| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the volume of the solid, we need to integrate the area of the square cross-sections with respect to y. First, let's find the limits of integration. The base of the solid is bounded by the parabola x^2 = 4y and the line y = 2. To find the intersection points, we can substitute y = 2 into the equation of the parabola: x^2 = 4(2) x^2 = 8 x = ±√8 x = ±2√2 So, the limits of integration are from y = 0 to y = 2. Now, let's find the area of the square cross-sections. Since each cross-section is a square, the area is equal to the side length squared. The side length of the square is equal to the distance between the two intersection points of the parabola and the line y = y0, where y0 is a value between 0 and 2. The equation of the parabola is x^2 = 4y, so the x-coordinates of the intersection points are x = ±√(4y0). The distance between these two points is 2√(4y0) = 4√y0. So, the area of the square cross-section is (4√y0)^2 = 16y0. Now, we can integrate the area of the square cross-sections with respect to y: V = ∫[0,2] 16y dy = 16 ∫[0,2] y dy = 16 [y^2/2] from 0 to 2 = 16 [(2^2)/2 - 0] = 16 [4/2] = 16 [2] = 32 Therefore, the volume of the solid is 32 cubic units. The correct answer is D. 32. |
D |
| deepseekr1 | The base of the solid is the region bounded by \( x^2 = 4y \) and \( y = 2 \). For a given \( y \), the horizontal width along the x-axis is determined by the parabola, which spans from \( x = -2\sqrt{y} \) to \( x = 2\sqrt{y} \), giving a length of \( 4\sqrt{y} \). Since each cross-section perpendicular to the y-axis is a square, the side length of the square equals this width. Thus, the area of each cross-section is \( (4\sqrt{y})^2 = 16y \). To find the volume, integrate this area from \( y = 0 \) to \( y = 2 \): \[ \text{Volume} = \int_{0}^{2} 16y \, dy = 16 \left[ \frac{y^2}{2} \right]_{0}^{2} = 16 \cdot \frac{4}{2} = 32. \] The volume of the solid is **32**, corresponding to option **D**. **Answer:** D |
D |
| gpto1 | To determine the volume of the solid, we follow these steps: 1. **Find the limits of integration along the \( y \)-axis**: - The region is bounded by \( x^2 = 4y \) (a parabola opening upwards) and \( y = 2 \). - The limits for \( y \) are from \( y = 0 \) to \( y = 2 \). 2. **Express the side length of each square cross-section as a function of \( y \)**: - At any given \( y \), the width of the base is from \( x = -2\sqrt{y} \) to \( x = 2\sqrt{y} \). - The side length \( s \) of the square is \( s = 4\sqrt{y} \). 3. **Calculate the area of each square cross-section**: - Area \( A(y) = s^2 = [4\sqrt{y}]^2 = 16y \). 4. **Integrate the area from \( y = 0 \) to \( y = 2 \) to find the volume**: \[ V = \int_{0}^{2} A(y)\, dy = \int_{0}^{2} 16y\, dy = 16 \left[ \frac{y^2}{2} \right]_0^2 = 16 \left( \frac{2^2}{2} - 0 \right) = 16 \times 2 = 32. \] **Answer:** D |
D |
| GPT-4o | To find the volume of the solid, we first need to understand the shape and area of the cross-sections perpendicular to the y-axis. Given: - The base of the solid is bounded by the parabola \( x^2 = 4y \) and the line \( y = 2 \). - Each cross-section perpendicular to the y-axis is a square. 1. **Find the limits for y:** The parabola \( x^2 = 4y \) intersects the line \( y = 2 \). So, the limits for y are from \( y = 0 \) to \( y = 2 \). 