| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the problem using the Divergence Theorem, we first need to calculate the divergence of the vector field \(\vec{F} = \sin(\pi x) \vec{i} + (z y^3)\vec{j} + (z^2 + 4x)\vec{k}\). The divergence of a vector field \(\vec{F} = F_1 \vec{i} + F_2 \vec{j} + F_3 \vec{k}\) is given by: \[ \text{div} \vec{F} = \nabla \cdot \vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \] For \(\vec{F} = \sin(\pi x) \vec{i} + (z y^3)\vec{j} + (z^2 + 4x)\vec{k}\), we have: \[ \text{div} \vec{F} = \frac{\partial}{\partial x}(\sin(\pi x)) + \frac{\partial}{\partial y}(z y^3) + \frac{\partial}{\partial z}(z^2 + 4x) \] \[ = \pi \cos(\pi x) + 3zy^2 + 2z \] Next, we need to integrate the divergence of \(\vec{F}\) over the volume \(V\) enclosed by the surface \(S\): \[ \iiint_V (\pi \cos(\pi x) + 3zy^2 + 2z) dV \] Since the volume \(V\) is a rectangular box with \(-1 \le x \le 2\), \(0 \le y \le 1\), and \(1 \le z \le 4\), we can evaluate the integral as follows: \[ \int_{-1}^{2} \int_0^1 \int_1^4 (\pi \cos(\pi x) + 3zy^2 + 2z) dz dy dx \] \[ = \int_{-1}^{2} \int_0^1 \left[ \pi z \cos(\pi x) + 3zy^2z + 2z^2 \right]_1^4 dy dx \] \[ = \int_{-1}^{2} \int_0^1 \left( 4\pi \cos(\pi x) + 12y^2 - \pi \cos(\pi x) - 3y^2 + 32 - 2 \right) dy dx \] \[ = \int_{-1}^{2} \int_0^1 \left( 3\pi \cos(\pi x) + 9y^2 + 30 \right) dy dx \] \[ = \int_{-1}^{2} \left[ 3\pi y \cos(\pi x) + 3y^3 + 30y \right]_0^1 dx \] \[ = \int_{-1}^{2} \left( 3\pi \cos(\pi x) + 3 + 30 \right) dx \] \[ = \int_{-1}^{2} \left( 3\pi \cos(\pi x) + 33 \right) dx \] \[ = \left[ 3\pi \frac{\sin(\pi x)}{\pi} + 33x \right]_{-1}^{2} \] \[ = \left[ 3\sin(\pi x) + 33x \right]_{-1}^{2} \] \[ = \left( 3\sin(2\pi) + 33(2) \right) - \left( 3\sin(-\pi) + 33(-1) \right) \] \[ = 66 - (-33) \] \[ = 99 \] So, the value of \(\iint_S \vec{F} \cdot d \vec{S}\) using the Divergence Theorem is \(\boxed{99}\). |
67.5 |
| deepseekr1 | To evaluate the surface integral \(\iint_S \vec{F} \cdot d \vec{S}\) using the divergence theorem, we first compute the divergence of the vector field \(\vec{F} = \sin(\pi x) \vec{i} + z y^3 \vec{j} + (z^2 + 4x) \vec{k}\). The divergence of \(\vec{F}\) is given by: \[ \text{div} \vec{F} = \frac{\partial}{\partial x} (\sin(\pi x)) + \frac{\partial}{\partial y} (z y^3) + \frac{\partial}{\partial z} (z^2 + 4x) \] Calculating each component: \[ \frac{\partial}{\partial x} (\sin(\pi x)) = \pi \cos(\pi x) \] \[ \frac{\partial}{\partial y} (z y^3) = 3 z y^2 \] \[ \frac{\partial}{\partial z} (z^2 + 4x) = 2 z \] Thus, the divergence is: \[ \text{div} \vec{F} = \pi \cos(\pi x) + 3 z y^2 + 2 z \] Using the divergence theorem, we convert the surface integral into a triple integral over the volume \(V\) of the box defined by \(-1 \le x \le 2\), \(0 \le y \le 1\), and \(1 \le z \le 4\): \[ \iint_S \vec{F} \cdot d \vec{S} = \iiint_V (\pi \cos(\pi x) + 3 z y^2 + 2 z) \, dV \] We split this into three separate integrals: \[ I_1 = \iiint_V \pi \cos(\pi x) \, dV \] \[ I_2 = \iiint_V 3 z y^2 \, dV \] \[ I_3 = \iiint_V 2 z \, dV \] For \(I_1\): \[ I_1 = \pi \int_{-1}^{2} \cos(\pi x) \, dx \int_{0}^{1} dy \int_{1}^{4} dz \] The integrals over \(y\) and \(z\) are 1 and 3 respectively. The integral over \(x\) is zero because \(\int_{-1}^{2} \cos(\pi x) \, dx = 0\). Thus, \(I_1 = 0\). For \(I_2\): \[ I_2 = 3 \int_{-1}^{2} dx \int_{0}^{1} y^2 \, dy \int_{1}^{4} z \, dz \] Calculating each integral: \[ \int_{-1}^{2} dx = 3, \quad \int_{0}^{1} y^2 \, dy = \frac{1}{3}, \quad \int_{1}^{4} z \, dz = \frac{15}{2} \] Thus, \[ I_2 = 3 \cdot 3 \cdot \frac{1}{3} \cdot \frac{15}{2} = \frac{45}{2} \] For \(I_3\): \[ I_3 = 2 \int_{-1}^{2} dx \int_{0}^{1} dy \int_{1}^{4} z \, dz \] Calculating each integral: \[ \int_{-1}^{2} dx = 3, \quad \int_{0}^{1} dy = 1, \quad \int_{1}^{4} z \, dz = \frac{15}{2} \] Thus, \[ I_3 = 2 \cdot 3 \cdot 1 \cdot \frac{15}{2} = 45 \] Summing the results: \[ I_1 + I_2 + I_3 = 0 + \frac{45}{2} + 45 = \frac{135}{2} \] The value of the surface integral is \(\boxed{\dfrac{135}{2}}\). |
67.5 |
| gpto1 | To solve the problem using the **Divergence Theorem**, we need to evaluate the surface integral \(\iint_S \vec{F} \cdot d\vec{S}\) by converting it into a triple integral over the volume \(V\) enclosed by the surface \(S\). **Given:** \[ \vec{F} = \sin(\pi x)\, \vec{i} + z y^3\, \vec{j} + (z^2 + 4x)\, \vec{k} \] \(S\) is the surface of the box defined by: - \(x\) ranges from \(-1\) to \(2\) - \(y\) ranges from \(0\) to \(1\) - \(z\) ranges from \(1\) to \(4\) **Steps:** 1. **Compute the Divergence of \(\vec{F}\):** The divergence of a vector field \(\vec{F} = F_x\, \vec{i} + F_y\, \vec{j} + F_z\, \vec{k}\) is given by: \[ \nabla \cdot \vec{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} \] Compute each partial derivative: - \(\dfrac{\partial F_x}{\partial x} = \dfrac{\partial}{\partial