正方体容器水量计算

Creado el: 23 de enero de 2025

Respondido usando O1 por Chat01

Pregunta

用你的代码能力计算下面的题目：

棱长为2的正方体容器ABCD-A1B1C1D1中，E,F分别是棱AB,BC的中点，在E,F,C1处各有1个小孔（孔的大小忽略不计），则该容器可装水的最大体积是？

Pensamiento

Assessing hole impacts

I'm piecing together the volume of water a cube can hold, considering holes at specific points and how it affects the water level.

Assessing water dynamics

I’ve pinpointed the coordinates of the cube’s vertices and defined the locations of the holes. Analyzing the water dynamics, I’m determining the volume that won’t leak through these points.

Crafting the solution

Hmm, I’m thinking about how the water surface aligns with holes at specific heights, ensuring it doesn’t exceed the holes’ altitudes, allowing the container to hold the maximum water volume.

Establishing spatial coordinates

OK, let me see. I’ve pinpointed the positions of E, F, and C1 in 3D space, ensuring the water surface stays below these points to prevent leaks.

Considering leak dynamics

I'm evaluating how the water surface aligns with holes E, F, and C1. If the water level exceeds these points, leakage occurs, influenced by atmospheric pressure and external factors.

Evaluating bottom hole

Considering the bottom hole, any water inside the container potentially leaks out, leaving no water. This indicates the volume is likely zero.

Evaluating orientations

I’m considering if tilting the container could keep the water surface below the holes. Aligning the plane through E, F, and C1 horizontally might let us fill the container more.

Maximizing volume

I'm figuring out the plane through E, F, and C1, ensuring the water level reaches this plane. This approach helps pinpoint the maximum water volume that fits within the cube.

Mapping the shape

I'm pulling together the volume under the plane through E, F, and C1, focusing on the cubical region opposite to the normal vector, ensuring the water surface aligns with the holes.

Piecing together

I’m analyzing vectors EF and EC1 to find the normal vector. Now, I’m using the cross product to determine the plane equation passing through E, F, and C1.

Determining the plane

I formulated the plane equation through E, F, and C1 as 2x - 2y + z = 2. Testing point A shows it lies on the side where the plane equation holds true.

Assessing the region

OK, let me see. The positive z-component of the normal vector indicates the plane points upwards, suggesting the water needs to fill the area below the plane equation 2x - 2y + z ≤ 2.

Determining plane relationship

I’m testing point C1 on the plane, which matches the plane equation. Meanwhile, point D1 falls beneath the plane threshold, suggesting it lies below the specified reference plane.

Identifying water region

Considering point A indicates the water region is likely below the plane 2x - 2y + z = 2. I’m focusing on how this intersects with the cube to determine the water volume.

Identifying volume range

I’m mapping out the volume as a polyhedron within the cube, with z restricted by 2 - 2x + 2y and z ≤ 2. Integrating over x and y helps determine the volume.

Recast the solution

OK, let's see. Ensuring the volume is between x and y constraints, z ranges from 0 to 2 - 2x + 2y, and y is at least x - 1, all within a 2x2 grid.

Mapping constraints

I’m thinking about the xy-plane constraints: 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, and y ≥ 0, ensuring the volume falls within 0 to 2 - 2x + 2y.

Setting integration limits

I'm focusing on the integration limits: x ranges from 0 to 2, y from max(0, x-1) to 2, and z from 0 to 2 - 2x + 2y.

Piecing together

I’m working through two integrals for volume, adjusting my focus based on x range. The first integral simplifies to 8 - 4x, integrating y and z within specific limits.

Calculating volume

I’m mapping out the volume for x in [1, 2], integrating y and z within specific limits. This involves piecing together the integrals and considering the specified interval endpoints for x.

Reassessing the integral

I’m piecing together the final integral, recalculating to ensure the accuracy of 9 - 6x + x^2 for x in [1, 2], integrating over y.