2. **Find the side length of the square cross-section at a given y:** For a given value of y, the parabola \( x^2 = 4y \) gives \( x = \pm \sqrt{4y} \). Therefore, the length of the side of the square at any \( y \) is: \[ 2\sqrt{4y} = 4\sqrt{y} \] 3. **Compute the area of the square cross-section:** The area A(y) of the square cross-section at a given y is: \[ A(y) = \left(4\sqrt{y}\right)^2 = 16y \] 4. **Integrate the area to find the volume:** The volume V of the solid is the integral of the area of the cross-sections from \( y = 0 \) to \( y = 2 \): \[ V = \int_{0}^{2} 16y \, dy \] 5. **Evaluate the integral:** \[ \int_{0}^{2} 16y \, dy = 16 \left[ \frac{y^2}{2} \right]_0^2 = 16 \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = 16 \left( \frac{4}{2} \right) = 16 \cdot 2 = 32 \] So, the volume of the solid is: **Answer: D. 32** |
D |
| llama3.1-70B-Instruct | B |
D |
| llama3.1-405B-Instruct | B |
D |
| DeepSeek-Coder-V2 | B. 16 | D |
| Qwen2-72B-Instruct | B. 16 | D |
| baichuan4 | To solve this problem, we need to find the volume of the solid whose base is the region bounded by the parabola \(x^2 = 4y\) and the line \(y = 2\). Since each plane section perpendicular to the \(y\)-axis is a square, the length of each side of the square at any height \(y\) is given by the difference between the two roots of the equation \(x^2 = 4y\), which is \(2\sqrt{y}\). The volume of the solid can be found by integrating the area of the square cross-sections over the height of the solid. The area of each square is \((2\sqrt{y})^2 = 4y\). The height of the solid goes from \(y = 0\) (where the parabola intersects the \(y\)-axis) to \(y = 2\) (the line defining the upper boundary of the solid). Thus, the volume \(V\) is given by: \[ V = \int_{0}^{2} 4y \, dy \] Integrating gives us: \[ V = 4 \left[ \frac{1}{2}y^2 \right]_{0}^{2} \] \[ V = 2 \left[ 2^2 - 0^2 \right] \] \[ V = 2 \times 4 \] \[ V = 8 \] So, the correct answer is: A. 8 |
D |
| doubao | B The region bounded by x^2 = 4y and y = 2 gives us the limits for y as 0 to 2. For each y, the side of the square cross-section perpendicular to the y-axis is 2x, where x = ±2√y. So the area of each cross-section is (2*2√y)^2 = 16y. Integrating this from 0 to 2 gives the volume as 16. |
D |
| minimax | To find the volume of the solid, we need to determine the area of the base and then multiply it by the height of the solid. The base of the solid is bounded by the parabola \( x^2 = 4y \) and the line \( y = 2 \). To find the points of intersection, we substitute \( y = 2 \) into the parabola's equation: \( x^2 = 4(2) \) \( x^2 = 8 \) \( x = \pm\sqrt{8} \) \( x = \pm2\sqrt{2} \) So the base is a parabolic region bounded by the line \( y = 2 \) and the parabola \( x^2 = 4y \), with the points of intersection being \( (-2\sqrt{2}, 2) \) and \( (2\sqrt{2}, 2) \). Since each plane section perpendicular to the y-axis is a square, the side length of the square at any height \( y \) is the same as the width of the base at that height. The width of the base at any height \( y \) is the distance between the two points where the plane intersects the parabola, which is \( 2x \) (since the plane intersects the parabola at \( x \) and \( -x \)). The width of the base at height \( y \) is \( 2\sqrt{4y} = 4\sqrt{y} \). At the top of the base, where \( y = 2 \), the width is \( 4\sqrt{2} \), which is the side length of the square at the top of the solid. The area of the base is the area of the square at the top, which is \( (4\sqrt{2})^2 = 16 \cdot 2 = 32 \). The height of the solid is the distance from \( y = 0 \) to \( y = 2 \), which is 2 units. Therefore, the volume of the solid is the area of the base times the height: \( \text{Volume} = \text{Area of base} \times \text{Height} \) \( \text{Volume} = 32 \times 2 \) \( \text{Volume} = 64 \) However, none of the options given (A, B, C, D) match the calculated volume of 64. It's possible there may be a mistake in the question or the options provided. If the question and options are correct as given, then the correct answer is not listed. |
