u - v \end{cases} \] Where \(u \geq 0\), \(v \geq 0\), and \(u + v \leq 1\). This parameterization covers the triangular region. Compute the tangent vectors: \[ \vec{r}_u = \frac{\partial \vec{r}}{\partial u} = \left(1, 0, -1\right), \quad \vec{r}_v = \frac{\partial \vec{r}}{\partial v} = \left(0, 1, -1\right) \] Compute the normal vector using the cross product \(\vec{n} = \vec{r}_u \times \vec{r}_v\): \[ \vec{n} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & -1 \\ 0 & 1 & -1 \\ \end{vmatrix} = \left(1( -1) - (-1)(1)\right)\vec{i} - \left(1( -1) - (-1)(0)\right)\vec{j} + \left(1(1) - (0)(0)\right)\vec{k} = (1)\vec{i} + (1)\vec{j} + (1)\vec{k} \] Thus, \(\vec{n} = (1,1,1)\). The differential of surface area element is \(dS = |\vec{n}|\, du\, dv\), but since \(\vec{n}\) is constant, \( \vec{n}\, dS = \left(1,1,1\right) du\, dv \). **Step 3: Evaluate the Surface Integral** We have: \[ \int_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot \vec{n}\, dS = \iint_D (\nabla \times \vec{F}) \cdot \vec{n}\, du\, dv \] Compute the dot product \((\nabla \times \vec{F}) \cdot \vec{n}\): \[ (\nabla \times \vec{F}) \cdot \vec{n} = [0, \, 2z - 1, \, 0] \cdot [1, \, 1, \, 1] = (0)(1) + (2z - 1)(1) + (0)(1) = 2z - 1 \] Recall that \(z = 1 - u - v\), so: \[ 2z - 1 = 2(1 - u - v) - 1 = 2 - 2u - 2v - 1 = 1 - 2u - 2v \] Thus, the integrand becomes \(1 - 2u - 2v\). The limits of integration are: - \(u\) ranges from \(0\) to \(1\) - For each \(u\), \(v\) ranges from \(0\) to \(1 - u\) Set up the integral: \[ \int_C \vec{F} \cdot d\vec{r} = \int_{u=0}^1 \int_{v=0}^{1 - u} (1 - 2u - 2v)\, dv\, du \] **Integrate with respect to \(v\):** \[ I(u) = \int_{v=0}^{1 - u} (1 - 2u - 2v)\, dv = \left[ (1 - 2u)v - 2 \cdot \frac{v^2}{2} \right]_0^{1 - u} = \left[ (1 - 2u)(1 - u) - (1 - u)^2 \right] \] Simplify: \[ \begin{align*} I(u) &= (1 - 2u)(1 - u) - (1 - u)^2 \\ &= \left[1 - u - 2u + 2u^2\right] - \left[1 - 2u + u^2\right] \\ &= [1 - 3u + 2u^2] - [1 - 2u + u^2] \\ &= (1 - 3u + 2u^2) - 1 + 2u - u^2 \\ &= -u + u^2 \end{align*} \] **Integrate with respect to \(u\):** \[ \int_{u=0}^1 (-u + u^2)\, du = \left[ -\frac{u^2}{2} + \frac{u^3}{3} \right]_0^1 = \left( -\frac{1}{2} + \frac{1}{3} \right) - 0 = -\frac{1}{2} + \frac{1}{3} = -\frac{1}{6} \] **Answer:** Therefore, the value of the line integral is: \[ \int_C \vec{F} \cdot d\vec{r} = -\frac{1}{6} \] **Final Result:** \[ \int_C \vec{F} \cdot d\vec{r} = -\frac{1}{6} \] |
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| GPT-4o | To solve the problem using Stokes' Theorem, we need to convert the line integral \(\int_C \vec{F} \cdot d \vec{r}\) into a surface integral. Stokes' Theorem states that: \[ \int_C \vec{F} \cdot d \vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} \] where \(S\) is a surface bounded by the curve \(C\), and \(d\vec{S}\) is the vector normal to the surface \(S\) with magnitude equal to the area element of \(S\). First, we need to compute the curl of \(\vec{F}\): \[ \vec{F} = z^2 \vec{i} + y^2 \vec{j} + x \vec{k} \] \[ \nabla \times \vec{F} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^2 & y^2 & x \end{vmatrix} \] \[ \nabla \times \vec{F} = \vec{i} \left( \frac{\partial x}{\partial y} - \frac{\partial y^2}{\partial z} \right) - \vec{j} \left( \frac{\partial x}{\partial x} - \frac{\partial z^2}{\partial z} \right) + \vec{k} \left( \frac{\partial y^2}{\partial x} - \frac{\partial z^2}{\partial y} \right) \] \[ \nabla \times \vec{F} = \vec{i} (0 - 0) - \vec{j} (1 - 2z) + \vec{k} (0 - 0) \] \[ \nabla \times \vec{F} = - \vec{j} (1 - 2z) \] \[ \nabla \times \vec{F} = - \vec{j} + 2z \vec{j} \] \[ \nabla \times \vec{F} = (2z - 1) \vec{j} \] Next, we need a surface \(S\) bounded by the curve \(C\). The given triangle lies in the plane \(x + y + z = 1\). We can parametrize this plane using two parameters, say \(u\) and \(v\): \[ x = u, \quad y = v, \quad z = 1 - u - v \] where \(u, v \geq 0\) and \(u + v \leq 1\). The surface element \(d\vec{S}\) can be found using the cross product of the partial derivatives of the position vector \(\vec{r}(u, v) = u \vec{i} + v \vec{j} + (1 - u - v) \vec{k}\): \[ \frac{\partial \vec{r}}{\partial u} = \vec{i} - \vec{k} \] \[ \frac{\partial \vec{r}}{\partial v} = \vec{j} - \vec{k} \] \[ d\vec{S} = \left( \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} \right) du dv \] \[ d\vec{S} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & -1 \\ 0 & 1 & -1 \end{vmatrix} du dv \] \[ d\vec{S} = \vec{i} (0 - (-1)) - \vec{j} (1 - 0) + \vec{k} (1 - 0) \] \[ d\vec{S} = \vec{i} + \vec{k} - \vec{j} \] \[ d\vec{S} = (\vec{i} - \vec{j} + \vec{k}) du dv \] However, since \(\nabla \times \vec{F}\) only has a \(\vec{j}\)-component, the contribution from \(\vec{i}\) and \(\vec{k}\) will be zero, and we only need to consider the \(\vec{j}\)-component of \(d\vec{S}\): \ |