Calculating the volume

I’m integrating 9 - 6x + x^2 from 1 to 2, ensuring all variables are included. The final step confirmed the volume to be 7/3.

Breaking down the volume

Taking another look, the total volume from integrating two separate regions—one fixed and one variable—is 8.3333..., confirming the calculated volume as appropriate for this problem.

Adjusting my approach

OK, let me see. The water volume mentioned surpasses the cube's capacity, suggesting a miscalculation or potential error in the volume integration process that needs reconsideration.

Reevaluating constraints

I’m focusing on z's upper bound to ensure no miscalculations. The water volume should remain within the cube's 8-unit capacity.

Balancing constraints

I’m thinking through how to maintain the water level within 2 - 2x + 2y ≤ 2, ensuring the volume fits within the cube's limits without exceeding 8 units.

Taking a closer look

I’m dividing the domain into subregions A and B to refine the volume calculation. Subregion A’s z-bound is 2 - 2x + 2y, while subregion B’s z-bound is z ≤ 2.

Refining the boundaries

I’m focusing on the domain constraints: x and y range from 0 to 2. To ensure water fits within the cube, y must be at least x - 1.

Ensuring containment

OK, let me just think: I'll confirm that the plane EFC1 stays above the water level to avoid leaks, making sure the water remains contained within the cube.

Ensuring accurate integration

I’m refining the integration limits to include only z regions where 2 - 2x + 2y is positive. This leads me to adjust the volume calculation for precise containment.

Mapping out constraints

I’m detailing y’s range to ensure accurate volume computation under the plane equation, considering x and y constraints to prevent miscalculations.

Mapping y's range

For x in [0,1], y ranges from 0 to x. For x in [1,2], y ranges from x-1 to x. This approach efficiently trims down the integration limits.

Piecing together

For x in [0,1], y ranges from 0 to x, naturally satisfying y ≥ x-1. For x in [1,2], y ranges from x-1 to x. This ensures comprehensive integration limits.

Mapping integration limits

I’m refining the subregions for integration to ensure accurate volume calculation. For x in [0,1], y ranges from 0 to x. For x in [1,2], y spans from x-1 to x, with z limited to 2.

Mapping constraints

Piecing together integration limits, y ranges from x to 2, and z spans from 0 to 2 - 2x + 2y. This ensures precise volume calculation.

Mapping constraints

I detailed the xy-plane partition by the line y = x, ensuring accurate volume calculation by narrowing down z limits based on x and y constraints.

Mapping out constraints

I’m pinpointing the valid region in the xy-plane for integration, ensuring the volume calculation aligns with the constraints of y and x, leading to a precise z boundary.

Calculating volume

I'm integrating over triangle region B where x < y, determining the area to ensure the valid volume, considering the constraints and height in the z-direction.

Determining volume

Calculating the volume of region A involves integrating a triangular area with a varying height limit, applying constraints to ensure the water is fully contained within the cube.

Integrating for volume

I'm calculating the volume of region A by integrating over the defined limits, considering constraints for x and y to ensure the water stays contained within the cube.

Summing up regions

I’ve combined the regions, ensuring water containment within the cube by checking integration limits and constraints, resulting in a calculated volume of 6.

Calculating total volume

I'm collating the volumes from regions A and B. Confirming they fit within 2x - 2y + z ≤ 2, I determined the combined volume to be 16/3.

Validating volume inclusion

I checked region B’s inclusion in 2x - 2y + z ≤ 2 by verifying its boundaries and ensuring all points fit within these constraints.

Reflecting on volume calculations

I’m working through the volume calculation for the cube segment bound by the plane 2x - 2y + z ≤ 2, ensuring it stays within the cube’s volume limit.

Ensuring region alignment

I’m mapping out that the calculated volume corresponds to the water region beneath the plane EFC1, ensuring the water level does not surpass the containers' boundaries.

Ensuring accuracy

I’m making sure the water level aligns with the plane through E, F, and C1, without allowing water to leak from any other points in the container.

