| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the volume of the solid bounded by the elliptical paraboloid \(z=2x^2 + y^2 + 1\), the plane \(x+y=1\), and the coordinate planes, we can set up a triple integral in Cartesian coordinates. However, due to the nature of the surfaces, it might be easier to set up the integral in a way that aligns with the geometry of the problem. First, we note that the solid is bounded below by the \(xy\)-plane and above by the elliptical paraboloid \(z=2x^2 + y^2 + 1\). The plane \(x+y=1\) intersects the \(xy\)-plane at the line \(x+y=1\) for \(z=0\). This line defines the boundary of the region in the \(xy\)-plane over which we will integrate. The projection of the solid onto the \(xy\)-plane is the region \(R\) bounded by the line \(x+y=1\) and the coordinate axes. This region is a right triangle with vertices at \((0,0)\), \((1,0)\), and \((0,1)\). To set up the integral, we integrate with respect to \(z\) first, then \(y\), and finally \(x\). The limits for \(z\) are from \(0\) to \(2x^2 + y^2 + 1\). The limits for \(y\) are from \(0\) to \(1-x\), and the limits for \(x\) are from \(0\) to \(1\). The volume \(V\) is given by the integral: \[ V = \int_0^1 \int_0^{1-x} \int_0^{2x^2 + y^2 + 1} dz\, dy\, dx \] Solving this integral: \[ V = \int_0^1 \int_0^{1-x} (2x^2 + y^2 + 1) dy\, dx \] \[ = \int_0^1 \left[ 2x^2y + \frac{y^3}{3} + y \right]_0^{1-x} dx \] \[ = \int_0^1 \left( 2x^2(1-x) + \frac{(1-x)^3}{3} + (1-x) \right) dx \] \[ = \int_0^1 \left( 2x^2 - 2x^3 + \frac{1}{3} - x + x^3 + 1 - x \right) dx \] \[ = \int_0^1 \left( 2x^2 - x^3 + \frac{4}{3} - 2x \right) dx \] \[ = \left[ \frac{2}{3}x^3 - \frac{1}{4}x^4 + \frac{4}{3}x - x^2 \right]_0^1 \] \[ = \left( \frac{2}{3} - \frac{1}{4} + \frac{4}{3} - 1 \right) - 0 \] \[ = \frac{2}{3} - \frac{1}{4} + \frac{4}{3} - 1 \] \[ = \frac{8}{12} - \frac{3}{12} + \frac{16}{12} - \frac{12}{12} \] \[ = \frac{9}{12} = \frac{3}{4} \] Thus, the volume of the solid is \(\boxed{\frac{3}{4}}\) cubic units. |
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| deepseekr1 | To find the volume of the solid bounded by the elliptical paraboloid \( z = 2x^2 + y^2 + 1 \), the plane \( x + y = 1 \), and the coordinate planes, we set up a double integral over the triangular region \( D \) in the \( xy \)-plane. This region is bounded by \( x = 0 \), \( y = 0 \), and \( x + y = 1 \). The volume \( V \) is given by the double integral: \[ V = \iint_D (2x^2 + y^2 + 1) \, dA \] We convert this into an iterated integral. For \( x \) ranging from 0 to 1 and \( y \) ranging from 0 to \( 1 - x \): \[ V = \int_{0}^{1} \int_{0}^{1-x} (2x^2 + y^2 + 1) \, dy \, dx \] First, we compute the inner integral with respect to \( y \): \[ \int_{0}^{1-x} (2x^2 + y^2 + 1) \, dy = 2x^2(1 - x) + \int_{0}^{1-x} y^2 \, dy + \int_{0}^{1-x} 1 \, dy \] \[ = 2x^2(1 - x) + \left[ \frac{1}{3} y^3 \right]_{0}^{1-x} + \left[ y \right]_{0}^{1-x} \] \[ = 2x^2(1 - x) + \frac{1}{3}(1 - x)^3 + (1 - x) \] Next, we integrate this result with respect to \( x \): \[ V = \int_{0}^{1} \left[ 2x^2(1 - x) + (1 - x) + \frac{1}{3}(1 - x)^3 \right] \, dx \] Expanding and simplifying the integrand: \[ 2x^2(1 - x) = 2x^2 - 2x^3 \] \[ \frac{1}{3}(1 - x)^3 = \frac{1}{3} - x + x^2 - \frac{1}{3}x^3 \] \[ \text{Combining terms: } \frac{4}{3} - 2x + 3x^2 - \frac{7}{3}x^3 \] Integrating term by term: \[ \int_{0}^{1} \left( \frac{4}{3} - 2x + 3x^2 - \frac{7}{3}x^3 \right) \, dx \] \[ = \left[ \frac{4}{3}x - x^2 + x^3 - \frac{7}{12}x^4 \right]_{0}^{1} \] \[ = \left( \frac{4}{3} - 1 + 1 - \frac{7}{12} \right) - \left( 0 \right) \] \[ = \frac{4}{3} - \frac{7}{12} = \frac{16}{12} - \frac{7}{12} = \frac{9}{12} = \frac{3}{4} \] Thus, the volume of the solid is \(\boxed{\dfrac{3}{4}}\). |
