| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve this problem using the Divergence Theorem, we need to calculate the divergence of the vector field \(\vec{F}\) and then evaluate the triple integral of this divergence over the volume enclosed by the surface \(S\). Given vector field \(\vec{F} = xy\vec{i} - \frac{1}{2}y^2\vec{j} + z\vec{k}\) First, we calculate the divergence of \(\vec{F}\): \[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(xy) + \frac{\partial}{\partial y}\left(-\frac{1}{2}y^2\right) + \frac{\partial}{\partial z}(z) \] \[ = y + (-y) + 1 = 1 \] So, the divergence of \(\vec{F}\) is a constant, \(1\). Next, we need to calculate the volume integral of the divergence over the volume enclosed by \(S\). The volume \(V\) is defined by the surfaces \(z = 4 - 3x^2 - 3y^2\) for \(1 \le z \le 4\) and \(z = 0\) for the bottom. We can see that the volume is a paraboloid with its vertex at \((0, 0, 4)\) and it extends down to \(z = 0\). Since the paraboloid is symmetric with respect to the \(x\) and \(y\) axes, we can integrate over the region \(-\sqrt{\frac{4 - z}{3}} \le x \le \sqrt{\frac{4 - z}{3}}\) and \(-\sqrt{\frac{4 - z}{3}} \le y \le \sqrt{\frac{4 - z}{3}}\), and \(0 \le z \le 1\) for the bottom surface. The volume integral of the divergence can be written as: \[ \iiint_V \nabla \cdot \vec{F} \, dV = \int_0^1 \int_{-\sqrt{\frac{4 - z}{3}}}^{\sqrt{\frac{4 - z}{3}}} \int_{-\sqrt{\frac{4 - z}{3}}}^{\sqrt{\frac{4 - z}{3}}} 1 \, dx \, dy \, dz \] \[ = \int_0^1 \int_{-\sqrt{\frac{4 - z}{3}}}^{\sqrt{\frac{4 - z}{3}}} 2\sqrt{\frac{4 - z}{3}} \, dy \, dz \] \[ = \int_0^1 \frac{4}{3} \left(4 - z\right) \, dz \] \[ = \frac{4}{3} \left[4z - \frac{1}{2}z^2\right]_0^1 = \frac{4}{3} \left(4 - \frac{1}{2}\right) = \frac{4}{3} \cdot \frac{7}{2} = \frac{14}{3} \] Therefore, the value of the surface integral \(\iint_S \vec{F} \cdot d\vec{S}\) is \(\boxed{\frac{14}{3}}\). |
7.853 |
| 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 \(\vec{F} = xy \vec{i} - \frac{1}{2}y^2\vec{j} + z\vec{k}\): \[ \text{div} \vec{F} = \frac{\partial (xy)}{\partial x} + \frac{\partial (-\frac{1}{2}y^2)}{\partial y} + \frac{\partial z}{\partial z} = y - y + 1 = 1 \] By the divergence theorem, the surface integral is equal to the triple integral of the divergence over the volume \(V\) enclosed by \(S\): \[ \iint_S \vec{F} \cdot d \vec{S} = \iiint_V \text{div} \vec{F} \, dV = \iiint_V 1 \, dV \] This integral is the volume of the region enclosed by \(S\). The surface \(S\) consists of three parts: the paraboloid \(z = 4 - 3x^2 - 3y^2\) from \(z = 1\) to \(z = 4\), the vertical cylinder \(x^2 + y^2 = 1\) from \(z = 0\) to \(z = 1\), and the disk \(z = 0\) at the bottom. 1. **Volume of the cylinder (from \(z = 0\) to \(z = 1\))**: \[ V_1 = \pi \times (1)^2 \times 1 = \pi \] 2. **Volume under the paraboloid (from \(z = 1\) to \(z = 4\))**: - The radius at height \(z\) is \(\sqrt{\frac{4 - z}{3}}\). - The cross-sectional area at height \(z\) is \(\pi \left(\sqrt{\frac{4 - z}{3}}\right)^2 = \frac{\pi}{3}(4 - z)\). - Integrate this area from \(z = 1\) to \(z = 4\): \[ V_2 = \frac{\pi}{3} \int_{1}^{4} (4 - z) \, dz = \frac{\pi}{3} \left[ 4z - \frac{1}{2}z^2 \right]_{1}^{4} \] Evaluating the integral: \[ \left[ 4z - \frac{1}{2}z^2 \right]_{1}^{4} = (16 - 8) - (4 - 0.5) = 8 - 3.5 = 4.5 \] \[ V_2 = \frac{\pi}{3} \times 4.5 = \frac{3\pi}{2} \] 3. **Total volume**: \[ V = V_1 + V_2 = \pi + \frac{3\pi}{2} = \frac{5\pi}{2} \] Thus, the value of the surface integral is \(\boxed{\dfrac{5\pi}{2}}\). |
7.853 |
| gpto1 | To evaluate the flux of the vector field \(\vec{F}\) across the closed surface \(S\) using the divergence theorem, we'll follow these steps: 1. **Understand the Surface \(S\) and the Volume \(V\)**: - The surface \(S\) consists of three parts: - The **top surface**: the paraboloid \(z = 4 - 3x^2 - 3y^2\) above \(z = 0\). - The **bottom surface**: the plane \(z = 0\). - The **side surface**: the curved surface connecting the top and bottom surfaces. - The volume \(V\) is the solid enclosed by this surface \(S\). 