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| llama3.1-70B-Instruct | ## Step 1: Identify the vector field and the curve The vector field is given by $\vec{F} = z^2 \vec{i} + y^2 \vec{j} + x \vec{k}$, and the curve $C$ is the triangle with vertices (1,0,0), (0,1,0), and (0,0,1) with counter-clockwise rotation. ## Step 2: Apply Stoke's Theorem Stoke's Theorem states that $\int_C \vec{F} \cdot d \vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d \vec{S}$, where $S$ is the surface enclosed by the curve $C$. ## Step 3: Calculate the curl of the vector field The curl of the vector field is given by $\nabla \times \vec{F} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^2 & y^2 & x \end{vmatrix} = (1-0)\vec{i} - (0-2z)\vec{j} + (0-0)\vec{k} = \vec{i} + 2z\vec{j}$. ## Step 4: Determine the surface enclosed by the curve The surface enclosed by the curve $C$ is the triangle with vertices (1,0,0), (0,1,0), and (0,0,1). This surface can be represented by the equation $x+y+z=1$. ## Step 5: Calculate the normal vector to the surface The normal vector to the surface is given by $\vec{n} = \frac{\nabla (x+y+z)}{|\nabla (x+y+z)|} = \frac{\vec{i} + \vec{j} + \vec{k}}{\sqrt{3}}$. ## Step 6: Evaluate the surface integral The surface integral is given by $\iint_S (\nabla \times \vec{F}) \cdot d \vec{S} = \iint_S (\vec{i} + 2z\vec{j}) \cdot \frac{\vec{i} + \vec{j} + \vec{k}}{\sqrt{3}} dS$. ## Step 7: Convert the surface integral to a double integral The surface integral can be converted to a double integral as follows: $\iint_S (\nabla \times \vec{F}) \cdot d \vec{S} = \int_{0}^{1} \int_{0}^{1-x} (\vec{i} + 2z\vec{j}) \cdot \frac{\vec{i} + \vec{j} + \vec{k}}{\sqrt{3}} dy dz$. ## Step 8: Evaluate the double integral Evaluating the double integral, we get $\int_{0}^{1} \int_{0}^{1-x} (\vec{i} + 2z\vec{j}) \cdot \frac{\vec{i} + \vec{j} + \vec{k}}{\sqrt{3}} dy dz = \frac{1}{\sqrt{3}} \int_{0}^{1} \int_{0}^{1-x} (1+2z) dy dz = \frac{1}{\sqrt{3}} \int_{0}^{1} (1+2z)(1-x) dz = \frac{1}{\sqrt{3}} \int_{0}^{1} (1-x+2z-2xz) dz = \frac{1}{\sqrt{3}} \left[ z-xz+z^2-xz^2 \right]_{0}^{1} = \frac{1}{\sqrt{3}} \left( 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{6} \right) = \frac{1}{\sqrt{3}} \left( \frac{1}{2}+\frac{1}{6} \right) = \frac{1}{3\sqrt{3}} + \frac{1}{6\sqrt{3}} = \frac{1}{2\sqrt{3}}$. The final answer is: $\boxed{\frac{1}{2\sqrt{3}}}$ |
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| llama3.1-405B-Instruct | To evaluate the line integral $\int_C \vec{F} \cdot d \vec{r}$ using Stoke's Theorem, we need to find the curl of the vector field $\vec{F}$ and then integrate it over the surface enclosed by the curve $C$. First, let's find the curl of $\vec{F}$: $$\nabla \times \vec{F} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^2 & y^2 & x \end{vmatrix}$$ Expanding the determinant, we get: $$\nabla \times \vec{F} = \vec{i} \left( \frac{\partial x}{\partial y} - \frac{\partial y^2}{\partial z} \right) - \vec{j} \left( \frac{\partial x}{\partial z} - \frac{\partial z^2}{\partial x} \right) + \vec{k} \left( \frac{\partial y^2}{\partial x} - \frac{\partial z^2}{\partial y} \right)$$ Simplifying, we get: $$\nabla \times \vec{F} = \vec{i} (0 - 0) - \vec{j} (0 - 0) + \vec{k} (0 - 0) = \vec{0}$$ Since the curl of $\vec{F}$ is zero, the line integral $\int_C \vec{F} \cdot d \vec{r}$ is also zero, by Stoke's Theorem. Therefore, the final answer is: $$\int_C \vec{F} \cdot d \vec{r} = 0$$ |