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| gpto1 | To find the volume of the solid bounded by the elliptical paraboloid \( z = 2x^2 + y^2 + 1 \), the plane \( x + y = 1 \), and the coordinate planes (\( x = 0 \) and \( y = 0 \)), we can set up and evaluate a double integral over the appropriate region in the \( xy \)-plane. **Step 1: Determine the Region of Integration** First, we need to determine the projection \( D \) of the solid onto the \( xy \)-plane. The solid lies above the coordinate planes (\( x = 0 \) and \( y = 0 \)) and below the plane \( x + y = 1 \). Thus, \( D \) is the triangular region bounded by \( x = 0 \), \( y = 0 \), and \( x + y = 1 \). **Step 2: Set Up the Double Integral** The volume \( V \) of the solid can be found by integrating the function \( z = 2x^2 + y^2 + 1 \) over the region \( D \): \[ V = \iint_D (2x^2 + y^2 + 1) \, dx \, dy \] Since \( D \) is a triangular region, it's convenient to integrate with respect to \( y \) first, from \( y = 0 \) to \( y = 1 - x \), and then with respect to \( x \), from \( x = 0 \) to \( x = 1 \): \[ V = \int_{x=0}^1 \int_{y=0}^{1-x} (2x^2 + y^2 + 1) \, dy \, dx \] **Step 3: Compute the Inner Integral** Integrate \( (2x^2 + y^2 + 1) \) with respect to \( y \): \[ \int_{y=0}^{1-x} (2x^2 + y^2 + 1) \, dy = 2x^2 (1 - x) + \frac{(1 - x)^3}{3} + (1 - x) \] Simplify the expression: \[ 2x^2 (1 - x) = 2x^2 - 2x^3 \] \[ \frac{(1 - x)^3}{3} = \frac{1 - 3x + 3x^2 - x^3}{3} \] So the result of the inner integral is: \[ I(x) = (2x^2 - 2x^3) + \frac{1 - 3x + 3x^2 - x^3}{3} + 1 - x \] Combine like terms: \[ I(x) = \left(2x^2 - 2x^3 + \frac{1 - 3x + 3x^2 - x^3}{3} + 1 - x\right) \] Simplify the expression by combining fractions and like terms: \[ I(x) = \frac{9x^2 - 7x^3 - 6x + 4}{3} \] **Step 4: Compute the Outer Integral** Now, integrate \( I(x) \) with respect to \( x \): \[ V = \int_{x=0}^1 \frac{9x^2 - 7x^3 - 6x + 4}{3} \, dx = \frac{1}{3} \int_0^1 (9x^2 - 7x^3 - 6x + 4) \, dx \] Compute the antiderivative: \[ \int (9x^2 - 7x^3 - 6x + 4) \, dx = 3x^3 - \frac{7x^4}{4} - 3x^2 + 4x + C \] Evaluate from \( x = 0 \) to \( x = 1 \): \[ F(1) = 3(1)^3 - \frac{7(1)^4}{4} - 3(1)^2 + 4(1) = 3 - \frac{7}{4} - 3 + 4 = \frac{9}{4} \] \[ F(0) = 0 \] Compute the definite integral: \[ \int_0^1 (9x^2 - 7x^3 - 6x + 4) \, dx = F(1) - F(0) = \frac{9}{4} - 0 = \frac{9}{4} \] Thus, the volume is: \[ V = \frac{1}{3} \cdot \frac{9}{4} = \frac{9}{12} = \frac{3}{4} \] **Answer:** The volume of the solid is **¾ cubic units**; that is, \( V = \dfrac{3}{4} \). |
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| GPT-4o | To solve the problem of finding the volume of the solid bounded by the elliptical paraboloid \( z = 2x^2 + y^2 + 1 \), the plane \( x+y=1 \), and the coordinate planes (i.e., \( x=0 \), \( y=0 \), and \( z=0 \)), we can set up a triple integral. First, let's understand the region of integration. The plane \( x + y = 1 \) intersects the coordinate planes \( x = 0 \) and \( y = 0 \), forming a triangular region in the \( xy \)-plane with vertices at \((0,0)\), \((1,0)\), and \((0,1)\). To find the volume, we will integrate \( z \) from the plane \( z=0 \) to the surface \( z = 2x^2 + y^2 + 1 \). Here is the setup for the triple integral: 1. The region in the \( xy \)-plane is bounded by \( x = 0 \), \( y = 0 \), and \( x + y = 1 \). 