2. **Compute the Divergence of \(\vec{F}\)**: - The divergence theorem states: \[ \iint_{S} \vec{F} \cdot d\vec{S} = \iiint_{V} (\nabla \cdot \vec{F}) \, dV \] - Compute the divergence \(\nabla \cdot \vec{F}\): \[ \nabla \cdot \vec{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} \] Given \(\vec{F} = xy\, \vec{i} - \dfrac{1}{2} y^2\, \vec{j} + z\, \vec{k}\): - \(\frac{\partial F_x}{\partial x} = \frac{\partial}{\partial x}(xy) = y\) - \(\frac{\partial F_y}{\partial y} = \frac{\partial}{\partial y}\left(-\dfrac{1}{2} y^2\right) = -y\) - \(\frac{\partial F_z}{\partial z} = \frac{\partial}{\partial z}(z) = 1\) - So, \(\nabla \cdot \vec{F} = y - y + 1 = 1\) 3. **Set Up the Triple Integral Over Volume \(V\)**: - Since \(\nabla \cdot \vec{F} = 1\), the flux simplifies to the volume of \(V\): \[ \iiint_{V} (\nabla \cdot \vec{F}) \, dV = \iiint_{V} 1 \, dV = \text{Volume of } V \] - The projection of \(V\) onto the \(xy\)-plane is a circle defined by the intersection of the paraboloid with \(z = 0\): \[ z = 4 - 3x^2 - 3y^2 = 0 \implies x^2 + y^2 = \frac{4}{3} \] - We'll use cylindrical coordinates \((r, \theta, z)\) since the region is circularly symmetric: - \(x = r\cos\theta\) - \(y = r\sin\theta\) - \(r^2 = x^2 + y^2\), so \(r\) ranges from \(0\) to \(\sqrt{\dfrac{4}{3}}\). - \(\theta\) ranges from \(0\) to \(2\pi\). - For each \((r, \theta)\), \(z\) ranges from \(0\) to \(4 - 3r^2\). 4. **Compute the Volume \(V\)**: - Set up the triple integral: \[ \text{Volume} = \int_{\theta=0}^{2\pi} \int_{r=0}^{\sqrt{4/3}} \int_{z=0}^{4 - 3r^2} r\, dz\, dr\, d\theta \] - Note the extra \(r\) in the integrand due to the Jacobian determinant when converting to cylindrical coordinates. 5. **Evaluate the Integral**: - Integrate with respect to \(z\): \[ \int_{z=0}^{4 - 3r^2} dz = [z]_{0}^{4 - 3r^2} = 4 - 3r^2 \] - The integral becomes: \[ \int_{\theta=0}^{2\pi} \int_{r=0}^{\sqrt{4/3}} (4 - 3r^2) r\, dr\, d\theta \] - Simplify the integrand: \[ (4 - 3r^2) r = 4r - 3r^3 \] - Integrate with respect to \(r\): \[ \int_{r=0}^{\sqrt{4/3}} (4r - 3r^3)\, dr = \left[2r^2 - \frac{3}{4} r^4\right]_0^{\sqrt{4/3}} \] Evaluate at the upper limit \(r = \sqrt{\dfrac{4}{3}}\): \[ r^2 = \frac{4}{3}, \quad r^4 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \] So, \[ 2r^2 - \frac{3}{4} r^4 = 2\left(\frac{4}{3}\right) - \frac{3}{4}\left(\frac{16}{9}\right) = \frac{8}{3} - \frac{4}{3} = \frac{4}{3} \] - The integral over \(r\) is: \[ \int_{r=0}^{\sqrt{4/3}} (4r - 3r^3)\, dr = \frac{4}{3} \] - Integrate with respect to \(\theta\): \[ \int_{\theta=0}^{2\pi} \left(\frac{4}{3}\right)\, d\theta = \frac{4}{3} \left[ \theta \right]_0^{2\pi} = \frac{4}{3} (2\pi) = \frac{8\pi}{3} \] 6. **Conclusion**: - The flux of \(\vec{F}\) across \(S\) is equal to the volume of \(V\): \[ \iint_{S} \vec{F} \cdot d\vec{S} = \frac{8\pi}{3} \] **Answer**: \(\displaystyle \iint_{S} \vec{F} \cdot d\vec{S} = \frac{8\pi}{3}\) |
7.853 |
| GPT-4o | To solve the problem using the Divergence Theorem, we need to evaluate the flux of the vector field \(\vec{F}\) through the closed surface \(S\). The Divergence Theorem relates the flux of \(\vec{F}\) through \(S\) to the volume integral of the divergence of \(\vec{F}\) over the region \(V\) enclosed by \(S\). Mathematically, the Divergence Theorem is stated as: \[ \iint_{S} \vec{F} \cdot d\vec{S} = \iiint_{V} (\nabla \cdot \vec{F}) \, dV \] Given \(\vec{F} = xy \vec{i} - \frac{1}{2}y^2 \vec{j} + z \vec{k}\), we first need to compute the divergence of \(\vec{F}\): \[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(xy) + \frac{\partial}{\partial y} \left(-\frac{1}{2} y^2 \right) + \frac{\partial}{\partial z}(z) \] Calculating each term separately: \[ \frac{\partial}{\partial x}(xy) = y \] \[ \frac{\partial}{\partial y} \left(-\frac{1}{2} y^2 \right) = -y \] \[ \frac{\partial}{\partial z}(z) = 1 \] Adding these together: \[ \nabla \cdot \vec{F} = y - y + 1 = 1 \] Thus, \(\nabla \cdot \vec{F} = 1\). Now, we need to integrate this result over the volume \(V\) enclosed by the surface \(S\). The surface \(S\) consists of the paraboloid \(z = 4 - 3x^2 - 3y^2\) for \(0 \le z \le 1\), and the disk at the bottom, \(z = 0\) and bounded by the circle in the \(xy\)-plane. To set up the volume integral, we convert to cylindrical coordinates: \[ x = r \cos \theta, \quad y = r \sin \theta, \quad z = z \] The volume element in cylindrical coordinates is \(dV = r \, dz \, dr \, d\theta\). The bounds for \(r\) and \(\theta\) are: - \(0 \le r \le 1\) - \(0 \le \theta \le 2\pi\) The bounds for \(z\) are: - \(0 \le z \le 4 - 3r^2\) Thus, the volume integral becomes: \[ \iiint_{V} 1 \, dV = \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{4 - 3r^2} r \, dz \, dr \, d\theta \] We first integrate with respect to \(z\): \[ \int_{0}^{4 - 3r^2} dz = 4 - 3r^2 \] So the integral now is: \[ \int_{0}^{2\pi} \int_{0}^{1} (4 - 3r^2) r \, dr \, d\theta \] Simplify the integrand: \[ \int_{0}^{2\pi} \int_{0}^{1} (4r - 3r^3) \, dr \, d\theta \] Now, integrate with respect to \(r\): \[ \int_{0}^{1} (4r - 3r^3) \, dr = \left[ 2r^2 - \frac{3}{4}r^4 \right]_{0}^{1} = \left( 2(1)^2 - \frac{3}{4}(1)^4 \right) - (0) = 2 - \frac{3}{4} = \frac{8}{4} - \frac{3}{4} = \frac{5}{4} \] Finally, integrate with respect to \(\theta\): \[ \int_{0}^{2\pi} \frac{5}{4} \, d\theta = \frac{5}{4} \theta \bigg|_{0}^{2\pi} = \frac{5}{4} (2\pi) = \frac{10\pi}{4} = \frac{5\pi}{2} \] Thus, the |