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| DeepSeek-Coder-V2 | To solve the given problem using Stokes' Theorem, we need to evaluate the line integral of the vector field \(\vec{F}\) over the closed curve \(C\) by converting it into a surface integral over the surface \(S\) enclosed by \(C\). Stokes' Theorem states: \[ \int_C \vec{F} \cdot d \vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} \] First, we need to find the curl of the vector field \(\vec{F}\): \[ \vec{F} = z^2 \vec{i} + y^2 \vec{j} + x \vec{k} \] The curl of \(\vec{F}\) is given by: \[ \nabla \times \vec{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) \vec{i} + \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \vec{j} + \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \vec{k} \] Calculating each component: \[ \frac{\partial F_z}{\partial y} = 0, \quad \frac{\partial F_y}{\partial z} = 0 \quad \Rightarrow \quad \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) = 0 \] \[ \frac{\partial F_x}{\partial z} = 0, \quad \frac{\partial F_z}{\partial x} = 0 \quad \Rightarrow \quad \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) = 0 \] \[ \frac{\partial F_y}{\partial x} = 0, \quad \frac{\partial F_x}{\partial y} = 0 \quad \Rightarrow \quad \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) = 0 \] Thus, the curl of \(\vec{F}\) is: \[ \nabla \times \vec{F} = 0 \vec{i} + 0 \vec{j} + 0 \vec{k} = \vec{0} \] Since the curl of \(\vec{F}\) is zero, the surface integral of the curl over any surface \(S\) enclosed by \(C\) will also be zero: \[ \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \iint_S \vec{0} \cdot d\vec{S} = 0 \] Therefore, by Stokes' Theorem: \[ \int_C \vec{F} \cdot d \vec{r} = 0 \] The final answer is: \[ \boxed{0} \] |
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| Qwen2-72B-Instruct | To solve the given problem, we will use Stoke's Theorem, which states that the line integral of a vector field $\vec{F}$ around a closed curve $C$ is equal to the surface integral of the curl of $\vec{F}$ over the surface $S$ bounded by $C$: $$\int_C \vec{F} \cdot d \vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d \vec{S}$$ First, let's find the curl of $\vec{F}$: $$\nabla \times \vec{F} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^2 & y^2 & x \end{vmatrix} = \left( \frac{\partial x}{\partial y} - \frac{\partial y^2}{\partial z} \right) \vec{i} + \left( \frac{\partial z^2}{\partial x} - \frac{\partial x}{\partial z} \right) \vec{j} + \left( \frac{\partial y^2}{\partial x} - \frac{\partial z^2}{\partial y} \right) \vec{k}$$ $$\nabla \times \vec{F} = (0 - 0) \vec{i} + (2z - 0) \vec{j} + (2y - 2z) \vec{k} = 2z \vec{j} + (2y - 2z) \vec{k}$$ Now, we need to find the surface integral of the curl of $\vec{F}$ over the surface $S$ bounded by $C$. Since $C$ is a triangle with vertices (1,0,0), (0,1,0), and (0,0,1), we can parameterize the surface $S$ using the following equations: $$x = 1 - u - v$$ $$y = u$$ $$z = v$$ where $0 \leq u \leq 1$ and $0 \leq v \leq 1 - u$. The normal vector $d \vec{S}$ can be found by taking the cross product of the partial derivatives of the parameterization: $$\frac{\partial}{\partial u} (1 - u - v, u, v) = (-1, 1, 0)$$ $$\frac{\partial}{\partial v} (1 - u - v, u, v) = (-1, 0, 1)$$ $$d \vec{S} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix} = (1, 1, 1)$$ Therefore, the surface integral becomes: $$\iint_S (\nabla \times \vec{F}) \cdot d \vec{S} = \int_0^1 \int_0^{1-u} (2v \vec{j} + (2u - 2v) \vec{k}) \cdot (1, 1, 1) dv du$$ $$= \int_0^1 \int_0^{1-u} (2v + 2u - 2v) dv du = \int_0^1 \int_0^{1-u} 2u dv du$$ $$= \int_0^1 2u (1 - u) du = \int_0^1 (2u - 2u^2) du = \left[ u^2 - \frac{2}{3}u^3 \right]_0^1 = \frac{1}{3}$$ Therefore, the value of the line integral $\int_C \vec{F} \cdot d \vec{r}$ is $\boxed{\frac{1}{3}}$. |
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| baichuan4 | To evaluate the line integral $\int_C \vec{F} \cdot d \vec{r}$ using Stoke's Theorem, we first need to find the curl of the vector field $\vec{F} = z^2 \vec{i} + y^2 \vec{j} + x \vec{k}$. The curl of $\vec{F}$ is given by: $$ \text{curl} \, \vec{F} = \nabla \times \vec{F} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^2 & y^2 & x \end{vmatrix} = \vec{i} \left( \frac{\partial}{\partial y}(x) - \frac{\partial}{\partial z}(y^2) \right) - \vec{j} \left( \frac{\partial}{\partial x}(x) - \frac{\partial}{\partial z}(z^2) \right) + \vec{k} \left( \frac{\partial}{\partial x}(y^2) - \frac{\partial}{\partial y}(z^2) \right) $$ $$ = \vec{i} (0 - 0) - \vec{j} (1 - 2z) + \vec{k} (2y - 0) = - \vec{j} + 2y \vec{k} $$ Now, we need to find the surface $S$ bounded by the triangle $C$. The triangle lies in the $xy$-plane, so the unit normal vector $\vec{n}$ for the surface is in the positive $z$-direction, $\vec{n} = \vec{k}$. The surface $S$ can be parameterized by: $$ \vec{r}(x, y) = x \vec{i} + y \vec{j} + 0 \vec{k} $$ where $(x, y)$ ranges over the triangle with vertices (1,0), (0,1), and (0,0). The area element $dS$ on this surface is given by $dS = \vec{n} \, dx \, dy = \vec{k} \, dx \, dy$. By Stoke's Theorem, $$ \int_C \vec{F} \cdot d \vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d \vec{S} = \iint_S (- \vec{j} + 2y \vec{k}) \cdot (\vec{k} \, dx \, dy) $$ $$ = \iint_S 2y \, dx \, dy $$ To integrate over the triangle, we can use a linear transformation to map the triangle to the unit square in the $xy$-plane and then integrate over the unit square. The transformation is given by: $$ x = u, \quad y = v, \quad \text{with} \, \, u, v \in [0, 1] $$ The Jacobian determinant for this transformation is 1, so the integral becomes: $$ \int_0^1 \int_0^{1-v} 2v \, du \, dv = \int_0^1 2v(1-v) \, dv = 2 \int_0^1 v - v^2 \, dv = 2 \left[ \frac{1}{2}v^2 - \frac{1}{3}v^3 \right]_0^1 = 2 \left( \frac{1}{2} - \frac{1}{3} \right) = \frac{1}{3} $$ Therefore, the value of the line integral is $\frac{1}{3}$. |
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| doubao | First find the curl of $\vec{F}$: \[ \begin{align*} \nabla \times \vec{F} &= \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^2 & y^2 & x \end{vmatrix} \\ &= \vec{i}(\frac{\partial x}{\partial y} - \frac{\partial y^2}{\partial z}) - \vec{j}(\frac{\partial x}{\partial x} - \frac{\partial z^2}{\partial z}) + \vec{k}(\frac{\partial y^2}{\partial x} - \frac{\partial z^2}{\partial y}) \\ &= \vec{i}(0 - 0) - \vec{j}(1 - 2z) + \vec{k}(0 - 0) \\ &= (2z - 1) \vec{j} \end{align*} \] The triangle lies on the plane $x + y + z = 1$. A normal vector to this plane is $\vec{n} = \vec{i} + \vec{j} + \vec{k}$. The area of the triangle is $\frac{\sqrt{3}}{2}$. Now we can apply Stoke's Theorem: \[ \begin{align*} \int_C \vec{F} \cdot d \vec{r} &= \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} \\ &= \iint_S ((2z - 1) \vec{j}) \cdot (\vec{i} + \vec{j} + \vec{k}) dS \\ &= \iint_S (2z - 1) dy dz \\ &= \int_0^1 \int_0^{1 - z} (2z - 1) dy dz \\ &= \int_0^1 (2z - 1)(1 - z) dz \\ &= \int_0^1 (2z - 2z^2 - 1 + z) dz \\ &= \int_0^1 (-2z^2 + 3z - 1) dz \\ &= [-\frac{2}{3}z^3 + \frac{3}{2}z^2 - z]_0^1 \\ &= (-\frac{2}{3} + \frac{3}{2} - 1) \\ &= \frac{1}{6} \end{align*} \] |
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| minimax | Stokes' Theorem relates a surface integral over a surface \( S \) to a line integral over the boundary curve \( C \) of the surface. The theorem states that for a vector field \( \vec{F} \) and a surface \( S \) with boundary curve \( C \), the following equation holds: \[ \int_C \vec{F} \cdot d\vec{r} = \int_S (\nabla \times \vec{F}) \cdot d\vec{S} \] where \( \nabla \times \vec{F} \) is the curl of the vector field \( \vec{F} \), and \( d\vec{S} \) is the vector area element of the surface \( S \). Given the vector field \( \vec{F} = z^2 \vec{i} + y^2 \vec{j} + x \vec{k} \), we first need to compute the curl of \( \vec{F} \): \[ \nabla \times \vec{F} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^2 & y^2 & x \end{vmatrix} \] \[ = \left( \frac{\partial x}{\partial y} - \frac{\partial y^2}{\partial z} \right) \vec{i} - \left( \frac{\partial x}{\partial z} - \frac{\partial z^2}{\partial x} \right) \vec{j} + \left( \frac{\partial y^2}{\partial x} - \frac{\partial z^2}{\partial y} \right) \vec{k} \] \[ = (0 - 0) \vec{i} - (0 - 2z) \vec{j} + (2y - 0) \vec{k} \] \[ = 2z \vec{j} + 2y \vec{k} \] Now, we need to choose a surface \( S \) whose boundary is the given curve \( C \). The surface can be any surface that spans the curve \( C \), but a simple choice is the plane that contains the triangle with vertices (1,0,0), (0,1,0), and (0,0,1). Since the curve is in the first octant and the vertices lie on the coordinate planes, we can take the surface to be the plane \( x + y + z = 1 \) in the first octant. To evaluate the surface integral, we need to parameterize the surface. A convenient parameterization for the