x} [\sin(\pi x)] = \pi \cos(\pi x)\) - \(\dfrac{\partial F_y}{\partial y} = \dfrac{\partial}{\partial y} [z y^3] = z \cdot 3 y^2\) - \(\dfrac{\partial F_z}{\partial z} = \dfrac{\partial}{\partial z} [z^2 + 4x] = 2z\) So the divergence is: \[ \nabla \cdot \vec{F} = \pi \cos(\pi x) + 3 z y^2 + 2z \] 2. **Set Up the Triple Integral over the Volume \(V\):** The Divergence Theorem states: \[ \iint_S \vec{F} \cdot d\vec{S} = \iiint_V (\nabla \cdot \vec{F})\, dV \] Our volume \(V\) is the box defined by the given ranges of \(x\), \(y\), and \(z\). The triple integral becomes: \[ \int_{x=-1}^{2} \int_{y=0}^{1} \int_{z=1}^{4} \left[ \pi \cos(\pi x) + 3 z y^2 + 2z \right] dz\, dy\, dx \] 3. **Evaluate the Triple Integral:** **Split the integral** into two parts for easier computation: \[ I = I_1 + I_2 \quad \text{where} \quad \begin{cases} I_1 = \int_{x=-1}^{2} \int_{y=0}^{1} \int_{z=1}^{4} \pi \cos(\pi x)\, dz\, dy\, dx \\ I_2 = \int_{x=-1}^{2} \int_{y=0}^{1} \int_{z=1}^{4} (3 z y^2 + 2z)\, dz\, dy\, dx \end{cases} \] **Compute \(I_1\):** Since \(\pi \cos(\pi x)\) does not depend on \(y\) or \(z\), and the integrals over \(y\) and \(z\) can be separated: \[ I_1 = \pi \left( \int_{x=-1}^{2} \cos(\pi x)\, dx \right) \left( \int_{y=0}^{1} dy \right) \left( \int_{z=1}^{4} dz \right) \] **Compute the \(x\)-integral of \(\cos(\pi x)\):** \[ \int_{x=-1}^{2} \cos(\pi x)\, dx = \frac{1}{\pi} \left[ \sin(\pi x) \right]_{x=-1}^{2} = \frac{1}{\pi} [\sin(2\pi) - \sin(-\pi)] = \frac{1}{\pi} (0 - 0) = 0 \] **Compute the \(y\) and \(z\) integrals:** \[ \int_{y=0}^{1} dy = 1, \quad \int_{z=1}^{4} dz = 4 - 1 = 3 \] **Thus, \(I_1 = 0\).** **Compute \(I_2\):** **First, compute the \(z\)-integral:** \[ \int_{z=1}^{4} (3 z y^2 + 2z)\, dz = (3 y^2 + 2) \int_{z=1}^{4} z\, dz = (3 y^2 + 2) \left[ \frac{1}{2} z^2 \right]_{z=1}^{4} = (3 y^2 + 2) \left( \frac{1}{2}(16 - 1) \right) = (3 y^2 + 2) \left( \frac{15}{2} \right) \] **Next, compute the \(y\)-integral:** \[ \int_{y=0}^{1} (3 y^2 + 2) \left( \frac{15}{2} \right)\, dy = \frac{15}{2} \int_{y=0}^{1} (3 y^2 + 2)\, dy = \frac{15}{2} \left[ y^3 + 2 y \right]_{y=0}^{1} = \frac{15}{2} (1 + 2) = \frac{15}{2} \times 3 = \frac{45}{2} \] **Finally, compute the \(x\)-integral:** \[ \int_{x=-1}^{2} dx = 2 - (-1) = 3 \] **Thus, \(I_2 = 3 \times \frac{45}{2} = \frac{135}{2}\).** 4. **Add \(I_1\) and \(I_2\):** \[ I = I_1 + I_2 = 0 + \frac{135}{2} = \frac{135}{2} \] **Answer:** To find the flux of \(\vec{F}\) across the surface \(S\) using the Divergence Theorem, we have: \[ \iint_S \vec{F} \cdot d\vec{S} = \frac{135}{2} \] |
67.5 |
| GPT-4o | To use the Divergence Theorem to evaluate the surface integral \(\iint_S \vec{F} \cdot d\vec{S}\) where \(\vec{F} = \sin(\pi x) \vec{i} + (z y^3)\vec{j} + (z^2 + 4x)\vec{k}\) and \(S\) is the surface of the box with \(-1 \le x \le 2\), \(0 \le y \le 1\), and \(1 \le z \le 4\), we follow these steps: 1. **State the Divergence Theorem:** The Divergence Theorem states that for a vector field \(\vec{F}\) and a closed surface \(S\) enclosing a volume \(V\), \[ \iint_S \vec{F} \cdot d\vec{S} = \iiint_V (\nabla \cdot \vec{F}) \, dV. \] 2. **Compute the divergence of \(\vec{F}\):** We need to find \(\nabla \cdot \vec{F}\), the divergence of \(\vec{F}\). \[ \vec{F} = \sin(\pi x) \vec{i} + (z y^3) \vec{j} + (z^2 + 4x) \vec{k} \] The divergence \(\nabla \cdot \vec{F}\) is given by: \[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x} (\sin(\pi x)) + \frac{\partial}{\partial y} (z y^3) + \frac{\partial}{\partial z} (z^2 + 4x). \] Compute each partial derivative: \[ \frac{\partial}{\partial x} (\sin(\pi x)) = \pi \cos(\pi x), \] \[ \frac{\partial}{\partial y} (z y^3) = 3z y^2, \] \[ \frac{\partial}{\partial z} (z^2 + 4x) = 2z. \] Therefore, \[ \nabla \cdot \vec{F} = \pi \cos(\pi x) + 3z y^2 + 2z. \] 3. **Set up the volume integral:** Now we need to evaluate the volume integral of \(\nabla \cdot \vec{F}\) over the volume \(V\) of the box. \[ \iiint_V (\pi \cos(\pi x) + 3z y^2 + 2z) \, dV. \] The limits for \(x\), \(y\), and \(z\) are given by the box: \[ -1 \le x \le 2, \quad 0 \le y \le 1, \quad 1 \le z \le 4. \] The volume integral is: \[ \iiint_V (\pi \cos(\pi x) + 3z y^2 + 2z) \, dx \, dy \, dz. \] 4. **Evaluate the integral:** We can separate the integral into three parts: \[ \iiint_V \pi \cos(\pi x) \, dx \, dy \, dz + \iiint_V 3z y^2 \, dx \, dy \, dz + \iiint_V 2z \, dx \, dy \, dz. \] Compute each integral separately: - For \(\pi \cos(\pi x)\): \[ \int_{-1}^{2} \pi \cos(\pi x) \, dx = \left[ \sin(\pi x) \right]_{-1}^{2} = \sin(2\pi) - \sin(-\pi) = 0 - 0 = 0. \] Since this term does not depend on \(y\) or \(z\), the integral over \(y\) and \(z\) simply multiplies by the volume of the \(y\)-\(z\) space, which is \(1 \times 3 = 3\). The whole term is zero. - For \(3z y^2\): \[ \iiint_V 3z y^2 \, dx \, dy \, dz = \int_{1}^{4} \int_{0}^{1} \int_{-1}^{2 |