7.853 |
| llama3.1-70B-Instruct | ## Step 1: Define the vector field and the surface The vector field is given by $\vec{F} = xy \vec{i} - \frac{1}{2}y^2\vec{j} + z\vec{k}$ and the surface $S$ consists of three parts: the top surface $z=4 - 3*x^2 - 3y^2$, the sides $1 \le z \le 1$, and the bottom surface $z=0$. ## Step 2: 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 $S$. We need to calculate the divergence of $\vec{F}$. ## Step 3: Calculate the divergence of the vector field The divergence of $\vec{F}$ is given by $\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(xy) + \frac{\partial}{\partial y}(-\frac{1}{2}y^2) + \frac{\partial}{\partial z}(z) = y - y + 1 = 1$. ## Step 4: Define the limits of integration for the volume integral The volume $V$ is bounded by the surfaces $z=4 - 3*x^2 - 3y^2$, $z=1$, and $z=0$. The projection of $V$ onto the $xy$-plane is a circle of radius 1, centered at the origin. The limits of integration for the volume integral are $-1 \le x \le 1$, $-\sqrt{1-x^2} \le y \le \sqrt{1-x^2}$, and $0 \le z \le 4 - 3*x^2 - 3y^2$. ## Step 5: Evaluate the volume integral The volume integral is $\iiint_V \nabla \cdot \vec{F} dV = \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{0}^{4 - 3*x^2 - 3y^2} 1 dz dy dx$. ## Step 6: Evaluate the innermost integral The innermost integral is $\int_{0}^{4 - 3*x^2 - 3y^2} 1 dz = 4 - 3*x^2 - 3y^2$. ## Step 7: Evaluate the middle integral The middle integral is $\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (4 - 3*x^2 - 3y^2) dy = 2 \int_{0}^{\sqrt{1-x^2}} (4 - 3*x^2 - 3y^2) dy = 2(4\sqrt{1-x^2} - 3x^2\sqrt{1-x^2} - (1-x^2)^{3/2})$. ## Step 8: Evaluate the outermost integral The outermost integral is $\int_{-1}^{1} 2(4\sqrt{1-x^2} - 3x^2\sqrt{1-x^2} - (1-x^2)^{3/2}) dx = 2 \int_{0}^{1} 2(4\sqrt{1-x^2} - 3x^2\sqrt{1-x^2} - (1-x^2)^{3/2}) dx = 2(\frac{4\pi}{2} - \frac{3\pi}{4} - \frac{2\pi}{3}) = \frac{7\pi}{6}$. The final answer is: $\boxed{\frac{7\pi}{6}}$ |
7.853 |
| 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 $\vec{F} = xy \vec{i} - \frac{1}{2}y^2\vec{j} + z\vec{k}$. ## 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 $\vec{F} = xy \vec{i} - \frac{1}{2}y^2\vec{j} + z\vec{k}$, we have $P = xy$, $Q = -\frac{1}{2}y^2$, and $R = z$. Therefore, $\nabla \cdot \vec{F} = y - y + 1 = 1$. ## Step 2: Define the region $E$ enclosed by $S$. The surface $S$ consists of the three surfaces defined by $z=4 - 3x^2 - 3y^2$ for $1 \le z \le 1$ on the sides and $z=0$ on the bottom. The region $E$ enclosed by $S$ is, therefore, the volume between these surfaces. ## Step 3: Apply the Divergence Theorem. The Divergence Theorem states that $\iint_S \vec{F} \cdot d \vec{S} = \iiint_E \nabla \cdot \vec{F} \, dV$. Since $\nabla \cdot \vec{F} = 1$, the integral simplifies to $\iiint_E 1 \, dV$, which is the volume of the region $E$. ## Step 4: Calculate the volume of $E$. The region $E$ is defined by the surfaces $z=4 - 3x^2 - 3y^2$ and $z=0$. To find the volume, we integrate over the area in the $xy$-plane defined by the intersection of these surfaces, which is a circle of radius $\sqrt{\frac{4}{3}}$ since at $z=0$, $4 - 3x^2 - 3y^2 = 0$ simplifies to $x^2 + y^2 = \frac{4}{3}$. ## Step 5: Perform the volume integral. The volume integral is $\int_{-\sqrt{\frac{4}{3}}}^{\sqrt{\frac{4}{3}}} \int_{-\sqrt{\frac{4}{3} - x^2}}^{\sqrt{\frac{4}{3} - x^2}} \int_0^{4 - 3x^2 - 3y^2} 1 \, dz \, dy \, dx$. This is a standard integral for the volume of a paraboloid and evaluates to $\frac{8\pi}{3}$. The final answer is: $\boxed{\frac{8\pi}{3}}$ |
7.853 |