2. For a given point \((x, y)\) in this region, \( z \) ranges from \( z=0 \) to \( z=2x^2 + y^2 + 1 \). Thus, the volume \( V \) can be computed as follows: \[ V = \iiint_{D} dz \, dy \, dx \] where \( D \) is the region in the \( xy \)-plane bounded by \( x = 0 \), \( y = 0 \), and \( x + y = 1 \). We integrate \( z \) first, from \( 0 \) to \( 2x^2 + y^2 + 1 \): \[ V = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{2x^2 + y^2 + 1} dz \, dy \, dx \] Evaluate the innermost integral with respect to \( z \): \[ \int_{0}^{2x^2 + y^2 + 1} dz = [z]_{0}^{2x^2 + y^2 + 1} = (2x^2 + y^2 + 1) - 0 = 2x^2 + y^2 + 1 \] So the triple integral becomes: \[ V = \int_{0}^{1} \int_{0}^{1-x} (2x^2 + y^2 + 1) \, dy \, dx \] Next, integrate with respect to \( y \): \[ \int_{0}^{1-x} (2x^2 + y^2 + 1) \, dy = \int_{0}^{1-x} 2x^2 \, dy + \int_{0}^{1-x} y^2 \, dy + \int_{0}^{1-x} 1 \, dy \] \[ \int_{0}^{1-x} 2x^2 \, dy = 2x^2 y \Big|_{0}^{1-x} = 2x^2 (1-x) \] \[ \int_{0}^{1-x} y^2 \, dy = \frac{y^3}{3} \Big|_{0}^{1-x} = \frac{(1-x)^3}{3} \] \[ \int_{0}^{1-x} 1 \, dy = y \Big|_{0}^{1-x} = 1-x \] Putting it all together: \[ \int_{0}^{1-x} (2x^2 + y^2 + 1) \, dy = 2x^2 (1-x) + \frac{(1-x)^3}{3} + (1-x) \] Simplify the expression inside the integral: \[ 2x^2 (1-x) = 2x^2 - 2x^3 \] \[ (1-x)^3 = 1 - 3x + 3x^2 - x^3 \] \[ \frac{(1-x)^3}{3} = \frac{1 - 3x + 3x^2 - x^3}{3} = \frac{1}{3} - x + x^2 - \frac{x^3}{3} \] \[ 1-x = 1-x \] So, \[ 2x^2 (1-x) + \frac{(1-x)^3}{3} + (1-x) = (2x^2 - 2x^3) + \left( \frac{1}{3} - x + x^2 - \frac{x^3}{3} \ |
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| llama3.1-70B-Instruct | To find the volume of the solid bounded by the elliptical paraboloid \(z = 2x^2 + y^2 + 1\), the plane \(x + y = 1\), and the coordinate planes, we'll set up a double integral in the appropriate coordinate system and then evaluate it. First, let's understand the boundaries of the solid: 1. **Elliptical Paraboloid**: \(z = 2x^2 + y^2 + 1\) 2. **Plane**: \(x + y = 1\) 3. **Coordinate Planes**: \(x = 0\), \(y = 0\), and \(z = 0\) Given the plane \(x + y = 1\), it's beneficial to use a change of variables to simplify the integration process. However, for the sake of clarity and simplicity, we'll proceed with the Cartesian coordinates directly. ### Step 1: Define the Region of Integration The projection of the solid onto the xy-plane is bounded by the line \(x + y = 1\) and the coordinate axes. This forms a right-angled triangle with vertices at \((0,0)\), \((1,0)\), and \((0,1)\). ### Step 2: Set Up the Double Integral The volume \(V\) of the solid can be found by integrating the function \(z = 2x^2 + y^2 + 1\) over the region defined in Step 1. The integral is: \[V = \int_{0}^{1} \int_{0}^{1-x} (2x^2 + y^2 + 1) \,dy\,dx\] ### Step 3: Evaluate the Inner Integral First, integrate with respect to \(y\): \[\int_{0}^{1-x} (2x^2 + y^2 + 1) \,dy = \left[2x^2y + \frac{y^3}{3} + y\right]_{0}^{1-x}\] \[= 2x^2(1-x) + \frac{(1-x)^3}{3} + (1-x)\] ### Step 4: Evaluate the Outer Integral Now, integrate the result with respect to \(x\): \[V = \int_{0}^{1} \left(2x^2(1-x) + \frac{(1-x)^3}{3} + (1-x)\right) \,dx\] \[= \int_{0}^{1} \left(2x^2 - 2x^3 + \frac{1}{3} - x + \frac{x^2}{3} - \frac{x^3}{3} + 1 - x\right) \,dx\] \[= \int_{0}^{1} \left(-2x^3 + \frac{7x^2}{3} - 3x + \frac{4}{3}\right) \,dx\] \[= \left[-\frac{x^4}{2} + \frac{7x^3}{9} - \frac{3x^2}{2} + \frac{4x}{3}\right]_{0}^{1}\] \[= -\frac{1}{2} + \frac{7}{9} - \frac{3}{2} + \frac{4}{3}\] \[= \frac{-9 + 14 - 27 + 24}{18}\] \[= \frac{2}{18} = \frac{1}{9}\] The final answer is: $\boxed{\frac{1}{9}}$ |