67.5 |
| llama3.1-70B-Instruct | ## Step 1: Define the vector field F The vector field F is given by $\vec{F} = sin(\pi x) \vec{i} + (z y^3)\vec{j} + (z^2 + 4x)\vec{k}$. ## Step 2: Compute the divergence of F To apply the divergence theorem, we need to compute the divergence of F, which is given by $\nabla \cdot \vec{F} = \frac{\partial}{\partial x} (sin(\pi x)) + \frac{\partial}{\partial y} (z y^3) + \frac{\partial}{\partial z} (z^2 + 4x)$. ## Step 3: Evaluate the partial derivatives Evaluating the partial derivatives, we get $\frac{\partial}{\partial x} (sin(\pi x)) = \pi cos(\pi x)$, $\frac{\partial}{\partial y} (z y^3) = 3z y^2$, and $\frac{\partial}{\partial z} (z^2 + 4x) = 2z$. ## Step 4: Compute the divergence Substituting the partial derivatives into the expression for the divergence, we get $\nabla \cdot \vec{F} = \pi cos(\pi x) + 3z y^2 + 2z$. ## Step 5: Define the region of integration The region of integration is the box with $-1 \le x \le 2, 0 \le y \le 1$ and $1 \le z \le 4$. ## Step 6: Apply the divergence theorem The divergence theorem states that $\iint_S \vec{F} \cdot d \vec{S} = \iiint_V \nabla \cdot \vec{F} dV$, where $V$ is the volume of the box. ## Step 7: Evaluate the triple integral Evaluating the triple integral, we get $\iiint_V \nabla \cdot \vec{F} dV = \int_{-1}^2 \int_0^1 \int_1^4 (\pi cos(\pi x) + 3z y^2 + 2z) dz dy dx$. ## Step 8: Evaluate the innermost integral Evaluating the innermost integral, we get $\int_1^4 (\pi cos(\pi x) + 3z y^2 + 2z) dz = \pi cos(\pi x) (4-1) + \frac{3}{2} y^2 (4^2 - 1^2) + 2 \frac{1}{2} (4^2 - 1^2)$. ## Step 9: Simplify the expression Simplifying the expression, we get $\pi cos(\pi x) (3) + \frac{3}{2} y^2 (15) + 2 \frac{1}{2} (15) = 3 \pi cos(\pi x) + \frac{45}{2} y^2 + 15$. ## Step 10: Evaluate the middle integral Evaluating the middle integral, we get $\int_0^1 (3 \pi cos(\pi x) + \frac{45}{2} y^2 + 15) dy = 3 \pi cos(\pi x) (1-0) + \frac{45}{2} \frac{1}{3} (1^3 - 0^3) + 15 (1-0)$. ## Step 11: Simplify the expression Simplifying the expression, we get $3 \pi cos(\pi x) + \frac{45}{6} + 15 = 3 \pi cos(\pi x) + \frac{15}{2} + 15$. ## Step 12: Evaluate the outermost integral Evaluating the outermost integral, we get $\int_{-1}^2 (3 \pi cos(\pi x) + \frac{15}{2} + 15) dx = 3 \pi sin(\pi x) |_{-1}^2 + \frac{15}{2} (2-(-1)) + 15 (2-(-1))$. ## Step 13: Simplify the expression Simplifying the expression, we get $3 \pi (sin(2 \pi) - sin(-\pi)) + \frac{15}{2} (3) + 15 (3) = 3 \pi (0 - 0) + \frac{45}{2} + 45$. ## Step 14: Simplify the expression Simplifying the expression, we get $0 + \frac{45}{2} + 45 = \frac{45}{2} + 45 = \frac{135}{2}$. The final answer is: $\boxed{\frac{135}{2}}$ |
67.5 |
| llama3.1-405B-Instruct | To evaluate the surface integral $\iint_S \vec{F} \cdot d \vec{S}$ using the divergence theorem, we first need to find the divergence of the vector field $\vec{F} = \sin(\pi x) \vec{i} + (z y^3)\vec{j} + (z^2 + 4x)\vec{k}$. ## Step 1: Calculate the divergence of $\vec{F}$. The divergence of a vector field $\vec{F} = P\vec{i} + Q\vec{j} + R\vec{k}$ is given by $\nabla \cdot \vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$. For $\vec{F}$ given, we have $P = \sin(\pi x)$, $Q = zy^3$, and $R = z^2 + 4x$. Therefore, $\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(\sin(\pi x)) + \frac{\partial}{\partial y}(zy^3) + \frac{\partial}{\partial z}(z^2 + 4x)$. ## Step 2: Compute each partial derivative for the divergence. - $\frac{\partial}{\partial x}(\sin(\pi x)) = \pi \cos(\pi x)$ - $\frac{\partial}{\partial y}(zy^3) = 3zy^2$ - $\frac{\partial}{\partial z}(z^2 + 4x) = 2z$ ## Step 3: Combine the partial derivatives to get the divergence. $\nabla \cdot \vec{F} = \pi \cos(\pi x) + 3zy^2 + 2z$ ## Step 4: Apply the Divergence Theorem. The Divergence Theorem states that $\iint_S \vec{F} \cdot d \vec{S} = \iiint_V \nabla \cdot \vec{F} \, dV$, where $V$ is the volume enclosed by the surface $S$. In this case, we integrate $\nabla \cdot \vec{F}$ over the volume of the box with $-1 \le x \le 2, 0 \le y \le 1$, and $1 \le z \le 4$. ## Step 5: Evaluate the triple integral. $\iiint_V \nabla \cdot \vec{F} \, dV = \int_{-1}^{2} \int_{0}^{1} \int_{1}^{4} (\pi \cos(\pi x) + 3zy^2 + 2z) \, dz \, dy \, dx$ ## Step 6: Integrate with respect to $z$. $\int_{1}^{4} (\pi \cos(\pi x) + 3zy^2 + 2z) \, dz = \pi \cos(\pi x)z + \frac{3}{2}z^2y^2 + z^2 \Big|_{1}^{4}$ ## Step 7: Evaluate the result of the $z$ integration. $= \pi \cos(\pi x)(4-1) + \frac{3}{2}(4^2-1^2)y^2 + (4^2-1^2)$ $= 3\pi \cos(\pi x) + \frac{45}{2}y^2 + 15$ ## Step 8: Integrate with respect to $y$. $\int_{0}^{1} (3\pi \cos(\pi x) + \frac{45}{2}y^2 + 15) \, dy = 3\pi \cos(\pi x)y + \frac{15}{2}y^3 + 15y \Big|_{0}^{1}$ ## Step 9: Evaluate the result of the $y$ integration. $= 3\pi \cos(\pi x) + \frac{15}{2} + 15$ ## Step 9: Integrate with respect to $x$. $\int_{-1}^{2} (3\pi \cos(\pi x) + \frac{45}{2}) \, dx = 3\sin(\pi x) + \frac{45}{2}x \Big|_{-1}^{2}$ ## Step 10: Evaluate the final integral. $= (3\sin(2\pi) + 45) - (3\sin(-\pi) + \frac{-45}{2})$ $= 45 + \frac{45}{2}$ $= \frac{135}{2}$ The final answer is: $\boxed{\frac{135}{2}}$ |