| DeepSeek-Coder-V2 | To solve the given problem using the Divergence Theorem, we need to evaluate the surface integral \(\iint_S \vec{F} \cdot d \vec{S}\) where \(\vec{F} = xy \vec{i} - \frac{1}{2}y^2\vec{j} + z\vec{k}\) and the surface \(S\) consists of the three surfaces: \(z = 4 - 3x^2 - 3y^2\) for \(1 \le z \le 4\) on the sides, and \(z = 0\) on the bottom. 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 compute the divergence of \(\vec{F}\): \[ \nabla \cdot \vec{F} = \frac{\partial (xy)}{\partial x} + \frac{\partial \left(-\frac{1}{2}y^2\right)}{\partial y} + \frac{\partial z}{\partial z} \] \[ = y + (-y) + 1 = 1 \] Thus, \(\nabla \cdot \vec{F} = 1\). Next, we need to evaluate the volume integral: \[ \iiint_V 1 \, dV \] The volume \(V\) is bounded by the surfaces \(z = 4 - 3x^2 - 3y^2\) for \(1 \le z \le 4\) and \(z = 0\). The region in the \(xy\)-plane is the projection of the surface \(z = 4 - 3x^2 - 3y^2\) onto the \(xy\)-plane, which is a circle centered at the origin with radius 1. We can set up the volume integral in cylindrical coordinates: \[ \iiint_V 1 \, dV = \int_0^{2\pi} \int_0^1 \int_0^{4 - 3r^2} r \, dz \, dr \, d\theta \] Evaluating the integral step by step: 1. Integrate with respect to \(z\): \[ \int_0^{4 - 3r^2} r \, dz = r \left[ z \right]_0^{4 - 3r^2} = r (4 - 3r^2) \] 2. Integrate with respect to \(r\): \[ \int_0^1 r (4 - 3r^2) \, dr = \int_0^1 (4r - 3r^3) \, dr \] \[ = \left[ 2r^2 - \frac{3r^4}{4} \right]_0^1 = 2(1)^2 - \frac{3(1)^4}{4} = 2 - \frac{3}{4} = \frac{5}{4} \] 3. Integrate with respect to \(\theta\): \[ \int_0^{2\pi} \frac{5}{4} \, d\theta = \frac{5}{4} \left[ \theta \right]_0^{2\pi} = \frac{5}{4} \cdot 2\pi = \frac{5\pi}{2} \] Thus, the value of the surface integral is: \[ \iint_S \vec{F} \cdot d \vec{S} = \frac{5\pi}{2} \] Therefore, the final answer is: \[ \boxed{\frac{5\pi}{2}} \] |
7.853 |
| Qwen2-72B-Instruct | To solve this problem using the Divergence Theorem, we need to calculate the divergence of the vector field $\vec{F}$ and then evaluate the triple integral of this divergence over the volume enclosed by the surface $S$. Given vector field $\vec{F} = xy\vec{i} - \frac{1}{2}y^2\vec{j} + z\vec{k}$ First, we calculate the divergence of $\vec{F}$: \[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(xy) + \frac{\partial}{\partial y}\left(-\frac{1}{2}y^2\right) + \frac{\partial}{\partial z}(z) \] \[ = y + (-y) + 1 = 1 \] So, the divergence of $\vec{F}$ is a constant, $1$. Next, we need to calculate the volume integral of the divergence over the volume enclosed by $S$. The volume $V$ is defined by the surfaces $z = 4 - 3x^2 - 3y^2$ for $1 \le z \le 4$ and $z = 0$ for the bottom. We can see that the volume is a paraboloid with its vertex at $(0, 0, 4)$ and it extends down to $z = 0$. Since the paraboloid is symmetric with respect to the $x$ and $y$ axes, we can integrate over the region $-\sqrt{\frac{4 - z}{3}} \le x \le \sqrt{\frac{4 - z}{3}}$ and $-\sqrt{\frac{4 - z}{3}} \le y \le \sqrt{\frac{4 - z}{3}}$, and $0 \le z \le 1$ for the bottom surface. The volume integral of the divergence can be written as: \[ \iiint_V \nabla \cdot \vec{F} \, dV = \int_0^1 \int_{-\sqrt{\frac{4 - z}{3}}}^{\sqrt{\frac{4 - z}{3}}} \int_{-\sqrt{\frac{4 - z}{3}}}^{\sqrt{\frac{4 - z}{3}}} 1 \, dx \, dy \, dz \] \[ = \int_0^1 \int_{-\sqrt{\frac{4 - z}{3}}}^{\sqrt{\frac{4 - z}{3}}} 2\sqrt{\frac{4 - z}{3}} \, dy \, dz \] \[ = \int_0^1 \frac{4}{3} \left(4 - z\right) \, dz \] \[ = \frac{4}{3} \left[4z - \frac{1}{2}z^2\right]_0^1 = \frac{4}{3} \left(4 - \frac{1}{2}\right) = \frac{4}{3} \cdot \frac{7}{2} = \frac{14}{3} \] Therefore, the value of the surface integral $\iint_S \vec{F} \cdot d\vec{S}$ is $\boxed{\frac{14}{3}}$. |
7.853 |
| 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 $\vec{F} = xy \vec{i} - \frac{1}{2}y^2\vec{j} + z\vec{k}$. The divergence is given by: $$\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(xy) - \frac{\partial}{\partial y}\left(-\frac{1}{2}y^2\right) + \frac{\partial}{\partial z}(z) = y + y + 1 = 2y + 1$$ Now, by the divergence theorem, we have: $$\iint_S \vec{F} \cdot d \vec{S} = \iiint_E (2y + 1) dV$$ where $E$ is the volume enclosed by the surface $S$. In this case, $E$ is a cylindrical volume with the top surface given by $z = 4 - 3x^2 - 3y^2$, the bottom surface given by $z = 0$, and the sides given by $1 \leq z \leq 1$ which implies that the lateral surface is actually a constant $z$ surface at $z=1$. We can set up the triple integral over the volume $E$. The volume $E$ is bounded by: - $0 \leq z \leq 4 - 3x^2 - 3y^2$ (from the top surface equation) - $-1 \leq x \leq 1$ (since $z=1$ on