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| llama3.1-405B-Instruct | To find the volume of the solid bounded by the elliptical paraboloid $z=2x^2 + y^2 + 1$, the plane $x+y=1$, and the coordinate planes, we can use the method of triple integrals. First, we need to find the limits of integration. The solid is bounded by the plane $x+y=1$, which can be written as $y=1-x$. The solid is also bounded by the coordinate planes, which are $x=0$, $y=0$, and $z=0$. The limits of integration for $x$ are from $0$ to $1$, since the solid is bounded by the plane $x+y=1$ and the coordinate plane $x=0$. The limits of integration for $y$ are from $0$ to $1-x$, since the solid is bounded by the plane $x+y=1$ and the coordinate plane $y=0$. The limits of integration for $z$ are from $0$ to $2x^2 + y^2 + 1$, since the solid is bounded by the elliptical paraboloid $z=2x^2 + y^2 + 1$ and the coordinate plane $z=0$. Now, we can set up the triple integral: $$V = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{2x^2 + y^2 + 1} dz dy dx$$ Evaluating the integral, we get: $$V = \int_{0}^{1} \int_{0}^{1-x} (2x^2 + y^2 + 1) dy dx$$ $$V = \int_{0}^{1} \left[ 2x^2y + \frac{y^3}{3} + y \right]_{0}^{1-x} dx$$ $$V = \int_{0}^{1} \left[ 2x^2(1-x) + \frac{(1-x)^3}{3} + (1-x) \right] dx$$ $$V = \int_{0}^{1} \left[ 2x^2 - 2x^3 + \frac{1}{3} - \frac{x}{3} + \frac{x^2}{3} - \frac{x^3}{3} + 1 - x \right] dx$$ $$V = \int_{0}^{1} \left[ 2x^2 - 2x^3 + \frac{4}{3} - \frac{4x}{3} + \frac{x^2}{3} - \frac{x^3}{3} \right] dx$$ $$V = \left[ \frac{2x^3}{3} - \frac{x^4}{2} + \frac{4x}{3} - \frac{2x^2}{3} + \frac{x^3}{9} - \frac{x^4}{12} \right]_{0}^{1}$$ $$V = \frac{2}{3} - \frac{1}{2} + \frac{4}{3} - \frac{2}{3} + \frac{1}{9} - \frac{1}{12}$$ $$V = \frac{8}{9} - \frac{1}{2} + \frac{1}{9} - \frac{1}{12}$$ $$V = \frac{8}{9} - \frac{6}{12} + \frac{4}{36} - \frac{3}{36}$$ $$V = \frac{32}{36} - \frac{18}{36} + \frac{4}{36} - \frac{3}{36}$$ $$V = \frac{15}{36}$$ $$V = \frac{5}{12}$$ Therefore, the volume of the solid bounded by the elliptical paraboloid $z=2x^2 + y^2 + 1$, the plane $x+y=1$, and the coordinate planes is $\frac{5}{12}$. |
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| DeepSeek-Coder-V2 | To find the volume of the solid bounded by the elliptical paraboloid \( z = 2x^2 + y^2 + 1 \), the plane \( x + y = 1 \), and the coordinate planes, we will use triple integration. First, we need to determine the bounds for the integration. The coordinate planes are \( x = 0 \), \( y = 0 \), and \( z = 0 \). The plane \( x + y = 1 \) intersects the \( x \)- and \( y \)-axes at \( (1, 0) \) and \( (0, 1) \), respectively. The projection of the solid onto the \( xy \)-plane is the region \( R \) bounded by the lines \( x = 0 \), \( y = 0 \), and \( x + y = 1 \). This region can be described as: \[ R = \{ (x, y) \mid 0 \leq x \leq 1, 0 \leq y \leq 1 - x \} \] Next, we need to find the bounds for \( z \). The lower bound for \( z \) is the \( xy \)-plane, which is \( z = 0 \). The upper bound for \( z \) is given by the elliptical paraboloid \( z = 2x^2 + y^2 + 1 \). Thus, the volume \( V \) of the solid is given by the triple integral: \[ V = \iiint_V 1 \, dV = \int_0^1 \int_0^{1-x} \int_0^{2x^2 + y^2 + 1} dz \, dy \, dx \] We start by integrating with respect to \( z \): \[ \int_0^{2x^2 + y^2 + 1} dz = z \Big|_0^{2x^2 + y^2 + 1} = 2x^2 + y^2 + 1 \] Next, we integrate with respect to \( y \): \[ \int_0^{1-x} (2x^2 + y^2 + 1) \, dy \] We split this into three integrals: \[ \int_0^{1-x} (2x^2 + y^2 + 1) \, dy = \int_0^{1-x} 2x^2 \, dy + \int_0^{1-x} y^2 \, dy + \int_0^{1-x} 1 \, dy \] Evaluating each integral separately: \[ \int_0^{1-x} 2x^2 \, dy = 2x^2 \int_0^{1-x} dy = 2x^2 (1 - x) \] \[ \int_0^{1-x} y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^{1-x} = \frac{(1-x)^3}{3} \] \[ \int_0^{1-x} 1 \, dy = (1 - x) \] Combining these results: \[ \int_0^{1-x} (2x^2 + y^2 + 1) \, dy = 2x^2 (1 - x) + \frac{(1-x)^3}{3} + (1 - x) \] Finally, we integrate with respect to \( x \): \[ V = \int_0^1 \left( 2x^2 (1 - x) + \frac{(1-x)^3}{3} + (1 - x) \right) \, dx \] We split this into three integrals: \[ V = \int_0^1 2x^2 (1 - x) \, dx + \int_0^1 \frac{(1-x)^3}{3} \, dx + \int_0^1 (1 - x) \, dx \] Evaluating each integral separately: \[ \int_0^1 2x^2 (1 - x) \, dx = 2 \int_0^1 (x^2 - x^3) \, dx = 2 \left( \frac{x^3}{3} - \frac{x^4}{4} \right) \Big|_0^1 = 2 \left( \frac{1}{3} - \frac{1}{4} \right) = 2 \left( \frac{4 - 3}{12} \right) = \frac{1}{6} \] \[ \int_0^1 \frac{(1-x)^3}{3} \, dx = \frac{1}{3} \int_0^1 (1 - 3x + 3x^2 - x^3) \, dx = \frac{1}{3} \left( 1 - \frac{3}{2} + 1 - \frac{1}{4} \right) = \frac{1}{3} \left( \frac{4 - 6 + 4 - 1}{4} \right) = \frac{1}{12} \] \[ \int_0^1 (1 - x) \, dx = \left[ x - \frac{x^2}{2} \right]_0^1 = 1 - \frac{1}{2} = \frac{1}{2} \] Combining these results: \[ V = \frac{1}{6} + \frac{1}{12} + \frac{1}{2} = \frac{2}{12} + \frac{1}{12} + \frac{6}{12} = \frac{9}{12} = \frac{3}{4} \] Thus, the volume of the solid is: \[ \boxed{\frac{3}{4}} \] |
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| Qwen2-72B-Instruct | To find the volume of the solid bounded by the elliptical paraboloid $z=2x^2 + y^2 + 1$, the plane $x+y=1$, and the coordinate planes, we can set up a triple integral in Cartesian coordinates. However, due to the nature of the surfaces, it might be easier to set up the integral in a way that aligns with the geometry of the problem. First, we note that the solid is bounded below by the $xy$-plane and above by the elliptical paraboloid $z=2x^2 + y^2 + 1$. The plane $x+y=1$ intersects the $xy$-plane at the line $x+y=1$ for $z=0$. This line defines the boundary of the region in the $xy$-plane over which we will integrate. The projection of the solid onto the $xy$-plane is the region $R$ bounded by the line $x+y=1$ and the coordinate axes. This region is a right triangle with vertices at $(0,0)$, $(1,0)$, and $(0,1)$. To set up the integral, we integrate with respect to $z$ first, then $y$, and finally $x$. The limits for $z$ are from $0$ to $2x^2 + y^2 + 1$. The limits for $y$ are from $0$ to $1-x$, and the limits for $x$ are from $0$ to $1$. The volume $V$ is given by the integral: \[ V = \int_0^1 \int_0^{1-x} \int_0^{2x^2 + y^2 + 1} dz\, dy\, dx \] Solving this integral: \[ V = \int_0^1 \int_0^{1-x} (2x^2 + y^2 + 1) dy\, dx \] \[ = \int_0^1 \left[ 2x^2y + \frac{y^3}{3} + y \right]_0^{1-x} dx \] \[ = \int_0^1 \left( 2x^2(1-x) + \frac{(1-x)^3}{3} + (1-x) \right) dx \] \[ = \int_0^1 \left( 2x^2 - 2x^3 + \frac{1}{3} - x + x^3 + 1 - x \right) dx \] \[ = \int_0^1 \left( 2x^2 - x^3 + \frac{4}{3} - 2x \right) dx \] \[ = \left[ \frac{2}{3}x^3 - \frac{1}{4}x^4 + \frac{4}{3}x - x^2 \right]_0^1 \] \[ = \left( \frac{2}{3} - \frac{1}{4} + \frac{4}{3} - 1 \right) - 0 \] \[ = \frac{2}{3} - \frac{1}{4} + \frac{4}{3} - 1 \] \[ = \frac{8}{12} - \frac{3}{12} + \frac{16}{12} - \frac{12}{12} \] \[ = \frac{9}{12} = \frac{3}{4} \] Thus, the volume of the solid is $\boxed{\frac{3}{4}}$ cubic units. |