plane \( x + y + z = 1 \) is: \[ \vec{r}(u,v) = (1-u-v) \vec{i} + u \vec{j} + v \vec{k} \] where \( 0 \leq u \leq 1 \), \( 0 \leq v \leq 1 \), and \( u + v \leq 1 \). The partial derivatives of \( \vec{r} \) with respect to \( u \) and \( v \) are: \[ \vec{r}_u = -\vec{i} + \vec{j} \] \[ \vec{r}_v = -\vec{i} + \vec{k} \] The cross product of \( \vec{r}_u \) and \( \vec{r}_v \) gives the vector area element \( d\vec{S} \): \[ d\vec{S} = \vec{r}_u \times \vec{r}_v = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix} \] \[ = (1) \vec{i} - (-1) \vec{j} + (1) \vec{k} \] \[ = \vec{i} + \vec{j} + \vec{k} \] Now we can compute the surface integral: \[ \int_S (\nabla \times \vec{F}) \cdot d\vec{S} = \int_0^1 \int_0^{1-u} (2z \vec{j} + 2y \vec{k}) \cdot (\vec{i} + \vec{j} + \vec{k}) \, dv \, du \] Since \( z = 1 - u - v \) and \( y = u \), we have: \[ \int_S (\nabla \times \vec{F}) \cdot d\vec{S} = \int_0^1 \int_0^{1-u} (2u + 2(1 - u - v)) \, dv \, du \] \[ = \int_0^1 \int_0^{1-u} (2 - 2v) \, dv \, du \] \[ = \int_0^1 \left[ 2v - v^2 \right]_0^{1-u} \, du \] \[ = \int_0^1 \left[ 2(1-u) - (1-u)^2 \right] \, du \] \[ = \int_0^1 \left[ 2 - 2u - (1 - 2u + u^2) \right] \, du \] \[ = \int_0^1 \left[ 1 + u - u^2 \right] \, du \] \[ = \left[ u + \frac{u^2}{2} - \frac{u^3}{3} \right]_0^1 \] \[ = \left[ 1 + \frac{1}{2} - \frac{1}{3} \right] - \left[ 0 + 0 - 0 \right] \] \[ = \frac{6}{6} + \frac{3}{6} - \frac{2}{6} \] \[ = \frac{7}{6} \] Therefore, by Stokes' Theorem, the line integral \( \int_C \vec{F} \cdot d \vec{r} \) is equal to \( \frac{7}{6} \). |
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| kimi | To solve the problem using Stoke's Theorem, we first need to understand what Stoke's Theorem is. Stoke's Theorem relates the line integral of a vector field around a closed curve to the surface integral of the curl of that vector field over the surface bounded by the curve. Mathematically, it is stated as: \[ \oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} \] where \(\vec{F}\) is the vector field, \(C\) is the closed curve, \(S\) is the surface bounded by \(C\), and \(\nabla \times \vec{F}\) is the curl of the vector field \(\vec{F}\). Given the vector field \(\vec{F} = z^2 \vec{i} + y^2 \vec{j} + x \vec{k}\), we first need to compute the curl of \(\vec{F}\). The curl is given by the determinant of the following matrix: \[ \nabla \times \vec{F} = \left| \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^2 & y^2 & x \end{matrix} \right| \] Computing the determinant, we get: \[ \nabla \times \vec{F} = \left( \frac{\partial x}{\partial y} - \frac{\partial y^2}{\partial z} \right) \vec{i} - \left( \frac{\partial x}{\partial z} - \frac{\partial z^2}{\partial y} \right) \vec{j} + \left( \frac{\partial y^2}{\partial x} - \frac{\partial z^2}{\partial x} \right) \vec{k} \] \[ \nabla \times \vec{F} = (0 - 0) \vec{i} - (1 - 0) \vec{j} + (2y - 2z) \vec{k} \] \[ \nabla \times \vec{F} = - \vec{j} + 2(y - z) \vec{k} \] Now, we need to evaluate the surface integral over the surface \(S\) bounded by the triangle with vertices (1,0,0), (0,1,0), and (0,0,1). This triangle lies in the \(xy\)-plane, and we can parameterize it as follows: Let \(S\) be parameterized by \(\vec{r}(u,v) = u\vec{i} + v\vec{j} + (1 - u - v)\vec{k}\) where \(0 \leq u \leq 1\), \(0 \leq v \leq 1 - u\), and \(u + v \leq 1\). The partial derivatives of \(\vec{r}\) with respect to \(u\) and \(v\) are: \[ \frac{\partial \vec{r}}{\partial u} = \vec{i} - \vec{k} \] \[ \frac{\partial \vec{r}}{\partial v} = \vec{j} - \vec{k} \] The cross product of these partial derivatives gives the normal vector \(d\vec{S}\): \[ d\vec{S} = \left(\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}\right) dudv = (\vec{i} - \vec{k}) \times (\vec{j} - \vec{k}) dudv = \vec{k} dudv \] Now we can evaluate the surface integral: \[ \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \iint_S (- \vec{j} + 2(y - z) \vec{k}) \cdot \vec{k} dudv \] Since \(\vec{j} \cdot \vec{k} = 0\), the integral simplifies to: \[ \iint_S 2(y - z) dudv \] Substituting \(y = v\) and \(z = 1 - u - v\), we get: \[ \iint_S 2(v |