the sides, we can infer the $x$ limits) - $-1 \leq y \leq 1$ (similarly for $y$ limits) Now we integrate the divergence over this volume: $$\iiint_E (2y + 1) dV = \int_{-1}^{1} \int_{-1}^{1} \int_{0}^{4 - 3x^2 - 3y^2} (2y + 1) \, dz \, dy \, dx$$ Let's compute this integral: $$ \begin{align*} &= \int_{-1}^{1} \int_{-1}^{1} \left[ (2y + 1)z \right]_{0}^{4 - 3x^2 - 3y^2} dy dx\\ &= \int_{-1}^{1} \int_{-1}^{1} (2y + 1)(4 - 3x^2 - 3y^2) dy dx\\ &= \int_{-1}^{1} \left[ \int_{-1}^{1} (8y - 6xy^2 - 6y^3 + 4 - 3x^2 - 3y^2) dy \right] dx\\ &= \int_{-1}^{1} \left[ 4y^2 - 6xy^3 - 3y^4 + 4y - 3x^2y - y^3 \right]_{-1}^{1} dx\\ &= \int_{-1}^{1} \left[ (4 - 6x - 3 + 4 - 3x^2 - 1) - (4 + 6x - 3 - 4 + 3x^2 + 1) \right] dx\\ &= \int_{-1}^{1} \left[ 4 - 6x - 3x^2 \right] dx\\ &= \left[ 4x - 3x^2 - 3x^3/3 \right]_{-1}^{1}\\ &= (4 - 3 - 1) - (-4 - 3 + 1)\\ &= 0 \end{align*} $$ So, the value of the surface integral $\iint_S \vec{F} \cdot d \vec{S}$ is $0$. |
7.853 |
| doubao | First, we need to find the divergence of $\vec{F}$: $\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(xy) + \frac{\partial}{\partial y}(-\frac{1}{2}y^2) + \frac{\partial}{\partial z}(z) = y - y + 1 = 1$. Now we use the divergence theorem to convert the surface integral to a volume integral: $\iint_S \vec{F} \cdot d\vec{S} = \iiint_V \nabla \cdot \vec{F} dV = \iiint_V 1 dV$. The region $V$ is bounded by the surface $z=4 - 3x^2 - 3y^2$ and $z=0$. Converting to cylindrical coordinates, we have $z=4 - 3r^2$ and $r^2 = x^2 + y^2$. The limits of integration for $r$ are from $0$ to $1$ (since the intersection of $z=4 - 3r^2$ and $z=1$ gives $3r^2 = 3 \Rightarrow r = 1$). The limits for $z$ are from $0$ to $4 - 3r^2$. And the limits for $\theta$ are from $0$ to $2\pi$. So the volume integral becomes: \[ \begin{align*} &\int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{4 - 3r^2} r dz dr d\theta\\ =&\int_{0}^{2\pi} \int_{0}^{1} [rz]_{0}^{4 - 3r^2} dr d\theta\\ =&\int_{0}^{2\pi} \int_{0}^{1} r(4 - 3r^2) dr d\theta\\ =&\int_{0}^{2\pi} [2r^2 - \frac{3r^4}{4}]_{0}^{1} d\theta\\ =&\int_{0}^{2\pi} (2 - \frac{3}{4}) d\theta\\ =&\int_{0}^{2\pi} \frac{5}{4} d\theta\\ =&\frac{5}{4} [\theta]_{0}^{2\pi}\\ =&\frac{5}{4} \times 2\pi\\ =&\frac{5\pi}{2}. \end{align*} \] |
7.853 |
| minimax | To use the divergence theorem to evaluate the surface integral $\iint_S \vec{F} \cdot d \vec{S}$, we first need to compute the divergence of the vector field $\vec{F}$ and then integrate it over the volume $V$ enclosed by the surface $S$. The divergence of $\vec{F}$ is given by: $$\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(xy) + \frac{\partial}{\partial y}\left(-\frac{1}{2}y^2\right) + \frac{\partial}{\partial z}(z)$$ Calculating the partial derivatives, we get: $$\nabla \cdot \vec{F} = y - y + 1 = 1$$ Now, we need to define the volume $V$ over which we will integrate the divergence. The surface $S$ consists of three parts: 1. The paraboloid $z = 4 - 3x^2 - 3y^2$ for $1 \le z \le 4$. 2. The rectangle on the plane $z = 1$ that is bounded by $-1 \le x \le 1$ and $-1 \le y \le 1$. 3. The rectangle on the plane $z = 0$ that is bounded by $-1 \le x \le 1$ and $-1 \le y \le 1$. The divergence theorem states that for a vector field $\vec{F}$ and a closed surface $S$ enclosing a volume $V$: $$\iint_S \vec{F} \cdot d \vec{S} = \iiint_V (\nabla \cdot \vec{F}) dV$$ Since the divergence of $\vec{F}$ is constant and equal to 1, the volume integral simplifies to: $$\iiint_V (\nabla \cdot \vec{F}) dV = \iiint_V 1 \, dV = \text{Volume of } V$$ The volume $V$ is the space between the paraboloid $z = 4 - 3x^2 - 3y^2$ and the plane $z = 1$, plus the volume of the rectangular prism below $z = 1$ and above $z = 0$. First, let's find the volume between the paraboloid and the plane $z = 1$. We need to set up the integral for the volume $V_1$: $$V_1 = \int_{-1}^{1} \int_{-1}^{1} \int_{1}^{4 - 3x^2 - 3y^2} 1 \, dz \, dy \, dx$$ Next, we find the volume of the rectangular prism below $z = 1$ and above $z = 0$, which is simply the area of the base times the height: $$V_2 = \int_{-1}^{1} \int_{-1}^{1} 1 \, dy \, dx \cdot (1 - 0) = \int_{-1}^{1} \int_{-1}^{1} 1 \, dy \, dx = 2 \cdot 2 = 4$$ Now, let's compute $V_1$: $$V_1 = \int_{-1}^{1} \int_{-1}^{1} (4 - 3x^2 - 3y^2 - 1) \, dy \, dx$$ $$V_1 = \int_{-1}^{1} \left[ 4y - 3x^2y - y^3 \right]_{-1}^{1} \, dx$$ $$V_1 = \int_{-1}^{1} \left[ (4 - 3x^2 - 1) - (-4 + 3x^2 + 1) \right] \, dx$$ $$V_1 = \int_{-1}^{1} \left[ 3 - 6x^2 \right] \, dx$$ $$V_1 = \left[ 3x - 2x^3 \right]_{-1}^{1}$$ $$V_1 = (3 - 2) - (-3 + 2)$$ $$V_1 = 1 - (-1)$$ $$V_1 = 2$$ The total volume $V$ is the sum of $V_1$ and $V_2$: $$V = V_1 + V_2 = 2 + 4 = 6$$ Therefore, the surface integral $\iint_S \vec{F} \cdot d \vec{S}$ evaluated using the divergence theorem is equal to the volume of $V$: $$\iint_S \vec{F} \cdot d \vec{S} = \iiint_V (\nabla \cdot \vec{F}) dV = \iiint_V 1 \, dV = V = 6$$ |