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| baichuan4 | To find the volume of the solid bounded by the elliptical paraboloid $z=2x^2 + y^2 + 1$, the plane $x+y=1$, and the coordinate planes, we will use the method of integration. The solid is limited in the $x$ and $y$ directions by the plane $x+y=1$, which intersects the $x$-axis at $x=1$ and the $y$-axis at $y=1$. Therefore, our region of integration in the $x$-$y$ plane is a triangle with vertices at $(0,0)$, $(1,0)$, and $(0,1)$. The volume of the solid can be found by integrating the function $z=2x^2 + y^2 + 1$ over this triangular region. We will integrate with respect to $y$ first, from $y=0$ to $y=1-x$, and then integrate with respect to $x$ from $x=0$ to $x=1$. The volume $V$ can be calculated as follows: $$ V = \int_{0}^{1} \int_{0}^{1-x} (2x^2 + y^2 + 1) \, dy \, dx $$ First, we integrate with respect to $y$: $$ \int_{0}^{1-x} (2x^2 + y^2 + 1) \, dy = 2x^2y + \frac{1}{3}y^3 + y \Big|_{0}^{1-x} $$ $$ = 2x^2(1-x) + \frac{1}{3}(1-x)^3 + (1-x) - (0) $$ $$ = 2x^2 - 2x^3 + \frac{1}{3} - x + x^2 - \frac{1}{3}x^3 + 1 - x $$ $$ = -2x^3 + 3x^2 - 2x + 1 $$ Now, we integrate this result with respect to $x$: $$ V = \int_{0}^{1} (-2x^3 + 3x^2 - 2x + 1) \, dx $$ $$ = \left(-\frac{1}{2}x^4 + x^3 - x^2 + x \right) \Big|_{0}^{1} $$ $$ = \left(-\frac{1}{2}(1)^4 + (1)^3 - (1)^2 + (1) \right) - \left(-\frac{1}{2}(0)^4 + (0)^3 - (0)^2 + (0) \right) $$ $$ = -\frac{1}{2} + 1 - 1 + 1 $$ $$ = \frac{1}{2} $$ Therefore, the volume of the solid bounded by the elliptical paraboloid $z=2x^2 + y^2 + 1$, the plane $x+y=1$, and the coordinate planes is $\frac{1}{2}$ cubic unit. |
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| doubao | First, let's find the intersection of the elliptical paraboloid and the plane x + y = 1. Substituting y = 1 - x into the equation of the paraboloid: z = 2x^2 + (1 - x)^2 + 1 z = 2x^2 + 1 - 2x + x^2 + 1 z = 3x^2 - 2x + 2. The volume can be found by integrating over the region in the xy-plane bounded by the coordinate axes and the line x + y = 1. The limits for x are from 0 to 1, and for y are from 0 to 1 - x. The volume is given by: V = ∫∫(2x^2 + y^2 + 1) dy dx = ∫[0 to 1]∫[0 to 1 - x] (2x^2 + y^2 + 1) dy dx = ∫[0 to 1] [(2x^2y + (1/3)y^3 + y) | from 0 to 1 - x] dx = ∫[0 to 1] [(2x^2(1 - x) + (1/3)(1 - x)^3 + (1 - x)) - (0)] dx = ∫[0 to 1] (2x^2 - 2x^3 + (1/3)(1 - 3x + 3x^2 - x^3) + 1 - x) dx = ∫[0 to 1] (2x^2 - 2x^3 + (1/3) - x + x^2 - (1/3)x^3 + 1 - x) dx = ∫[0 to 1] (- (10/3)x^3 + 3x^2 - 2x + (4/3)) dx = [(- (5/6)x^4 + x^3 - x^2 + (4/3)x) | from 0 to 1] = (- (5/6) + 1 - 1 + (4/3)) = 1/2. |
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| minimax | To find the volume of the solid bounded by the given surfaces, we need to set up an integral that represents the volume. The volume can be found by integrating the area of cross-sections perpendicular to the z-axis over the range of z-values that the solid occupies. First, let's find the limits of integration for x and y. The plane x + y = 1 can be rewritten as y = 1 - x. This gives us the boundary for x and y in the xy-plane. Next, we need to find the limits for z. The elliptical paraboloid $z = 2x^2 + y^2 + 1$ will intersect the xy-plane (z=0) when $2x^2 + y^2 + 1 = 0$. However, since $2x^2 + y^2$ is always non-negative, the smallest value of z will be at the vertex of the paraboloid, which is z = 1 when x = 0 and y = 0. Now, we can set up the integral for the volume. The volume element in terms of x and y is dV = z dA, where dA is the area element in the xy-plane. We can express z as a function of x and y using the equation of the paraboloid: $$ z = 2x^2 + y^2 + 1 $$ The area element dA is the differential of the region in the xy-plane, which is a triangle bounded by the lines x = 0, y = 0, and x + y = 1. The integral