Piecing it together

Checking if the volume calculation, 17/3 or 5.6667, is optimal. Considering fractions and decimals, I conclude that 17/3 is likely the correct answer.

Thought about 正方体容器的水容量，持续 4m 55s

Respuesta

下面给出一种比较“几何＋坐标”的思路来求解。结论先行：该容器可盛水的最大体积是 $\tfrac{17}{3}$ （即 $5\frac{2}{3}$ ）。

1. 题意解析与坐标设置

题目给出一个棱长为 2 的正方体 $ABCD - A_1 B_1 C_1 D_1$ ，并且指出：

$E,F$ 分别是棱 $AB,BC$ 的中点；
在 $E, F, C_1$ 处各有 1 个小孔（忽略孔大小）；
问此容器最多能盛多少体积的水。

典型做法是：

把正方体放在合适的坐标系里，找出经过这三个“漏孔”点 $E, F, C_1$ 的平面方程；
由于容器只在 $E, F, C_1$ 处开孔，若把该平面当作“水平水面”（通过适当旋转容器的想象），水面不能超过这个平面，否则水会从这些孔流走。
因而，容器中能盛的最大水量，就是**正方体在该平面以下（包含该平面）**那部分的体积。

1.1 坐标取法

不失一般性，可将底面 $ABCD$ 放在 $z=0$ 平面上，令：

\begin{aligned} & A = (0,0,0), \quad B = (2,0,0), \quad C = (2,2,0), \quad D = (0,2,0),\\ & A_1 = (0,0,2), \; B_1 = (2,0,2), \; C_1 = (2,2,2), \; D_1 = (0,2,2). \end{aligned}

则中点

E = \text{棱 }AB\text{ 的中点} = (1,0,0), \quad F = \text{棱 }BC\text{ 的中点} = (2,1,0).

孔的位置坐标即：

E(1,\,0,\,0),\quad F(2,\,1,\,0),\quad C_1(2,\,2,\,2).

2. 经过三孔的平面方程

要找平面 $EFC_1$ 。设平面方程为

\alpha x + \beta y + \gamma z + \delta = 0.

也可直接用“向量法”：

向量 $\overrightarrow{EF} = F - E = (1,\,1,\,0)$ 。
向量 $\overrightarrow{EC_1} = C_1 - E = (1,\,2,\,2)$ 。

它们的叉积给出平面法向量：

\overrightarrow{n} = \overrightarrow{EF} \times \overrightarrow{EC_1} = (1,1,0)\times(1,2,2).

用行列式/公式计算可得

\overrightarrow{n} = (2,\,-2,\,1).

于是，平面方程可写成

2(x - x_0) \;-\;2(y - y_0)\;+\;1(z - z_0)\;=\;0,

其中 $(x_0,y_0,z_0)$ 可取平面上任一点，比如 $E(1,0,0)$ 。简化后得到：

2(x-1)\;-\;2(y-0)\;+\;(z-0)\;=\;0 \;\;\Longrightarrow\;\; 2x -2 -2y + z \;=\;0 \;\;\Longrightarrow\;\; 2x - 2y + z \;=\; 2.

最终，平面 $EFC_1$ 的方程为

\boxed{\,2x \;-\;2y \;+\; z \;=\; 2.\,}

3. “盛水区域”即正方体中满足 $\;2x - 2y + z \le 2$ 的部分

如果把该平面当作“水平水面”，那么水位不可能高于此平面，否则就会从 $E,F,C_1$ 流出。故最大盛水区是

R = \Bigl\{(x,y,z)\;\bigm|\; 0 \le x,y,z \le 2,\; 2x -2y + z \;\le\; 2 \Bigr\}.

我们要计算的就是区域 $R$ 的体积。

3.1 留意到还需 $z \ge 0$

由于正方体只在 $0\le x,y,z \le 2$ 范围内，再加上不超过平面：

2x -2y + z \;\le\; 2,\quad \text{且}\quad z\ge 0.