7.853 |
| kimi | To evaluate the surface integral $\iint_S \vec{F} \cdot d \vec{S}$ using the divergence theorem, we first need to convert the surface integral 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 $\vec{F}$ is the vector field, $S$ is the surface, $V$ is the volume enclosed by the surface, and $\nabla \cdot \vec{F}$ is the divergence of the vector field. Given vector field $\vec{F} = xy \vec{i} - \frac{1}{2}y^2\vec{j} + z\vec{k}$, let's compute its divergence: $$\nabla \cdot \vec{F} = \frac{\partial (xy)}{\partial x} + \frac{\partial \left(-\frac{1}{2}y^2\right)}{\partial y} + \frac{\partial z}{\partial z}$$ $$\nabla \cdot \vec{F} = y - y + 1$$ $$\nabla \cdot \vec{F} = 1$$ Now, we need to find the volume $V$ enclosed by the surface $S$. The surface $S$ consists of three parts: 1. The top surface: $z = 4 - 3x^2 - 3y^2$ for $1 \leq z \leq 1$. However, this is a typo in the problem statement. It should be $1 \leq z \leq 4$. The top surface is a paraboloid. 2. The bottom surface: $z = 0$. 3. The side surfaces: These are the vertical planes where $z$ is constant. The volume $V$ is the region in the $xy$-plane between the circle defined by $3x^2 + 3y^2 = 4$ (since $z = 4$ at the top) and the plane $z = 0$. The volume integral becomes: $$\iiint_V 1 \, dV = \int_0^{2\pi} \int_0^2 \int_0^{4-3r^2} r \, dz \, dr \, d\theta$$ Where $r^2 = x^2 + y^2$ and $r \, dz \, dr \, d\theta$ is the differential volume element in cylindrical coordinates. Now, we can evaluate the integral: $$\int_0^{2\pi} \int_0^2 \int_0^{4-3r^2} r \, dz \, dr \, d\theta = \int_0^{2\pi} \int_0^2 \left[ rz \right]_0^{4-3r^2} r \, dr \, d\theta$$ $$= \int_0^{2\pi} \int_0^2 (4r - 3r^3) \, dr \, d\theta$$ $$= \int_0^{2\pi} \left[ 2r^2 - \frac{3r^4}{4} \right]_0^2 \, d\theta$$ $$= \int_0^{2\pi} \left( 8 - 3 \cdot 16 \right) \, d\theta$$ $$= \int_0^{2\pi} (8 - 48) \, d\theta$$ $$= \int_0^{2\pi} -40 \, d\theta$$ $$= -40 \cdot 2\pi$$ $$= -80\pi$$ So, the value of the surface integral $\iint_S \vec{F} \cdot d \vec{S}$ is $-80\pi$. |
7.853 |
| wenxinyiyan | We need to evaluate the integral $\iint_S \vec{F} \cdot d\vec{S}$ where $\vec{F} = xy\vec{i} - \frac{1}{2}y^2\vec{j} + z\vec{k}$ and $S$ is the surface defined by $z = 4 - 3x^2 - 3y^2$ with $1 \le z \le 1$ on the sides and $z = 0$ on the bottom. First, let's 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: $$\text{div}\vec{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}$$So, for $\vec{F} = xy\vec{i} - \frac{1}{2}y^2\vec{j} + z\vec{k}$, we have: $$\text{div}\vec{F} = \frac{\partial}{\partial x}(xy) + \frac{\partial}{\partial y}\left(-\frac{1}{2}y^2\right) + \frac{\partial}{\partial z}(z) = y - y + 1 = 1$$Now, we apply the Divergence Theorem: $$\iint_S \vec{F} \cdot d\vec{S} = \iiint_V \text{div}\vec{F} \, dV = \iiint_V 1 \, dV$$where $V$ is the volume enclosed by the surface $S$. The volume $V$ is defined by the paraboloid $z = 4 - 3x^2 - 3y^2$ and the plane $z = 1$. The paraboloid intersects the plane when $4 - 3x^2 - 3y^2 = 1$, which gives $x^2 + y^2 = 1$. So, $V$ is the volume between the paraboloid and the plane $z = 1$ inside the cylinder $x^2 + y^2 = 1$. We can use cylindrical coordinates to describe $V$. Let $x = r\cos\theta$, $y = r\sin\theta$, and $z = z$. Then, the volume integral becomes: $$\iiint_V 1 \, dV = \int_0^{2\pi} \int_0^1 \int_1^{4-3r^2} r \, dz \, dr \, d\theta$$Evaluating this integral, we get: $$\begin{aligned} \iiint_V 1 \, dV &= \int_0^{2\pi} \int_0^1 \int_1^{4-3r^2} r \, dz \, dr \, d\theta \\ &= \int_0^{2\pi} \int_0^1 r(4-3r^2 - 1) \, dr \, d\theta \\ &= \int_0^{2\pi} \int_0^1 r(3-3r^2) \, dr \, d\theta \\ &= \int_0^{2\pi} \left[\frac{3}{2}r^2 - \frac{3}{4}r^4\right]_0^1 \, d\theta \\ &= \int_0^{2\pi} \left(\frac{3}{2} - \frac{3}{4}\right) \, d\theta \\ &= \int_0^{2\pi} \frac{3}{4} \, d\theta \\ &= \frac{3}{4} \times 2\pi \\ &= \frac{3\pi}{2} \end{aligned}$$Therefore, $\iint_S \vec{F} \cdot d\vec{S} = \frac{3\pi}{2}$. |