for the volume V is: $$ V = \int \int (2x^2 + y^2 + 1) dA $$ We can integrate over the triangle in the xy-plane using the limits for x and y. Since the region is defined by the line x + y = 1, we can use x as the variable of integration and express y in terms of x: $$ y = 1 - x $$ The limits for x will be from 0 to 1, and for y, they will be from 0 to 1 - x. The integral becomes: $$ V = \int_{0}^{1} \int_{0}^{1-x} (2x^2 + (1 - x)^2 + 1) dy dx $$ Now we can compute the integral: $$ V = \int_{0}^{1} \left[ \int_{0}^{1-x} (2x^2 + 1 - 2x + x^2 + 1) dy \right] dx $$ $$ V = \int_{0}^{1} \left[ \int_{0}^{1-x} (3x^2 - 2x + 2) dy \right] dx $$ $$ V = \int_{0}^{1} \left[ (3x^2 - 2x + 2)y \right]_{0}^{1-x} dx $$ $$ V = \int_{0}^{1} \left[ (3x^2 - 2x + 2)(1-x) - 0 \right] dx $$ $$ V = \int_{0}^{1} (3x^2 - 2x + 2 - 3x^3 + 2x^2 - x^2) dx $$ $$ V = \int_{0}^{1} (-3x^3 + 4x^2 - 2x + 2) dx $$ $$ V = \left[ -\frac{3}{4}x^4 + \frac{4}{3}x^3 - x^2 + 2x \right]_{0}^{1} $$ $$ V = \left( -\frac{3}{4} + \frac{4}{3} - 1 + 2 \right) - (0) $$ $$ V = -\frac{3}{4} + \frac{4}{3} + 1 $$ $$ V = -\frac{9}{12} + \frac{16}{12} + \frac{12}{12} $$ $$ V = \frac{19}{12} $$ So the volume of the solid bounded by the given surfaces is $\frac{19}{12}$ cubic units. |
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| kimi | To find the volume of the solid bounded by the elliptical paraboloid \( z = 2x^2 + y^2 + 1 \), the plane \( x + y = 1 \), and the coordinate planes, we can use a triple integral in the region defined by these boundaries. First, let's determine the region in the \( xy \)-plane where the solid is bounded: 1. The paraboloid intersects the \( xy \)-plane when \( z = 0 \). This gives us the equation \( 2x^2 + y^2 = -1 \), which has no real solutions since \( 2x^2 + y^2 \) is always non-negative. So the paraboloid does not intersect the \( xy \)-plane. 2. The plane \( x + y = 1 \) intersects the \( x \)-axis when \( y = 0 \) and the \( y \)-axis when \( x = 0 \). So the region in the \( xy \)-plane is a right triangle with vertices at \( (0,0) \), \( (1,0) \), and \( (0,1) \). Now, we can set up the triple integral for the volume \( V \) of the solid: \[ V = \int \int \int_V dV \] Since the solid is bounded by the paraboloid and the plane, we can use the paraboloid as the upper bound for \( z \) and the plane as the lower bound. The integral will be in the region \( D \) defined by \( x \) from 0 to 1 and \( y \) from 0 to \( 1 - x \). The bounds for \( z \) will be from 0 to \( 2x^2 + y^2 + 1 \). So the volume integral is: \[ V = \int_{x=0}^{1} \int_{y=0}^{1-x} \int_{z=0}^{2x^2 + y^2 + 1} dz \, dy \, dx \] Evaluating the inner integral with respect to \( z \): \[ \int_{z=0}^{2x^2 + y^2 + 1} dz = 2x^2 + y^2 + 1 \] Now we have: \[ V = \int_{x=0}^{1} \int_{y=0}^{1-x} (2x^2 + y^2 + 1) \, dy \, dx \] Evaluating the middle integral with respect to \( y \): \[ \int_{y=0}^{1-x} (2x^2 + y^2 + 1) \, dy = \left[ 2x^2y + \frac{y^3}{3} + y \right]_{y=0}^{y=1-x} \] \[ = 2x^2(1-x) + \frac{(1-x)^3}{3} + (1-x) \] Simplifying: \[ = 2x^2 - 2x^3 + \frac{1 - 3x + 3x^2 - x^3}{3} + 1 - x \] \[ = 2x^2 - 2x^3 + \frac{1}{3} - x + x^2 - \frac{x^3}{3} + 1 - x \] \[ = 3x^2 - 3x^3 + \frac{4}{3} - \frac{4x}{3} \] Finally, evaluating the outer integral with respect to \( x \): \[ V = \int_{x=0}^{1} \left( 3x^2 - 3x^3 + \frac{4}{3} - \frac{4x}{3} \right) \, dx \] \[ V = \left[ x^3 - \frac{3x^4}{4} + \frac{4x}{3} - \frac{4x^2}{6} \right]_{x=0}^{x=1} \] \[ = \left( 1 - \frac{3}{4} + \frac{4}{3} - \frac{4}{6} \right) - (0) \] \[ = \left( \frac{1}{4} + \frac{4}{3} - \frac{2}{3} \right) \] \[ = \frac{1}{4} + \frac{2}{3} \] \[ = \frac{3}{12} + |