67.5 |
| DeepSeek-Coder-V2 | To solve the given problem using the Divergence Theorem, we need to evaluate the surface integral \(\iint_S \vec{F} \cdot d \vec{S}\) by converting it into a volume integral. The Divergence Theorem states that: \[ \iint_S \vec{F} \cdot d \vec{S} = \iiint_V (\nabla \cdot \vec{F}) \, dV \] where \(V\) is the volume enclosed by the surface \(S\). First, we need to compute the divergence of the vector field \(\vec{F}\): \[ \vec{F} = \sin(\pi x) \vec{i} + z y^3 \vec{j} + (z^2 + 4x) \vec{k} \] The divergence of \(\vec{F}\) is given by: \[ \nabla \cdot \vec{F} = \frac{\partial (\sin(\pi x))}{\partial x} + \frac{\partial (z y^3)}{\partial y} + \frac{\partial (z^2 + 4x)}{\partial z} \] Calculating each partial derivative: \[ \frac{\partial (\sin(\pi x))}{\partial x} = \pi \cos(\pi x) \] \[ \frac{\partial (z y^3)}{\partial y} = 3z y^2 \] \[ \frac{\partial (z^2 + 4x)}{\partial z} = 2z \] Thus, the divergence is: \[ \nabla \cdot \vec{F} = \pi \cos(\pi x) + 3z y^2 + 2z \] Next, we set up the volume integral over the given box \(V\) with bounds \(-1 \le x \le 2\), \(0 \le y \le 1\), and \(1 \le z \le 4\): \[ \iiint_V (\nabla \cdot \vec{F}) \, dV = \int_{-1}^{2} \int_{0}^{1} \int_{1}^{4} (\pi \cos(\pi x) + 3z y^2 + 2z) \, dz \, dy \, dx \] We will evaluate this integral step by step. First, integrate with respect to \(z\): \[ \int_{1}^{4} (\pi \cos(\pi x) + 3z y^2 + 2z) \, dz = \pi \cos(\pi x) \int_{1}^{4} 1 \, dz + 3y^2 \int_{1}^{4} z \, dz + 2 \int_{1}^{4} z \, dz \] \[ = \pi \cos(\pi x) [z]_{1}^{4} + 3y^2 \left[\frac{z^2}{2}\right]_{1}^{4} + 2 \left[\frac{z^2}{2}\right]_{1}^{4} \] \[ = \pi \cos(\pi x) (4 - 1) + 3y^2 \left(\frac{16}{2} - \frac{1}{2}\right) + 2 \left(\frac{16}{2} - \frac{1}{2}\right) \] \[ = 3\pi \cos(\pi x) + 3y^2 \left(\frac{15}{2}\right) + 2 \left(\frac{15}{2}\right) \] \[ = 3\pi \cos(\pi x) + \frac{45}{2} y^2 + 15 \] Next, integrate with respect to \(y\): \[ \int_{0}^{1} \left(3\pi \cos(\pi x) + \frac{45}{2} y^2 + 15\right) \, dy = 3\pi \cos(\pi x) \int_{0}^{1} 1 \, dy + \frac{45}{2} \int_{0}^{1} y^2 \, dy + 15 \int_{0}^{1} 1 \, dy \] \[ = 3\pi \cos(\pi x) [y]_{0}^{1} + \frac{45}{2} \left[\frac{y^3}{3}\right]_{0}^{1} + 15 [y]_{0}^{1} \] \[ = 3\pi \cos(\pi x) (1 - 0) + \frac{45}{2} \left(\frac{1}{3}\right) + 15 (1 - 0) \] \[ = 3\pi \cos(\pi x) + \frac{15}{2} + 15 \] \[ = 3\pi \cos(\pi x) + \frac{45}{2} \] Finally, integrate with respect to \(x\): \[ \int_{-1}^{2} \left(3\pi \cos(\pi x) + \frac{45}{2}\right) \, dx = 3\pi \int_{-1}^{2} \cos(\pi x) \, dx + \frac{45}{2} \int_{-1}^{2} 1 \, dx \] \[ = 3\pi \left[\frac{\sin(\pi x)}{\pi}\right]_{-1}^{2} + \frac{45}{2} [x]_{-1}^{2} \] \[ = 3\pi \left(\frac{\sin(2\pi) - \sin(-\pi)}{\pi}\right) + \frac{45}{2} (2 - (-1)) \] \[ = 3\pi (0 - 0) + \frac{45}{2} (3) \] \[ = 0 + \frac{135}{2} \] \[ = \frac{135}{2} \] Thus, the value of the surface integral is: \[ \iint_S \vec{F} \cdot d \vec{S} = \frac{135}{2} \] |
67.5 |
| Qwen2-72B-Instruct | To solve the problem using the Divergence Theorem, we first need to calculate the divergence of the vector field $\vec{F} = \sin(\pi x) \vec{i} + (z y^3)\vec{j} + (z^2 + 4x)\vec{k}$. The divergence of a vector field $\vec{F} = F_1 \vec{i} + F_2 \vec{j} + F_3 \vec{k}$ is given by: \[ \text{div} \vec{F} = \nabla \cdot \vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \] For $\vec{F} = \sin(\pi x) \vec{i} + (z y^3)\vec{j} + (z^2 + 4x)\vec{k}$, we have: \[ \text{div} \vec{F} = \frac{\partial}{\partial x}(\sin(\pi x)) + \frac{\partial}{\partial y}(z y^3) + \frac{\partial}{\partial z}(z^2 + 4x) \] \[ = \pi \cos(\pi x) + 3zy^2 + 2z \] Next, we need to integrate the divergence of $\vec{F}$ over the volume $V$ enclosed by the surface $S$: \[ \iiint_V (\pi \cos(\pi x) + 3zy^2 + 2z) dV \] Since the volume $V$ is a rectangular box with $-1 \le x \le 2$, $0 \le y \le 1$, and $1 \le z \le 4$, we can evaluate the integral as follows: \[ \int_{-1}^{2} \int_0^1 \int_1^4 (\pi \cos(\pi x) + 3zy^2 + 2z) dz dy dx \] \[ = \int_{-1}^{2} \int_0^1 \left[ \pi z \cos(\pi x) + 3zy^2z + 2z^2 \right]_1^4 dy dx \] \[ = \int_{-1}^{2} \int_0^1 \left( 4\pi \cos(\pi x) + 12y^2 - \pi \cos(\pi x) - 3y^2 + 32 - 2 \right) dy dx \] \[ = \int_{-1}^{2} \int_0^1 \left( 3\pi \cos(\pi x) + 9y^2 + 30 \right) dy dx \] \[ = \int_{-1}^{2} \left[ 3\pi y \cos(\pi x) + 3y^3 + 30y \right]_0^1 dx \] \[ = \int_{-1}^{2} \left( 3\pi \cos(\pi x) + 3 + 30 \right) dx \] \[ = \int_{-1}^{2} \left( 3\pi \cos(\pi x) + 33 \right) dx \] \[ = \left[ 3\pi \frac{\sin(\pi x)}{\pi} + 33x \right]_{-1}^{2} \] \[ = \left[ 3\sin(\pi x) + 33x \right]_{-1}^{2} \] \[ = \left( 3\sin(2\pi) + 33(2) \right) - \left( 3\sin(-\pi) + 33(-1) \right) \] \[ = 66 - (-33) \] \[ = 99 \] So, the value of $\iint_S \vec{F} \cdot d \vec{S}$ using the Divergence Theorem is $\boxed{99}$. |