7.853 |
| zhipu | To use the Divergence Theorem to evaluate the surface integral \(\iint_S \vec{F} \cdot d \vec{S}\), we need to convert it into a volume integral of the divergence of the vector field \(\vec{F}\) over the volume \(V\) enclosed by the surface \(S\). The Divergence Theorem states: \[ \iint_S \vec{F} \cdot d \vec{S} = \iiint_V (\nabla \cdot \vec{F}) \, dV \] First, let's find the divergence of the vector field \(\vec{F} = xy \vec{i} - \frac{1}{2}y^2 \vec{j} + z \vec{k}\). \[ \nabla \cdot \vec{F} = \frac{\partial (xy)}{\partial x} + \frac{\partial \left(-\frac{1}{2}y^2\right)}{\partial y} + \frac{\partial z}{\partial z} \] Calculating each partial derivative: \[ \frac{\partial (xy)}{\partial x} = y \] \[ \frac{\partial \left(-\frac{1}{2}y^2\right)}{\partial y} = -y \] \[ \frac{\partial z}{\partial z} = 1 \] Adding these together: \[ \nabla \cdot \vec{F} = y - y + 1 = 1 \] Now, we need to set up the volume integral of the divergence over the volume \(V\) enclosed by the surface \(S\). The surface \(S\) is bounded by \(z = 4 - 3x^2 - 3y^2\) on the top, \(z = 0\) on the bottom, and the sides defined by \(1 \le z \le 1\) (which seems to be a typo, as it should likely be some bounds for \(x\) and \(y\)). Assuming the correct interpretation is that the surface is bounded in the \(xy\)-plane by the projection of \(z = 4 - 3x^2 - 3y^2\) onto \(z = 0\). To find the bounds for \(x\) and \(y\), set \(z = 0\): \[ 0 = 4 - 3x^2 - 3y^2 \implies 3x^2 + 3y^2 = 4 \implies x^2 + y^2 = \frac{4}{3} \] This is a circle of radius \(\sqrt{\frac{4}{3}}\) in the \(xy\)-plane. We can use cylindrical coordinates \((r, \theta, z)\) to simplify the integration, where \(x = r \cos \theta\), \(y = r \sin \theta\), and \(r\) ranges from \(0\) to \(\sqrt{\frac{4}{3}}\), and \(\theta\) ranges from \(0\) to \(2\pi\). The volume element in cylindrical coordinates is \(dV = r \, dr \, d\theta \, dz\). The integral becomes: \[ \iiint_V (\nabla \cdot \vec{F}) \, dV = \int_{0}^{2\pi} \int_{0}^{\sqrt{\frac{4}{3}}} \int_{0}^{4 - 3r^2} 1 \cdot r \, dz \, dr \, d\theta \] Since \(\nabla \cdot \vec{F} = 1\), the integral simplifies to: \[ \int_{0}^{2\pi} \int_{0}^{\sqrt{\frac{4}{3}}} \int_{0}^{4 - 3r^2} r \, dz \, dr \, d\theta \] First, integrate with respect to \(z\): \[ \int_{0}^{4 - 3r^2} dz = 4 - 3r^2 \] So the integral becomes: \[ \int_{0}^{2\pi} \int_{0}^{\sqrt{\frac{4}{3}}} r (4 - 3r^2) \, dr \, d\theta \] Now, integrate with respect to \(r\): \[ \int_{0}^{\sqrt{\frac{4}{3}}} r (4 - 3r^2) \, dr = \int_{0}^{\sqrt{\frac{4}{3}}} (4r - 3r^3) \, dr \] \[ = \left[ 2r^2 - \frac{3r^4}{4} \right]_{0}^{\sqrt{\frac{4}{3}}} \] \[ = \left( 2 \left(\sqrt{\frac{4}{3}}\right)^2 - \frac{3 \left(\sqrt{\frac{4}{3}}\right)^4}{4} \right) \] \[ = \left( 2 \cdot \frac{4}{3} - \frac{3 \cdot \left(\frac{16}{9}\right)}{4} \right) \] \[ = \left( \frac{8}{3} - \frac{48}{36} \right) \] \[ = \left( \frac{8}{3} - \frac{4}{3} \right) \] \[ = \frac{4}{3} \] Finally, integrate with respect to \(\theta\): \[ \int_{0}^{2\pi} \frac{4}{3} \, d\theta = \frac{4}{3} \cdot 2\pi = \frac{8\pi}{3} \] Therefore, the value of the surface integral \(\iint_S \vec{F} \cdot d \vec{S}\) is: \[ \boxed{\frac{8\pi}{3}} \] |
7.853 |
| gpto1 | To evaluate the flux of the vector field \(\vec{F}\) across the closed surface \(S\) using the divergence theorem, we'll follow these steps: 1. **Understand the Surface \(S\) and the Volume \(V\)**: - The surface \(S\) consists of three parts: - The **top surface**: the paraboloid \(z = 4 - 3x^2 - 3y^2\) above \(z = 0\). - The **bottom surface**: the plane \(z = 0\). - The **side surface**: the curved surface connecting the top and bottom surfaces. - The volume \(V\) is the solid enclosed by this surface \(S\). 2. **Compute the Divergence of \(\vec{F}\)**: - The divergence theorem states: \[ \iint_{S} \vec{F} \cdot d\vec{S} = \iiint_{V} (\nabla \cdot \vec{F}) \, dV \] - Compute the divergence \(\nabla \cdot \vec{F}\): \[ \nabla \cdot \vec{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} \] Given \(\vec{F} = xy\, \vec{i} - \dfrac{1}{2} y^2\, \vec{j} + z\, \vec{k}\): - \(\frac{\partial F_x}{\partial x} = \frac{\partial}{\partial x}(xy) = y\) - \(\frac{\partial F_y}{\partial y} = \frac{\partial}{\partial y}\left(-\dfrac{1}{2} y^2\right) = -y\) - \(\frac{\partial F_z}{\partial z} = \frac{\partial}{\partial z}(z) = 1\) - So, \(\nabla \cdot \vec{F} = y - y + 1 = 1\) 3. **Set Up the Triple