这是一个“立方体被平面切去一部分”的多面体。

4. 分区域积分法

一个常见思路：直接对不等式

z \;\le\; 2 - 2x + 2y \quad\text{且}\quad z \;\le\; 2 \quad\text{且}\quad z \;\ge\; 0

做三重积分，但要注意 $z$ 上界是 $\min\bigl(2,\;2 -2x +2y\bigr)$ ，并且还要保证这上界非负。为方便，可先在 $xy$ -平面上拆分：

若 $2 -2x +2y \le 2$ ，则 $\;z$ 上界是 $2 -2x +2y$ （只要它 $\ge 0$ ），即
$2 -2x +2y \;\le\;2 \;\;\Longleftrightarrow\;\; -2x +2y \;\le\;0 \;\;\Longleftrightarrow\;\; y \;\le\; x.$
同时要求 $2-2x+2y \ge 0 \implies y\ge x-1$ .
总之在 $y \le x$ 且 $y \ge x-1$ 之处， $z$ 可到 $2-2x+2y$ ，否则为 0。
若 $2 -2x +2y > 2$ ，即 $y>x$ ，那么平面在立方体顶面 $z=2$ 之上，对 $z$ 没有更低的限制，于是此时 $z$ 的上界就是立方体本身的上面 $z=2$ 。

再加上 $0 \le x,y \le 2$ 这些基本范围，便可把平面内的 $xy$ 区域分成两块来做。

4.1 上三角区（ $y > x$ ）对应整列高度 $z = 0 \to 2$

先看“ $y>x$ ”那块，它是单位方形 $[0,2]\times[0,2]$ 中对角线 $y=x$ 上方的三角形。那里的水可从 $z=0$ 盛到 $z=2$ ，整整高度为 2。

这块三角形在 $xy$ -平面上的面积 = 2（因为 $2\times 2$ 的正方形面积是 4，沿对角线分成两个面积为 2 的三角形）。于是这块体积就是

\text{体积}_1 = \bigl(\text{三角形面积}\bigr) \times \bigl(\text{高度 }2\bigr) = 2 \times 2 = 4.

4.2 下（含中）三角区（ $y \le x$ ）再细分

在 $0\le x\le 2$ 里， $y \le x$ 表示位于对角线下方。而我们还需 $y \ge x-1$ 以保证平面处的 $z$ -上界 $2-2x+2y$ 不为负数。这样又可细分：

当 $x \in [0,1]$ 时， $x-1 \le 0$ ，所以“ $y \ge x-1$ ”与“ $y \ge 0$ ”合并后就是 $y\ge 0$ 。再加上 $y \le x$ ，得到 $y$ 在区间 $[0,x]$ 。
当 $x \in [1,2]$ 时， $x-1 \ge 0$ ，要 $y \ge x-1$ 同时又 $y \le x$ 。故此时 $y\in [\,x-1,\,x\,]$ 。

在这两块里， $z$ 从 0 积到 $2 -2x +2y$ 。于是可以分两部分做二维积分。

（a） $x\in[0,1]$ ， $y\in[0,x]$

这里的体积

V_{0\le x \le 1} =\int_{x=0}^{1}\!\int_{y=0}^{x}\! \Bigl(2 -2x +2y\Bigr) \,dy\,dx.

先对 $y$ 积分：

\int_{y=0}^{x} (2 - 2x + 2y)\,dy = \int_{y=0}^{x} (2 - 2x)\,dy \;+\;\int_{y=0}^{x} 2y\,dy = (2-2x)\,x \;+\; \bigl[y^2\bigr]_{0}^{x} = x(2-2x) + x^2 = 2x -2x^2 + x^2 = 2x - x^2.