D |
| kimi | To find the volume of the solid, we first need to determine the dimensions of the solid. The base of the solid is bounded by the parabola \(x^2 = 4y\) and the line \(y = 2\). To find the points where the parabola intersects the line \(y = 2\), we substitute \(y = 2\) into the equation of the parabola: \[x^2 = 4(2)\] \[x^2 = 8\] \[x = \pm\sqrt{8}\] \[x = \pm2\sqrt{2}\] So the base of the solid is a rectangle with a width of \(4\sqrt{2}\) (from \(-2\sqrt{2}\) to \(2\sqrt{2}\)) and a height of 2 (from \(y = 0\) to \(y = 2\)). Since each plane section perpendicular to the y-axis is a square, the height of the solid is the same as the width of the base, which is \(4\sqrt{2}\). The volume \(V\) of the solid can be calculated using the formula for the volume of a rectangular prism: \[V = \text{width} \times \text{height} \times \text{length}\] In this case, the width and length are the same (the width of the base), and the height is the vertical distance from \(y = 0\) to \(y = 2\): \[V = (4\sqrt{2}) \times 2 \times (4\sqrt{2})\] \[V = 8 \times 4\sqrt{2}\] \[V = 32\] So the volume of the solid is 32. Answer: D. 32 |
D |
| wenxinyiyan | B Explanation: 1. **Determine the Base Area**: * The base is bounded by the parabola $x^2 = 4y$ and the line $y = 2$. * To find the intersection points, solve the system of equations: $\begin{cases} x^2 = 4y \\ y = 2 \end{cases}$ * Substituting $y = 2$ into the first equation gives $x^2 = 8$ or $x = \pm 2\sqrt{2}$. * The base area is thus the area between the parabola and the line, which can be calculated by integrating the width of the region (which is $4\sqrt{y}$ due to symmetry) from $y = 0$ to $y = 2$: $ \text{Base Area} = \int_{0}^{2} 4\sqrt{y} \, dy$ * Using the power rule for integration, we get: $ \text{Base Area} = 4 \left[ \frac{2}{3}y^{\frac{3}{2}} \right]_{0}^{2} = \frac{16}{3} \left[ 2^{\frac{3}{2}} - 0 \right] = \frac{32}{3}$ * Note: This is actually the area under the parabola and above the x-axis, up to $y = 2$. Since each cross-section is a square, the "actual" base area for the solid will be considered as the width of the square, not its area. 2. **Calculate the Volume**: * Since each plane section perpendicular to the y-axis is a square, the volume can be found by integrating the area of these squares from $y = 0$ to $y = 2$. * The side length of each square is $4\sqrt{y}$ (from the parabola's symmetry), so the area of each square is $(4\sqrt{y})^2 = 16y$. * The volume is then: $ \text{Volume} = \int_{0}^{2} 16y \, dy$ * Integrating, we get: $ \text{Volume} = 16 \left[ \frac{1}{2}y^{2} \right]_{0}^{2} = 8 \left[ 2^{2} - 0 \right] = 32$ * However, since we initially calculated the "base area" as the area under the parabola (which is actually the width of the squares), we need to adjust our final answer. Specifically, we divide by the "height" of the base region (from $y = 0$ to $y = 2$), which is 2: $ \text{Adjusted Volume} = \frac{32}{2} = 16$ |
D |