Integral Over Volume \(V\)**: - Since \(\nabla \cdot \vec{F} = 1\), the flux simplifies to the volume of \(V\): \[ \iiint_{V} (\nabla \cdot \vec{F}) \, dV = \iiint_{V} 1 \, dV = \text{Volume of } V \] - The projection of \(V\) onto the \(xy\)-plane is a circle defined by the intersection of the paraboloid with \(z = 0\): \[ z = 4 - 3x^2 - 3y^2 = 0 \implies x^2 + y^2 = \frac{4}{3} \] - We'll use cylindrical coordinates \((r, \theta, z)\) since the region is circularly symmetric: - \(x = r\cos\theta\) - \(y = r\sin\theta\) - \(r^2 = x^2 + y^2\), so \(r\) ranges from \(0\) to \(\sqrt{\dfrac{4}{3}}\). - \(\theta\) ranges from \(0\) to \(2\pi\). - For each \((r, \theta)\), \(z\) ranges from \(0\) to \(4 - 3r^2\). 4. **Compute the Volume \(V\)**: - Set up the triple integral: \[ \text{Volume} = \int_{\theta=0}^{2\pi} \int_{r=0}^{\sqrt{4/3}} \int_{z=0}^{4 - 3r^2} r\, dz\, dr\, d\theta \] - Note the extra \(r\) in the integrand due to the Jacobian determinant when converting to cylindrical coordinates. 5. **Evaluate the Integral**: - Integrate with respect to \(z\): \[ \int_{z=0}^{4 - 3r^2} dz = [z]_{0}^{4 - 3r^2} = 4 - 3r^2 \] - The integral becomes: \[ \int_{\theta=0}^{2\pi} \int_{r=0}^{\sqrt{4/3}} (4 - 3r^2) r\, dr\, d\theta \] - Simplify the integrand: \[ (4 - 3r^2) r = 4r - 3r^3 \] - Integrate with respect to \(r\): \[ \int_{r=0}^{\sqrt{4/3}} (4r - 3r^3)\, dr = \left[2r^2 - \frac{3}{4} r^4\right]_0^{\sqrt{4/3}} \] Evaluate at the upper limit \(r = \sqrt{\dfrac{4}{3}}\): \[ r^2 = \frac{4}{3}, \quad r^4 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \] So, \[ 2r^2 - \frac{3}{4} r^4 = 2\left(\frac{4}{3}\right) - \frac{3}{4}\left(\frac{16}{9}\right) = \frac{8}{3} - \frac{4}{3} = \frac{4}{3} \] - The integral over \(r\) is: \[ \int_{r=0}^{\sqrt{4/3}} (4r - 3r^3)\, dr = \frac{4}{3} \] - Integrate with respect to \(\theta\): \[ \int_{\theta=0}^{2\pi} \left(\frac{4}{3}\right)\, d\theta = \frac{4}{3} \left[ \theta \right]_0^{2\pi} = \frac{4}{3} (2\pi) = \frac{8\pi}{3} \] 6. **Conclusion**: - The flux of \(\vec{F}\) across \(S\) is equal to the volume of \(V\): \[ \iint_{S} \vec{F} \cdot d\vec{S} = \frac{8\pi}{3} \] **Answer**: \(\displaystyle \iint_{S} \vec{F} \cdot d\vec{S} = \frac{8\pi}{3}\) |
7.853 |
| deepseekv3 | To evaluate the surface integral \(\iint_S \vec{F} \cdot d \vec{S}\) using the **Divergence Theorem**, we follow these steps: --- ### **Step 1: Understand the Problem** We are given the vector field: \[ \vec{F} = xy \, \vec{i} - \frac{1}{2} y^2 \, \vec{j} + z \, \vec{k} \] and the surface \(S\) consists of: 1. The paraboloid \(z = 4 - 3x^2 - 3y^2\), where \(1 \le z \le 4\) (this is a typo in the problem statement; it should be \(1 \le z \le 4\) instead of \(1 \le z \le 1\)), 2. The disk \(z = 0\) at the bottom. 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 \(S\). --- ### **Step 2: Compute the Divergence** First, compute the divergence of \(\vec{F}\): \[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(xy) + \frac{\partial}{\partial y}\left(-\frac{1}{2} y^2\right) + \frac{\partial}{\partial z}(z) \] \[ \nabla \cdot \vec{F} = y - y + 1 = 1 \] --- ### **Step 3: Set Up the Volume Integral** The volume \(V\) is bounded by: 1. The paraboloid \(z = 4 - 3x^2 - 3y^2\) (top surface), 2. The plane \(z = 0\) (bottom surface). In cylindrical coordinates \((r, \theta, z)\), the paraboloid becomes: \[ z = 4 - 3r^2 \] The limits for \(r\) are determined by the intersection of the paraboloid with \(z = 1\): \[ 1 = 4 - 3r^2 \implies 3r^2 = 3 \implies r = 1 \] Thus, the volume integral becomes: \[ \iiint_V (\nabla \cdot \vec{F}) \, dV = \iiint_V 1 \, dV \] In cylindrical coordinates: \[ dV = r \, dz \, dr \, d\theta \] The limits are: \[ 0 \le r \le 1, \quad 0 \le \theta \le 2\pi, \quad 0 \le z \le 4 - 3r^2 \] --- ### **Step 4: Evaluate the Integral** \[ \iiint_V 1 \, dV = \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{4-3r^2} r \, dz \, dr \, d\theta \] First, integrate with respect to \(z\): \[ \int_{0}^{4-3r^2} dz = 4 - 3r^2 \] Now, integrate with respect to \(r\): \[ \int_{0}^{1} r(4 - 3r^2) \, dr = \int_{0}^{1} (4r - 3r^3) \, dr = \left[2r^2 - \frac{3}{4} r^4\right]_{0}^{1} = 2 - \frac{3}{4} = \frac{5}{4} \] Finally, integrate with respect to \(\theta\): \[ \int_{0}^{2\pi} \frac{5}{4} \, d\theta = \frac{5}{4} \cdot 2\pi = \frac{5\pi}{2} \] --- ### **Final Answer** \[ \iint_S \vec{F} \cdot d \vec{S} = \frac{5\pi}{2} \] |
7.853 |