67.5 |
| baichuan4 | To evaluate the surface integral $\iint_S \vec{F} \cdot d \vec{S}$ using the divergence theorem, we first need to compute the divergence of the vector field $\vec{F}$. The divergence measures how much a vector field "flows" outward from a point. The vector field $\vec{F}$ is given by: $$\vec{F} = \sin(\pi x) \vec{i} + (z y^3)\vec{j} + (z^2 + 4x)\vec{k}$$ The divergence of $\vec{F}$ is: $$\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(\sin(\pi x)) + \frac{\partial}{\partial y}(z y^3) + \frac{\partial}{\partial z}(z^2 + 4x)$$ Computing these partial derivatives: $$\frac{\partial}{\partial x}(\sin(\pi x)) = \pi \cos(\pi x)$$ $$\frac{\partial}{\partial y}(z y^3) = 3zy^2 \frac{\partial z}{\partial y} = 3zy^2 \cdot 0 = 0$$ $$\frac{\partial}{\partial z}(z^2 + 4x) = 2z + 0 = 2z$$ So the divergence is: $$\nabla \cdot \vec{F} = \pi \cos(\pi x) + 0 + 2z = \pi \cos(\pi x) + 2z$$ Now, by the divergence theorem, we can convert the surface integral into a volume integral: $$\iint_S \vec{F} \cdot d \vec{S} = \iiint_E (\nabla \cdot \vec{F}) dV$$ The volume $E$ is the box with dimensions: $-1 \leq x \leq 2$, $0 \leq y \leq 1$, and $1 \leq z \leq 4$. We integrate the divergence over this volume: $$\iiint_E (\pi \cos(\pi x) + 2z) dV = \int_{-1}^{2} \int_{0}^{1} \int_{1}^{4} (\pi \cos(\pi x) + 2z) \, dz \, dy \, dx$$ Computing the integral: $$\int_{-1}^{2} \int_{0}^{1} \left[ \int_{1}^{4} (\pi \cos(\pi x) + 2z) \, dz \right] dy \, dx = \int_{-1}^{2} \int_{0}^{1} \left[ \pi \cos(\pi x)(4 - 1) + 2\left(\frac{1}{2}z^2\right)_{1}^{4} \right] dy \, dx$$ $$= \int_{-1}^{2} \int_{0}^{1} \left[ 3\pi \cos(\pi x) + 3.5(4 - 1) \right] dy \, dx$$ $$= \int_{-1}^{2} \left[ 3\pi \cos(\pi x)y + 3.5(4 - 1)y \right]_{0}^{1} dx$$ $$= \int_{-1}^{2} \left[ 3\pi \cos(\pi x) + 10.5 \right] dx$$ $$= \left[ 3\sin(\pi x) + 10.5x \right]_{-1}^{2}$$ $$= 3\sin(2\pi) + 21 - (3\sin(-\pi) - 10.5)$$ $$= 3(0) + 21 - (3(0) - 10.5)$$ $$= 21 + 10.5$$ $$= 31.5$$ Therefore, the value of the surface integral $\iint_S \vec{F} \cdot d \vec{S}$ is $31.5$. |
67.5 |
| doubao | First, we find the divergence of $\vec{F}$: $\text{div}\vec{F} = \frac{\partial}{\partial x}(\sin(\pi x)) + \frac{\partial}{\partial y}(zy^3) + \frac{\partial}{\partial z}(z^2 + 4x) = \pi\cos(\pi x) + 3zy^2 + 2z$. Then we evaluate the triple integral over the volume of the box: $\iiint_V \text{div}\vec{F} dV = \int_{-1}^{2} \int_{0}^{1} \int_{1}^{4} (\pi\cos(\pi x) + 3zy^2 + 2z) dz dy dx$. Evaluating this integral step by step: $\int_{-1}^{2} \pi\cos(\pi x) dx = 0$ (since the integral of cosine over a full period is zero). $\int_{-1}^{2} \int_{0}^{1} \int_{1}^{4} (3zy^2 + 2z) dz dy dx = \int_{-1}^{2} \int_{0}^{1} [\frac{3}{2}z^2y^2 + z^2]_{1}^{4} dy dx = \int_{-1}^{2} \int_{0}^{1} (24y^2 + 15) dy dx$. Evaluating the inner integral: $\int_{0}^{1} (24y^2 + 15) dy = [8y^3 + 15y]_{0}^{1} = 8 + 15 = 23$. So the final result is $\int_{-1}^{2} 23 dx = 23 \times 3 = 69$. |
67.5 |
| minimax | To evaluate the surface integral $\iint_S \vec{F} \cdot d \vec{S}$ using the divergence theorem, we first need to compute the divergence of the vector field $\vec{F}$: $$\vec{F} = \sin(\pi x) \vec{i} + (z y^3) \vec{j} + (z^2 + 4x) \vec{k}$$ The divergence of $\vec{F}$, denoted as $\nabla \cdot \vec{F}$, is given by: $$\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(\sin(\pi x)) + \frac{\partial}{\partial y}(z y^3) + \frac{\partial}{\partial z}(z^2 + 4x)$$ Now, we compute the partial derivatives: $$\frac{\partial}{\partial x}(\sin(\pi x)) = \pi \cos(\pi x)$$ $$\frac{\partial}{\partial y}(z y^3) = 3z y^2$$ $$\frac{\partial}{\partial z}(z^2 + 4x) = 2z$$ So the divergence of $\vec{F}$ is: $$\nabla \cdot \vec{F} = \pi \cos(\pi x) + 3z y^2 + 2z$$ According to the divergence theorem, the surface integral of $\vec{F}$ over the closed surface $S$ of the box is equal to the volume integral of the divergence of $\vec{F}$ over the volume $V$ enclosed by $S$: $$\iint_S \vec{F} \cdot d \vec{S} = \iiint_V (\nabla \cdot \vec{F}) \, dV$$ Now we integrate the divergence over the volume $V$: $$\iiint_V (\nabla \cdot \vec{F}) \, dV = \iiint_V (\pi \cos(\pi x) + 3z y^2 + 2z) \, dV$$ The volume integral can be split into six separate integrals, one for each face of the box. However, since the divergence theorem states that the surface integral over the closed surface is equal to the volume integral, we can directly integrate over the volume $V$ defined by the box: $$\iiint_V (\pi \cos(\pi x) + 3z y^2 + 2z) \, dV = \int_{-1}^{2} \int_{0}^{1} \int_{1}^{4} (\pi \cos(\pi x) + 3z y^2 + 2z) \, dz \, dy \, dx$$ Now we perform the integration: $$\int_{-1}^{2} \int_{0}^{1} \int_{1}^{4} (\pi \cos(\pi x) + 3z y^2 + 2z) \, dz \, dy \, dx = \int_{-1}^{2} \int_{0}^{1} \left[ \pi \cos(\pi x) z + \frac{3}{2} z^2 y^2 + z^2 \right]_{1}^{4} \, dy \, dx$$ $$= \int_{-1}^{2} \int_{0}^{1} \left[ \pi \cos(\pi x) (4 - 1) + \frac{3}{2} (4^2 - 1^2) y^2 + (4^2 - 1^2) \right] \, dy \, dx$$ $$= \int_{-1}^{2} \int_{0}^{1} \left[ 3\pi \cos(\pi x) + \frac{3}{2} (15) y^2 + 15 \right] \, dy \, dx$$ $$= \int_{-1}^{2} \left[ 3\pi \cos(\pi x) y + \frac{3}{2} (15) \frac{y^3}{3} + 15y \right]_{0}^{1} \, dx$$ $$= \int_{-1}^{2} \left[ 3\pi \cos(\pi x) + 5 + 15 \right] \, dx$$ $$= \int_{-1}^{2} \left[ 3\pi \cos(\pi x) + 20 \right] \, dx$$ $$= \left[ \frac{3\pi}{\pi} \sin(\pi x) + 20x \right]_{-1}^{2}$$ $$= \left[ 3 \sin(\pi \cdot 2) + 20 \cdot 2 \right] - \left[ 3 \sin(\pi \cdot (-1)) + 20 \cdot (-1) \right]$$ $$= \left[ 3 \sin(2\pi) + 40 \right] - \left[ 3 \sin(-\pi) - 20 \right]$$ Since $\sin(2\pi) = \sin(-\pi) = 0$, we have: $$= 40 + 20$$ $$= 60$$ Therefore, the surface integral $\iint_S \vec{F} \cdot d \vec{S}$ evaluated using the divergence theorem is 60. |
67.5 |
| kimi | To solve this problem using the divergence theorem, we first need to compute the divergence of the vector field $\vec{F}$, and then integrate this divergence over the volume enclosed by the surface $S$. The divergence of a vector field $\vec{F}$ is given by: $$\nabla \cdot \vec{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}$$ For the given vector field $\vec{F} = \sin(\pi x) \vec{i} + (z y^3)\vec{j} + (z^2 + 4x)\vec{k}$, we compute the divergence: $$\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(\sin(\pi x)) + \frac{\partial}{\partial y}(z y^3) + \frac{\partial}{\partial z}(z^2 + 4x)$$ Let's compute each partial derivative: $$\frac{\partial}{\partial x}(\sin(\pi x)) = \pi \cos(\pi x)$$ $$\frac{\partial}{\partial y}(z y^3) = 3z y^2$$ $$\frac{\partial}{\partial z}(z^2 + 4x) = 2z$$ So the divergence is: $$\nabla \cdot \vec{F} = \pi \cos(\pi x) + 3z y^2 + 2z$$ Now, we integrate this divergence over the volume $V$ enclosed by the surface $S$: $$\iint_S \vec{F} \cdot d \vec{S} = \iiint_V (\nabla \cdot \vec{F}) dV$$ The limits for the integration are given by the bounds of the box: - For $x$, the limits are from $-1$ to $2$. - For $y$, the limits are from $0$ to $1$. - For $z$, the limits are from $1$ to $4$. So the integral becomes: $$\iiint_V (\pi \cos(\pi x) + 3z y^2 + 2z) dV = \int_{-1}^{2} \int_{0}^{1} \int_{1}^{4} (\pi \cos(\pi x) + 3z y^2 + 2z) dz dy dx$$ Now we perform the integration: 1. Integrate with respect to $z$: $$\int_{1}^{4} (\pi \cos(\pi x) + 3z y^2 + 2z) dz = \left[ \pi z \cos(\pi x) + y^2 z^2 + z^2 \right]_{1}^{4}$$ 2. Evaluate the integral at the limits: $$= \left( 4\pi \cos(\pi x) + 4y^2 + 16 \right) - \left( \pi \cos(\pi x) + y^2 + 1 \right)$$ $$= 3\pi \cos(\pi x) + 3y^2 + 15$$ 3. Integrate with respect to $y$: $$\int_{0}^{1} (3\pi \cos(\pi x) + 3y^2 + 15) dy = \left[ 3\pi y \cos(\pi x) + y^3 + 15y \right]_{0}^{1}$$ 4. Evaluate the integral at the limits: $$= \left( 3\pi \cos(\pi x) + 1 + 15 \right) - (0 + 0 + 0)$$ $$= 3\pi \cos(\pi x) + 16$$ 5. Integrate with respect to $x$: $$\int_{-1}^{2} (3\pi \cos(\pi x) + 16) dx = \left[ -\frac{3}{\pi} \sin(\pi x) + 16x \right]_{-1}^{2}$$ 6. Evaluate the integral at the limits: $$= \left( -\frac{3}{\pi} \sin(2\pi) + 32 \right) - \left( -\frac{3 |
67.5 |
| wenxinyiyan | To evaluate the surface integral $\iint_S \vec{F} \cdot d\vec{S}$ using the Divergence Theorem, we first need to find the divergence of the vector field $\vec{F}$. Given $\vec{F} = \sin(\pi x)\vec{i} + (zy^3)\vec{j} + (z^2 + 4x)\vec{k}$, the divergence of $\vec{F}$ is: $$\text{div}\vec{F} = \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(\sin(\pi x)) + \frac{\partial}{\partial y}(zy^3) + \frac{\partial}{\partial z}(z^2 + 4x)$$ Calculating each partial derivative: $$\frac{\partial}{\partial x}(\sin(\pi x)) = \pi \cos(\pi x)$$ $$\frac{\partial}{\partial y}(zy^3) = 3zy^2$$ $$\frac{\partial}{\partial z}(z^2 + 4x) = 2z$$ So, the divergence is: $$\text{div}\vec{F} = \pi \cos(\pi x) + 3zy^2 + 2z$$ Now, we use the Divergence Theorem, which states: $$\iint_S \vec{F} \cdot d\vec{S} = \iiint_V \text{div}\vec{F} \, dV$$ where $V$ is the volume enclosed by the surface $S$. In this case, $V$ is the box defined by $-1 \le x \le 2$, $0 \le y \le 1$, and $1 \le z \le 4$. We need to evaluate the triple integral: $$\iiint_V (\pi \cos(\pi x) + 3zy^2 + 2z) \, dV$$ This can be split into three integrals: $$\pi \iiint_V \cos(\pi x) \, dV + 3 \iiint_V zy^2 \, dV + 2 \iiint_V z \, dV$$ Evaluating each integral over the volume $V$: $$\pi \int_{-1}^{2} \int_{0}^{1} \int_{1}^{4} \cos(\pi x) \, dzdydx + 3 \int_{-1}^{2} \int_{0}^{1} \int_{1}^{4} zy^2 \, dzdydx + 2 \int_{-1}^{2} \int_{0}^{1} \int_{1}^{4} z \, dzdydx$$ Integrating with respect to $z$, $y$, and then $x$, we get: $$\pi \int_{-1}^{2} \int_{0}^{1} [z\cos(\pi x)]_{1}^{4} \, dydx + 3 \int_{-1}^{2} \int_{0}^{1} \left[\frac{z^2}{2}y^2\right]_{1}^{4} \, dydx + 2 \int_{-1}^{2} \int_{0}^{1} \left[\frac{z^2}{2}\right]_{1}^{4} \, dydx$$ $$\pi \int_{-1}^{2} \int_{0}^{1} (4\cos(\pi x) - \cos(\pi x)) \, dydx + 3 \int_{-1}^{2} \int_{0}^{1} \left(8y^2 - \frac{y^2}{2}\right) \, dydx + 2 \int_{-1}^{2} \int_{0}^{1} \left(8 - \frac{1}{2}\right) \, dydx$$ $$\pi \int_{-1}^{2} \int_{0}^{1} 3\cos(\pi x) \, dydx + 3 \int_{-1}^{2} \int_{0}^{1} \frac{15}{2}y^2 \, dydx + 2 \int_{-1}^{2} \int_{0}^{1} \frac{15}{2} \, dydx$$ $$\pi \int_{-1}^{2} 3\cos(\pi x) \, dx + \frac{45}{2} \int_{-1}^{2} \int_{0}^{1} y^2 \, dydx + 15 \int_{-1}^{2} \int_{0}^{1} \, dydx$$ $$\pi \left[\frac{3}{\pi}\sin(\pi x)\right]_{-1 |
67.5 |
| zhipu | To solve this problem using the divergence theorem, we need to follow these steps: ### Step 1: Understand the Divergence Theorem The divergence theorem states that for a vector field \(\vec{F}\) and a closed surface \(S\) enclosing a volume \(V\), \[ \iint_S \vec{F} \cdot d\vec{S} = \iiint_V (\nabla \cdot \vec{F}) \, dV \] where \(\nabla \cdot \vec{F}\) is the divergence of \(\vec{F}\). ### Step 2: Compute the Divergence of \(\vec{F}\) Given \(\vec{F} = \sin(\pi x) \vec{i} + (z y^3) \vec{j} + (z^2 + 4x) \vec{k}\), we find the divergence: \[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x} (\sin(\pi x)) + \frac{\partial}{\partial y} (z y^3) + \frac{\partial}{\partial z} (z^2 + 4x) \] Calculate each partial derivative: \[ \frac{\partial}{\partial x} (\sin(\pi x)) = \pi \cos(\pi x) \] \[ \frac{\partial}{\partial y} (z y^3) = 3z y^2 \] \[ \frac{\partial}{\partial z} (z^2 + 4x) = 2z \] Thus, \[ \nabla \cdot \vec{F} = \pi \cos(\pi x) + 3z y^2 + 2z \] ### Step 3: Set Up the Volume Integral We need to evaluate the volume integral of the divergence over the box defined by \(-1 \le x \le 2\), \(0 \le y \le 1\), and \(1 \le z \le 4\): \[ \iiint_V (\nabla \cdot \vec{F}) \, dV = \int_{-1}^{2} \int_{0}^{1} \int_{1}^{4} (\pi \cos(\pi x) + 3z y^2 + 2z) \, dz \, dy \, dx \] ### Step 4: Evaluate the Integral We break the integral into three separate integrals: \[ \iiint_V \pi \cos(\pi x) \, dV + \iiint_V 3z y^2 \, dV + \iiint_V 2z \, dV \] #### Integral 1: \(\iiint_V \pi \cos(\pi x) \, dV\) \[ \int_{-1}^{2} \int_{0}^{1} \int_{1}^{4} \pi \cos(\pi x) \, dz \, dy \, dx \] Since \(\pi \cos(\pi x)\) is independent of \(y\) and \(z\): \[ \pi \int_{-1}^{2} \cos(\pi x) \, dx \int_{0}^{1} dy \int_{1}^{4} dz \] \[ \pi \left[ \frac{\sin(\pi x)}{\pi} \right]_{-1}^{2} \cdot (1) \cdot (3) \] \[ \pi \left( \frac{\sin(2\pi)}{\pi} - \frac{\sin(-\pi)}{\pi} \right) \cdot 3 \] \[ \pi \left( 0 - 0 \right) \cdot 3 = 0 \] #### Integral 2: \(\iiint_V 3z y^2 \, dV\) \[ 3 \int_{-1}^{2} \int_{0}^{1} \int_{1}^{4} z y^2 \, dz \, dy \, dx \] Since \(z y^2\) is independent of \(x\): \[ 3 \int_{-1}^{2} dx \int_{0}^{1} y^2 dy \int_{1}^{4} z dz \] \[ 3 \cdot 3 \cdot \left[ \frac{y^3}{3} \right]_{0}^{1} \cdot \left[ \frac{z^2}{2} \right]_{1}^{4} \] \[ 3 \cdot 3 \cdot \frac{1}{3} \cdot \left( \frac{16}{2} - \frac{1}{2} \right) \] \[ 3 \cdot 1 \cdot \frac{15}{2} = \frac{45}{2} \] #### Integral 3: \(\iiint_V 2z \, dV\) \[ 2 \int_{-1}^{2} \int_{0}^{1} \int_{1}^{4} z \, dz \, dy \, dx \] Since \(z\) is independent of \(x\) and \(y\): \[ 2 \int_{-1}^{2} dx \int_{0}^{1} dy \int_{1}^{4} z dz \] \[ 2 \cdot 3 \cdot 1 \cdot \left[ \frac{z^2}{2} \right]_{1}^{4} \] \[ 2 \cdot 3 \cdot 1 \cdot \left( \frac{16}{2} - \frac{1}{2} \right) \] \[ 6 \cdot \frac{15}{2} = 45 \] ### Step 5: Sum the Results \[ 0 + \frac{45}{2} + 45 = \frac{45}{2} + \frac{90}{2} = \frac{135}{2} \] Thus, the value of the surface integral \(\iint_S \vec{F} \cdot d\vec{S}\) is: \[ \boxed{\frac{135}{2}} \] |
67.5 |
| gpto1 | To solve the problem using the **Divergence Theorem**, we need to evaluate the surface integral \(\iint_S \vec{F} \cdot d\vec{S}\) by converting it into a triple integral over the volume \(V\) enclosed by the surface \(S\). **Given:** \[ \vec{F} = \sin(\pi x)\, \vec{i} + z y^3\, \vec{j} + (z^2 + 4x)\, \vec{k} \] \(S\) is the surface of the box defined by: - \(x\) ranges from \(-1\) to \(2\) - \(y\) ranges from \(0\) to \(1\) - \(z\) ranges from \(1\) to \(4\) **Steps:** 1. **Compute the Divergence of \(\vec{F}\):** The divergence of a vector field \(\vec{F} = F_x\, \vec{i} + F_y\, \vec{j} + F_z\, \vec{k}\) is given by: \[ \nabla \cdot \vec{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} \] Compute each partial derivative: - \(\dfrac{\partial F_x}{\partial x} = \dfrac{\partial}{\partial x} [\sin(\pi x)] = \pi \cos(\pi x)\) - \(\dfrac{\partial F_y}{\partial y} = \dfrac{\partial}{\partial y} [z y^3] = z \cdot 3 y^2\) - \(\dfrac{\partial F_z}{\partial z} = \dfrac{\partial}{\partial