| zhipu | To determine the volume of the solid, we first need to understand the base and the shape of the cross-sections. The base of the solid is the region bounded by \(x^2 = 4y\) and the line \(y = 2\). 1. **Find the intersection points**: - The parabola \(x^2 = 4y\) intersects the line \(y = 2\) when \(x^2 = 4 \cdot 2 = 8\). Thus, \(x = \pm \sqrt{8} = \pm 2\sqrt{2}\). 2. **Determine the side length of the square cross-section**: - For a given \(y\), the equation \(x^2 = 4y\) implies \(x = \pm 2\sqrt{y}\). - The side length of the square at height \(y\) is the distance between \(x = -2\sqrt{y}\) and \(x = 2\sqrt{y}\), which is \(4\sqrt{y}\). 3. **Calculate the area of the square cross-section**: - The area of the square at height \(y\) is \((4\sqrt{y})^2 = 16y\). 4. **Integrate to find the volume**: - The volume \(V\) of the solid is the integral of the area of the cross-sections from \(y = 0\) to \(y = 2\): \[ V = \int_{0}^{2} 16y \, dy \] - Compute the integral: \[ V = 16 \int_{0}^{2} y \, dy = 16 \left[ \frac{y^2}{2} \right]_{0}^{2} = 16 \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = 16 \left( \frac{4}{2} \right) = 16 \cdot 2 = 32 \] Therefore, the volume of the solid is \(32\). Answer: D |
D |
| gpto1 | To determine the volume of the solid, we follow these steps: 1. **Find the limits of integration along the \( y \)-axis**: - The region is bounded by \( x^2 = 4y \) (a parabola opening upwards) and \( y = 2 \). - The limits for \( y \) are from \( y = 0 \) to \( y = 2 \). 2. **Express the side length of each square cross-section as a function of \( y \)**: - At any given \( y \), the width of the base is from \( x = -2\sqrt{y} \) to \( x = 2\sqrt{y} \). - The side length \( s \) of the square is \( s = 4\sqrt{y} \). 3. **Calculate the area of each square cross-section**: - Area \( A(y) = s^2 = [4\sqrt{y}]^2 = 16y \). 4. **Integrate the area from \( y = 0 \) to \( y = 2 \) to find the volume**: \[ V = \int_{0}^{2} A(y)\, dy = \int_{0}^{2} 16y\, dy = 16 \left[ \frac{y^2}{2} \right]_0^2 = 16 \left( \frac{2^2}{2} - 0 \right) = 16 \times 2 = 32. \] **Answer:** D |
D |
| deepseekv3 | To solve this problem, we need to calculate the volume of the solid whose base is bounded by the parabola \( x^2 = 4y \) and the line \( y = 2 \), and whose cross-sections perpendicular to the \( y \)-axis are squares. ### Step 1: Understand the base region The parabola \( x^2 = 4y \) opens upwards with its vertex at the origin. The line \( y = 2 \) intersects the parabola at the points where \( x^2 = 4 \times 2 = 8 \), so \( x = \pm 2\sqrt{2} \). Thus, the base of the solid lies between \( y = 0 \) (the vertex of the parabola) and \( y = 2 \), and for each \( y \) in this interval, the width of the base is the distance between the two points on the parabola at that \( y \)-value. ### Step 2: Determine the side length of the square cross-section For a given \( y \) in the interval \( [0, 2] \), the width of the base is the distance between \( x = -2\sqrt{y} \) and \( x = 2\sqrt{y} \), which is \( 4\sqrt{y} \). Since the cross-section perpendicular to the \( y \)-axis is a square, the side length of the square is also \( 4\sqrt{y} \). ### Step 3: Calculate the area of the square cross-section The area \( A(y) \) of the square cross-section at height \( y \) is: \[ A(y) = (4\sqrt{y})^2 = 16y \] ### Step 4: Integrate to find the volume The volume \( V \) of the solid is the integral of the cross-sectional area from \( y = 0 \) to \( y = 2 \): \[ V = \int_{0}^{2} 16y \, dy \] Compute the integral: \[ V = 16 \left[ \frac{y^2}{2} \right]_{0}^{2} = 16 \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = 16 \times 2 = 32 \] ### Conclusion The volume of the solid is \( 32 \), so the correct answer is: D. 32 |
D |