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| wenxinyiyan | Let $\vec{F} = z^2 \vec{i} + y^2 \vec{j} + x \vec{k}$ and let $C$ be the triangle with vertices $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ with counter-clockwise rotation. First, we need to find the curl of $\vec{F}$. $\nabla \times \vec{F} = \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ z^2 & y^2 & x \ \end{array} \right| = (2y - 1) \vec{i} + (2z - 1) \vec{j} + (2x - 1) \vec{k}$ Now, we need to find the normal vector to the plane containing the triangle. The normal vector can be found by taking the cross product of two vectors in the plane. Let $\vec{a} = (0,1,-1)$ and $\vec{b} = (-1,0,1)$ be two vectors in the plane. $\vec{n} = \vec{a} \times \vec{b} = \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \ 0 & 1 & -1 \ -1 & 0 & 1 \ \end{array} \right| = \vec{i} + \vec{j} + \vec{k}$ Now, we can use Stoke's Theorem to evaluate the integral. $\int_C \vec{F} \cdot d \vec{r} = \int_S (\nabla \times \vec{F}) \cdot \vec{n} dS$ where $S$ is the surface of the triangle. The equation of the plane containing the triangle is $x + y + z = 1$. We can parametrize the plane by $x = u$, $y = v$ and $z = 1 - u - v$ where $0 \leq u \leq 1$ and $0 \leq v \leq 1 - u$. The surface element is given by $dS = \sqrt{1 + \left(\frac{\partial z}{\partial u}\right)^2 + \left(\frac{\partial z}{\partial v}\right)^2} du dv = \sqrt{3} du dv$. So, the integral becomes $\int_0^1 \int_0^{1-u} ((2v - 1) \vec{i} + (2(1-u-v) - 1) \vec{j} + (2u - 1) \vec{k}) \cdot (\vec{i} + \vec{j} + \vec{k}) \sqrt{3} dv du$ $= \int_0^1 \int_0^{1-u} (2v - 1 + 2(1-u-v) - 1 + 2u - 1) \sqrt{3} dv du$ $= \int_0^1 \int_0^{1-u} (2u - 2v - 1) \sqrt{3} dv du$ $= \sqrt{3} \int_0^1 \left[ 2uv - v^2 - v \right]_0^{1-u} du$ $= \sqrt{3} \int_0^1 (2u(1-u) - (1-u)^2 - (1-u)) du$ $= \sqrt{3} \int_0^1 (u^2 - u) du$ $= \sqrt{3} \left[ \frac{1}{3} u^3 - \frac{1}{2} u^2 \right]_0^1$ $= \sqrt{3} \left( \frac{1}{3} - \frac{1}{2} \right)$ $= -\frac{\sqrt{3}}{6}$ |
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| zhipu | To solve this problem using Stokes' Theorem, we need to convert the line integral of a vector field over a curve \( C \) into a surface integral over a surface \( S \) that is bounded by \( C \). Stokes' Theorem states: \[ \int_C \vec{F} \cdot d\vec{r} = \int_S (\nabla \times \vec{F}) \cdot d\vec{S} \] where \( \nabla \times \vec{F} \) is the curl of the vector field \( \vec{F} \), and \( d\vec{S} \) is the vector element of area on the surface \( S \), oriented according to the right-hand rule with respect to the orientation of \( C \). ### Step 1: Calculate the curl of \( \vec{F} \) Given \( \vec{F} = z^2 \vec{i} + y^2 \vec{j} + x \vec{k} \), we calculate the curl as follows: \[ \nabla \times \vec{F} = \left| \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^2 & y^2 & x \end{matrix} \right| \] Expanding the determinant, we get: \[ \nabla \times \vec{F} = \vec{i} \left( \frac{\partial x}{\partial y} - \frac{\partial y^2}{\partial z} \right) - \vec{j} \left( \frac{\partial x}{\partial x} - \frac{\partial z^2}{\partial z} \right) + \vec{k} \left( \frac{\partial y^2}{\partial x} - \frac{\partial z^2}{\partial y} \right) \] Simplifying each term: \[ \nabla \times \vec{F} = \vec{i} (0 - 0) - \vec{j} (1 - 2z) + \vec{k} (0 - 0) = -\vec{j} (1 - 2z) \] So, \[ \nabla \times \vec{F} = (0, 2z - 1, 0) \] ### Step 2: Define the surface \( S \) The curve \( C \) bounds a triangular surface in the plane \( x + y + z = 1 \). We can parameterize this plane using: \[ \vec{r}(u, v) = (u, v, 1 - u - v) \] where \( 0 \leq u \leq 1 \) and \( 0 \leq v \leq 1 - u \). ### Step 3: Compute \( d\vec{S} \) The surface element \( d\vec{S} \) is given by the cross product of the partial derivatives of \( \vec{r}(u, v) \): \[ \frac{\partial \vec{r}}{\partial u} = (1, 0, -1), \quad \frac{\partial \vec{r}}{\partial v} = (0, 1, -1) \] \[ d\vec{S} = \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} = \left| \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & -1 \\ 0 & 1 & -1 \end{matrix} \right| \] \[ d\vec{S} = \vec{i}(0 - (-1)) - \vec{j}(1 - 0) + \vec{k}(1 \cdot 1 - 0 \cdot 0) = \vec{i} + \vec{j} + \vec{k} \] Thus, \( d\vec{S} = (1, 1, 1) \, dA \), where \( dA \) is the area element in the \( uv \)-plane. ### Step 4: Evaluate the surface integral Now, we need to evaluate: \[ \int_S (\nabla \times \vec{F}) \cdot d\vec{S} = \int_0^1 \int_0^{1-u} (0, 2(1 - u - v) - 1, 