z} [z^2 + 4x] = 2z\) So the divergence is: \[ \nabla \cdot \vec{F} = \pi \cos(\pi x) + 3 z y^2 + 2z \] 2. **Set Up the Triple Integral over the Volume \(V\):** The Divergence Theorem states: \[ \iint_S \vec{F} \cdot d\vec{S} = \iiint_V (\nabla \cdot \vec{F})\, dV \] Our volume \(V\) is the box defined by the given ranges of \(x\), \(y\), and \(z\). The triple integral becomes: \[ \int_{x=-1}^{2} \int_{y=0}^{1} \int_{z=1}^{4} \left[ \pi \cos(\pi x) + 3 z y^2 + 2z \right] dz\, dy\, dx \] 3. **Evaluate the Triple Integral:** **Split the integral** into two parts for easier computation: \[ I = I_1 + I_2 \quad \text{where} \quad \begin{cases} I_1 = \int_{x=-1}^{2} \int_{y=0}^{1} \int_{z=1}^{4} \pi \cos(\pi x)\, dz\, dy\, dx \\ I_2 = \int_{x=-1}^{2} \int_{y=0}^{1} \int_{z=1}^{4} (3 z y^2 + 2z)\, dz\, dy\, dx \end{cases} \] **Compute \(I_1\):** Since \(\pi \cos(\pi x)\) does not depend on \(y\) or \(z\), and the integrals over \(y\) and \(z\) can be separated: \[ I_1 = \pi \left( \int_{x=-1}^{2} \cos(\pi x)\, dx \right) \left( \int_{y=0}^{1} dy \right) \left( \int_{z=1}^{4} dz \right) \] **Compute the \(x\)-integral of \(\cos(\pi x)\):** \[ \int_{x=-1}^{2} \cos(\pi x)\, dx = \frac{1}{\pi} \left[ \sin(\pi x) \right]_{x=-1}^{2} = \frac{1}{\pi} [\sin(2\pi) - \sin(-\pi)] = \frac{1}{\pi} (0 - 0) = 0 \] **Compute the \(y\) and \(z\) integrals:** \[ \int_{y=0}^{1} dy = 1, \quad \int_{z=1}^{4} dz = 4 - 1 = 3 \] **Thus, \(I_1 = 0\).** **Compute \(I_2\):** **First, compute the \(z\)-integral:** \[ \int_{z=1}^{4} (3 z y^2 + 2z)\, dz = (3 y^2 + 2) \int_{z=1}^{4} z\, dz = (3 y^2 + 2) \left[ \frac{1}{2} z^2 \right]_{z=1}^{4} = (3 y^2 + 2) \left( \frac{1}{2}(16 - 1) \right) = (3 y^2 + 2) \left( \frac{15}{2} \right) \] **Next, compute the \(y\)-integral:** \[ \int_{y=0}^{1} (3 y^2 + 2) \left( \frac{15}{2} \right)\, dy = \frac{15}{2} \int_{y=0}^{1} (3 y^2 + 2)\, dy = \frac{15}{2} \left[ y^3 + 2 y \right]_{y=0}^{1} = \frac{15}{2} (1 + 2) = \frac{15}{2} \times 3 = \frac{45}{2} \] **Finally, compute the \(x\)-integral:** \[ \int_{x=-1}^{2} dx = 2 - (-1) = 3 \] **Thus, \(I_2 = 3 \times \frac{45}{2} = \frac{135}{2}\).** 4. **Add \(I_1\) and \(I_2\):** \[ I = I_1 + I_2 = 0 + \frac{135}{2} = \frac{135}{2} \] **Answer:** To find the flux of \(\vec{F}\) across the surface \(S\) using the Divergence Theorem, we have: \[ \iint_S \vec{F} \cdot d\vec{S} = \frac{135}{2} \] |
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| deepseekv3 | To evaluate the surface integral \(\iint_S \vec{F} \cdot d \vec{S}\) using the Divergence Theorem, follow these steps: ### 1. **Compute the Divergence of \(\vec{F}\)** The Divergence Theorem states: \[ \iint_S \vec{F} \cdot d \vec{S} = \iiint_V \nabla \cdot \vec{F} \, dV \] where \(V\) is the volume enclosed by the surface \(S\). Given \(\vec{F} = \sin(\pi x) \vec{i} + z y^3 \vec{j} + (z^2 + 4x) \vec{k}\), the divergence \(\nabla \cdot \vec{F}\) is: \[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x} (\sin(\pi x)) + \frac{\partial}{\partial y} (z y^3) + \frac{\partial}{\partial z} (z^2 + 4x) \] Compute each partial derivative: \[ \frac{\partial}{\partial x} (\sin(\pi x)) = \pi \cos(\pi x) \] \[ \frac{\partial}{\partial y} (z y^3) = 3 z y^2 \] \[ \frac{\partial}{\partial z} (z^2 + 4x) = 2z \] Thus, the divergence is: \[ \nabla \cdot \vec{F} = \pi \cos(\pi x) + 3 z y^2 + 2z \] ### 2. **Set Up the Triple Integral** The volume \(V\) is the box defined by \(-1 \le x \le 2\), \(0 \le y \le 1\), and \(1 \le z \le 4\). The triple integral becomes: \[ \iiint_V \nabla \cdot \vec{F} \, dV = \int_{-1}^2 \int_0^1 \int_1^4 (\pi \cos(\pi x) + 3 z y^2 + 2z) \, dz \, dy \, dx \] ### 3. **Evaluate the Integral** First, integrate with respect to \(z\): \[ \int_1^4 (\pi \cos(\pi x) + 3 z y^2 + 2z) \, dz = \pi \cos(\pi x) \cdot (4 - 1) + \frac{3 y^2}{2} (4^2 - 1^2) + \frac{2}{2} (4^2 - 1^2) \] Simplify: \[ = 3 \pi \cos(\pi x) + \frac{3 y^2}{2} \cdot 15 + 15 \] \[ = 3 \pi \cos(\pi x) + \frac{45 y^2}{2} + 15 \] Next, integrate with respect to \(y\): \[ \int_0^1 \left(3 \pi \cos(\pi x) + \frac{45 y^2}{2} + 15\right) \, dy = 3 \pi \cos(\pi x) \cdot 1 + \frac{45}{2} \cdot \frac{1}{3} + 15 \cdot 1 \] Simplify: \[ = 3 \pi \cos(\pi x) + \frac{15}{2} + 15 \] \[ = 3 \pi \cos(\pi x) + \frac{45}{2} \] Finally, integrate with respect to \(x\): \[ \int_{-1}^2 \left(3 \pi \cos(\pi x) + \frac{45}{2}\right) \, dx = 3 \pi \int_{-1}^2 \cos(\pi x) \, dx + \frac{45}{2} \int_{-1}^2 1 \, dx \] The integral of \(\cos(\pi x)\) over \([-1, 2]\) is zero because \(\cos(\pi x)\) is symmetric and integrates to zero over a full period. Thus: \[ = 0 + \frac{45}{2} \cdot (2 - (-1)) \] \[ = \frac{45}{2} \cdot 3 \] \[ = \frac{135}{2} \] ### 4. **Final Answer** \[ \iint_S \vec{F} \cdot d \vec{S} = \frac{135}{2} \] \[ \boxed{\dfrac{135}{2}} \] |
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