再对 $x$ 从 0 到 1 积分：

\int_{0}^{1} \bigl(2x - x^2\bigr)\,dx = \left[\,x^2 - \tfrac{x^3}{3}\right]_{0}^{1} = \Bigl(1 - \tfrac{1}{3}\Bigr) - 0 = \tfrac{2}{3}.

故

V_{0\le x\le 1} = \tfrac{2}{3}.

（b） $x\in[1,2]$ ， $y\in[x-1,\,x]$

体积

V_{1\le x \le 2} = \int_{x=1}^{2}\!\int_{y=x-1}^{x} \bigl(2 -2x +2y\bigr)\,dy\,dx.

先对 $y$ 积分：

\int_{y=x-1}^{x}(2 -2x + 2y)\,dy = (2 -2x)\,\bigl[x - (x-1)\bigr] \;+\; \Bigl[y^2\Bigr]_{y=x-1}^{y=x}.

这里 $\,x - (x-1) = 1,$
故前半部分是 $(2-2x)\times 1 = 2 -2x$ 。
后半部分 $\,x^2 - (x-1)^2 = x^2 - (x^2 -2x +1) = 2x -1.$
加一起得

(2 -2x) + (2x -1) = 1.

所以对任意 $x$ ，内层积分都是 1。于是外层对 $x$ 从 1 到 2 积分便是

\int_{1}^{2} 1\,dx = (2-1) = 1.

故

V_{1\le x\le 2} = 1.

小结下三角区体积

下三角区总体积

V_{\text{下}} \;=\; \tfrac{2}{3} \;+\; 1 \;=\; \tfrac{5}{3}.

4.3 总体积

上面我们拆成“上三角区”（ $y>x$ ）与“下三角区”（ $y \le x$ ) 两块来算：

上三角区（ $y>x$ ）体积 $= 4$ ；
下三角区（ $y \le x$ 且满足平面不等式所需）体积 $= \tfrac{5}{3}$ 。

二者相加，即为满足 $2x -2y +z \le 2$ 且 $0\le x,y,z\le 2$ 的全部区域体积：

V_R \;=\; 4 \;+\; \tfrac{5}{3} \;=\; \tfrac{12}{3} \;+\; \tfrac{5}{3} \;=\; \tfrac{17}{3} \;\approx\; 5.6667.

由于正方体的总体积是 8， $\tfrac{17}{3} \approx 5.67$ 小于 8，数值合理。

5. 答案

因此，这个有三个小孔（在 $E,F,C_1$ 处）的正方体容器，最多能盛的水量是：

\boxed{\;\frac{17}{3}\;}\quad (\text{即 }5\frac{2}{3}\text{ }).

Pregunta

Pensamiento

Respuesta

1. 题意解析与坐标设置

1.1 坐标取法

2. 经过三孔的平面方程

3. “盛水区域”即正方体中满足 2x−2y+z≤2\;2x - 2y + z \le 22x−2y+z≤2 的部分

3.1 留意到还需 z≥0z \ge 0z≥0

4. 分区域积分法

4.1 上三角区（y>xy > xy>x）对应整列高度 z=0→2z = 0 \to 2z=0→2

4.2 下（含中）三角区（y≤xy \le xy≤x）再细分

（a）x∈[0,1]x\in[0,1]x∈[0,1]，y∈[0,x]y\in[0,x]y∈[0,x]

（b）x∈[1,2]x\in[1,2]x∈[1,2]，y∈[x−1, x]y\in[x-1,\,x]y∈[x−1,x]

小结下三角区体积

4.3 总体积

5. 答案

Compartir este Q&A

3. “盛水区域”即正方体中满足 $\;2x - 2y + z \le 2$ 的部分

3.1 留意到还需 $z \ge 0$

4.1 上三角区（ $y > x$ ）对应整列高度 $z = 0 \to 2$

4.2 下（含中）三角区（ $y \le x$ ）再细分

（a） $x\in[0,1]$ ， $y\in[0,x]$

（b） $x\in[1,2]$ ， $y\in[x-1,\,x]$