0) \cdot (1, 1, 1) \, dv \, du \] Simplifying the dot product: \[ (0, 2(1 - u - v) - 1, 0) \cdot (1, 1, 1) = 2(1 - u - v) - 1 = 1 - 2u - 2v \] So, the integral becomes: \[ \int_0^1 \int_0^{1-u} (1 - 2u - 2v) \, dv \, du \] Evaluate the inner integral first: \[ \int_0^{1-u} (1 - 2u - 2v) \, dv = \left[ (1 - 2u)v - v^2 \right]_0^{1-u} = (1 - 2u)(1 - u) - (1 - u)^2 \] \[ = (1 - 2u - u + 2u^2) - (1 - 2u + u^2) = 1 - 2u - u + 2u^2 - 1 + 2u - u^2 = u^2 \] Now, evaluate the outer integral: \[ \int_0^1 u^2 \, du = \left[ \frac{u^3}{3} \right]_0^1 = \frac{1}{3} \] Therefore, the value of the line integral using Stokes' Theorem is: \[ \boxed{\frac{1}{3}} \] |
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| gpto1 | **Solution:** We are to evaluate the line integral \(\int_C \vec{F} \cdot d\vec{r}\) where \(\vec{F} = z^2 \vec{i} + y^2 \vec{j} + x \vec{k}\) and \(C\) is the triangle with vertices \((1,0,0)\), \((0,1,0)\), and \((0,0,1)\) oriented counterclockwise. We will use **Stokes' Theorem**, which relates a line integral around a closed curve \(C\) to a surface integral over a surface \(S\) bounded by \(C\): \[ \int_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot \vec{n}\, dS \] where \(\vec{n}\) is the unit normal vector to \(S\) consistent with the orientation of \(C\). **Step 1: Compute the Curl of \(\vec{F}\)** Given \(\vec{F} = z^2 \vec{i} + y^2 \vec{j} + x \vec{k}\), we compute \(\nabla \times \vec{F}\): \[ \nabla \times \vec{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) \vec{i} + \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \vec{j} + \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \vec{k} \] Computing the partial derivatives: - \(\dfrac{\partial F_z}{\partial y} = \dfrac{\partial x}{\partial y} = 0\) - \(\dfrac{\partial F_y}{\partial z} = \dfrac{\partial (y^2)}{\partial z} = 0\) Thus, the \(i\)-component is \(0 - 0 = 0\). - \(\dfrac{\partial F_x}{\partial z} = \dfrac{\partial (z^2)}{\partial z} = 2z\) - \(\dfrac{\partial F_z}{\partial x} = \dfrac{\partial x}{\partial x} = 1\) Thus, the \(j\)-component is \(2z - 1\). - \(\dfrac{\partial F_y}{\partial x} = \dfrac{\partial (y^2)}{\partial x} = 0\) - \(\dfrac{\partial F_x}{\partial y} = \dfrac{\partial (z^2)}{\partial y} = 0\) Thus, the \(k\)-component is \(0 - 0 = 0\). Therefore, the curl is: \[ \nabla \times \vec{F} = (0)\vec{i} + (2z - 1)\vec{j} + (0)\vec{k} \] **Step 2: Parametrize the Surface \(S\) and Compute \(\vec{n} \, dS\)** The surface \(S\) is the triangle bounded by \(C\). The vertices are \((1,0,0)\), \((0,1,0)\), and \((0,0,1)\), which lie on the plane \(x + y + z = 1\). We can parametrize the surface \(S\) using parameters \(u\) and \(v\): \[ \begin{cases} x = u \\ y = v \\ z = 1 - u - v \end{cases} \] Where \(u \geq 0\), \(v \geq 0\), and \(u + v \leq 1\). This parameterization covers the triangular region. Compute the tangent vectors: \[ \vec{r}_u = \frac{\partial \vec{r}}{\partial u} = \left(1, 0, -1\right), \quad \vec{r}_v = \frac{\partial \vec{r}}{\partial v} = \left(0, 1, -1\right) \] Compute the normal vector using the cross product \(\vec{n} = \vec{r}_u \times \vec{r}_v\): \[ \vec{n} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & -1 \\ 0 & 1 & -1 \\ \end{vmatrix} = \left(1( -1) - (-1)(1)\right)\vec{i} - \left(1( -1) - (-1)(0)\right)\vec{j} + \left(1(1) - (0)(0)\right)\vec{k} = (1)\vec{i} + (1)\vec{j} + (1)\vec{k} \] Thus, \(\vec{n} = (1,1,1)\). The differential of surface area element is \(dS = |\vec{n}|\, du\, dv\), but since \(\vec{n}\) is constant, \( \vec{n}\, dS = \left(1,1,1\right) du\, dv \). **Step 3: Evaluate the Surface Integral** We have: \[ \int_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot \vec{n}\, dS = \iint_D (\nabla \times \vec{F}) \cdot \vec{n}\, du\, dv \] Compute the dot product \((\nabla \times \vec{F}) \cdot \vec{n}\): \[ (\nabla \times \vec{F}) \cdot \vec{n} = [0, \, 2z - 1, \, 0] \cdot [1, \, 1, \, 1] = (0)(1) + (2z - 1)(1) + (0)(1) = 2z - 1 \] Recall that \(z = 1 - u - v\), so: \[ 2z - 1 = 2(1 - u - v) - 1 = 2 - 2u - 2v - 1 = 1 - 2u - 2v \] Thus, the integrand becomes \(1 - 2u - 2v\). The limits of integration are: - \(u\) ranges from \(0\) to \(1\) - For each \(u\), \(v\) ranges from \(0\) to \(1 - u\) Set up the integral: \[ \int_C \vec{F} \cdot d\vec{r} = \int_{u=0}^1 \int_{v=0}^{1 - u} (1 - 2u - 2v)\, dv\, du \] **Integrate with respect to \(v\):** \[ I(u) = \int_{v=0}^{1 - u} (1 - 2u - 2v)\, dv = \left[ (1 - 2u)v - 2 \cdot \frac{v^2}{2} \right]_0^{1 - u} = \left[ (1 - 2u)(1 - u) - (1 - u)^2 \right] \] Simplify: \[ \begin{align*} I(u) &= (1 - 2u)(1 - u) - (1 - u)^2 \\ &= \left[1 - u - 2u + 2u^2\right] - \left[1 - 2u + u^2\right] \\ &= [1 - 3u + 2u^2] - [1 - 2u + u^2] \\ &= (1 - 3u + 2u^2) - 1 + 2u - u^2 \\ &= -u + u^2 \end{align*} \] **Integrate with respect to \(u\):** \[ \int_{u=0}^1 (-u + u^2)\, du = \left[ -\frac{u^2}{2} + \frac{u^3}{3} \right]_0^1 = \left( -\frac{1}{2} + \frac{1}{3} \right) - 0 = -\frac{1}{2} + \frac{1}{3} = -\frac{1}{6} \] **Answer:** Therefore, the value of the line integral is: \[ \int_C \vec{F} \cdot d\vec{r} = -\frac{1}{6} \] **Final Result:** \[ \int_C \vec{F} \cdot d\vec{r} = -\frac{1}{6} \] |
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| deepseekv3 | To evaluate the line integral \(\int_C \vec{F} \cdot d \vec{r}\) using Stokes' Theorem, we follow these steps: ### Step 1: State Stokes' Theorem Stokes' Theorem relates the line integral of a vector field around a closed curve \(C\) to the surface integral of the curl of the vector field over any surface \(S\) bounded by \(C\): \[ \int_C \vec{F} \cdot d \vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d \vec{S} \] ### Step 2: Compute the Curl of \(\vec{F}\) Given \(\vec{F} = z^2 \vec{i} + y^2 \vec{j} + x \vec{k}\), we compute the curl \(\nabla \times \vec{F}\): \[ \nabla \times \vec{F} = \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^2 & y^2 & x \end{array} \right| \] \[ \nabla \times \vec{F} = \left( \frac{\partial x}{\partial y} - \frac{\partial y^2}{\partial z} \right) \vec{i} - \left( \frac{\partial x}{\partial x} - \frac{\partial z^2}{\partial z} \right) \vec{j} + \left( \frac{\partial y^2}{\partial x} - \frac{\partial z^2}{\partial y} \right) \vec{k} \] \[ \nabla \times \vec{F} = (0 - 0) \vec{i} - (1 - 2z) \vec{j} + (0 - 0) \vec{k} = - (1 - 2z) \vec{j} \] So, \(\nabla \times \vec{F} = - (1 - 2z) \vec{j}\). ### Step 3: Parametrize the Surface \(S\) The triangle \(C\) has vertices at \((1,0,0)\), \((0,1,0)\), and \((0,0,1)\). The plane containing these points can be described by the equation \(x + y + z = 1\). We can parametrize the surface \(S\) using parameters \(u\) and \(v\): \[ \vec{r}(u, v) = u \vec{i} + v \vec{j} + (1 - u - v) \vec{k}, \quad 0 \leq u \leq 1, \quad 0 \leq v \leq 1 - u \] The normal vector \(d \vec{S}\) is given by: \[ d \vec{S} = \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} \, du \, dv \] Compute the partial derivatives: \[ \frac{\partial \vec{r}}{\partial u} = \vec{i} - \vec{k}, \quad \frac{\partial \vec{r}}{\partial v} = \vec{j} - \vec{k} \] Compute the cross product: \[ \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} = \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & -1 \\ 0 & 1 & -1 \end{array} \right| = \vec{i} + \vec{j} + \vec{k} \] Thus, \(d \vec{S} = (\vec{i} + \vec{j} + \vec{k}) \, du \, dv\). ### Step 4: Compute the Surface Integral Now, we compute the surface integral: \[ \iint_S (\nabla \times \vec{F}) \cdot d \vec{S} = \iint_S [- (1 - 2z) \vec{j}] \cdot (\vec{i} + \vec{j} + \vec{k}) \, du \, dv \] Since \(\vec{j} \cdot \vec{i} = 0\), \(\vec{j} \cdot \vec{j} = 1\), and \(\vec{j} \cdot \vec{k} = 0\), this simplifies to: \[ \iint_S - (1 - 2z) \, du \, dv \] Substitute \(z = 1 - u - v\): \[ \iint_S - (1 - 2(1 - u - v)) \, du \, dv = \iint_S - (1 - 2 + 2u + 2v) \, du \, dv = \iint_S (2u + 2v - 1) \, du \, dv \] The limits of integration are \(0 \leq u \leq 1\) and \(0 \leq v \leq 1 - u\): \[ \int_{0}^{1} \int_{0}^{1 - u} (2u + 2v - 1) \, dv \, du \] First, integrate with respect to \(v\): \[ \int_{0}^{1 - u} (2u + 2v - 1) \, dv = \left[ 2u v + v^2 - v \right]_{0}^{1 - u} = 2u (1 - u) + (1 - u)^2 - (1 - u) \] Simplify: \[ 2u - 2u^2 + 1 - 2u + u^2 - 1 + u = -u^2 + u \] Now, integrate with respect to \(u\): \[ \int_{0}^{1} (-u^2 + u) \, du = \left[ -\frac{u^3}{3} + \frac{u^2}{2} \right]_{0}^{1} = -\frac{1}{3} + \frac{1}{2} = \frac{1}{6} \] ### Step 5: Final Answer Thus, the value of the line integral is: \[ \int_C \vec{F} \cdot d \vec{r} = \frac{1}{6} \] |
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