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| wenxinyiyan | 为了找到由椭圆抛物面$z = 2x^2 + y^2 + 1$，平面$x + y = 1$和坐标平面围成的固体的体积，我们可以按照以下步骤进行： 1. **确定积分区域**： * 首先，我们考虑平面$x + y = 1$与坐标平面$x = 0$和$y = 0$相交，形成一个三角形区域。 * 这个三角形区域的顶点在$(0,0)$，$(1,0)$和$(0,1)$。 2. **设置积分**： * 我们将使用二重积分来计算体积，积分区域是上述三角形。 * 积分函数是$z = 2x^2 + y^2 + 1$。 3. **计算积分**： * 积分可以表示为：$\int_{x=0}^{1} \int_{y=0}^{1-x} (2x^2 + y^2 + 1) \, dy \, dx$ 4. **求解积分**： * 首先对$y$积分：$\int_{y=0}^{1-x} (2x^2 + y^2 + 1) \, dy = [2x^2y + \frac{1}{3}y^3 + y]_{y=0}^{1-x}$ * 然后对$x$积分：$\int_{x=0}^{1} ([2x^2(1-x) + \frac{1}{3}(1-x)^3 + (1-x)] - [0]) \, dx$ * 积分后得到体积。 5. **计算结果**： * 经过计算，我们得到体积为$\frac{19}{12}$立方单位。 因此，由椭圆抛物面$z = 2x^2 + y^2 + 1$，平面$x + y = 1$和坐标平面围成的固体的体积是$\frac{19}{12}$立方单位。 |
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| zhipu | To find the volume of the solid bounded by the elliptical paraboloid \( z = 2x^2 + y^2 + 1 \), the plane \( x + y = 1 \), and the coordinate planes, we can set up a double integral in the \( xy \)-plane and integrate the height \( z \) over the region defined by the given boundaries. ### Step 1: Define the Region in the \( xy \)-Plane The plane \( x + y = 1 \) intersects the \( x \)-axis at \( (1,0) \) and the \( y \)-axis at \( (0,1) \). The coordinate planes include the \( x \)-axis (\( y = 0 \)) and the \( y \)-axis (\( x = 0 \)). Therefore, the region \( R \) in the \( xy \)-plane is a triangle with vertices at \( (0,0) \), \( (1,0) \), and \( (0,1) \). ### Step 2: Set Up the Double Integral The volume \( V \) of the solid can be found by integrating the function \( z = 2x^2 + y^2 + 1 \) over the region \( R \): \[ V = \iint_R (2x^2 + y^2 + 1) \, dA \] ### Step 3: Determine the Limits of Integration For the region \( R \), we can use the following limits of integration: - \( x \) ranges from 0 to 1. - For a fixed \( x \), \( y \) ranges from 0 to \( 1 - x \). Thus, the double integral becomes: \[ V = \int_{0}^{1} \int_{0}^{1-x} (2x^2 + y^2 + 1) \, dy \, dx \] ### Step 4: Evaluate the Inner Integral First, integrate with respect to \( y \): \[ \int_{0}^{1-x} (2x^2 + y^2 + 1) \, dy \] Break it down into separate integrals: \[ \int_{0}^{1-x} 2x^2 \, dy + \int_{0}^{1-x} y^2 \, dy + \int_{0}^{1-x} 1 \, dy \] Evaluate each part: 1. \[ \int_{0}^{1-x} 2x^2 \, dy = 2x^2 \left[ y \right]_{0}^{1-x} = 2x^2 (1-x) \] 2. \[ \int_{0}^{1-x} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{1-x} = \frac{(1-x)^3}{3} \] 3. \[ \int_{0}^{1-x} 1 \, dy = \left[ y \right]_{0}^{1-x} = 1-x \] Combine these results: \[ \int_{0}^{1-x} (2x^2 + y^2 + 1) \, dy = 2x^2 (1-x) + \frac{(1-x)^3}{3} + (1-x) \] ### Step 5: Evaluate the Outer Integral Now, integrate the result with respect to \( x \): \[ V = \int_{0}^{1} \left( 2x^2 (1-x) + \frac{(1-x)^3}{3} + (1-x) \right) \, dx \] Simplify the integrand: \[ V = \int_{0}^{1} \left( 2x^2 - 2x^3 + \frac{(1-x)^3}{3} + 1 - x \right) \, dx \] Expand \( (1-x)^3 \): \[ (1-x)^3 = 1 - 3x + 3x^2 - x^3 \] So, \[ \frac{(1-x)^3}{3} = \frac{1}{3} - x + x^2 - \frac{x^3}{3} \] Substitute back into the integral: \[ V = \int_{0}^{1} \left( 2x^2 - 2x^3 + \frac{1}{3} - x + x^2 - \frac{x^3}{3} + 1 - x \right) \, dx \] Combine like terms: \[ V = \int_{0}^{1} \left( 3x^2 - \frac{7x^3}{3} - 2x + \frac{4}{3} \right) \, dx \] ### Step 6: Integrate and Evaluate Integrate term by term: \[ \int_{0}^{1} 3x^2 \, dx = \left[ x^3 \right]_{0}^{1} = 1 \] \[ \int_{0}^{1} -\frac{7x^3}{3} \, dx = -\frac{7}{3} \left[ \frac{x^4}{4} \right]_{0}^{1} = -\frac{7}{12} \] \[ \int_{0}^{1} -2x \, dx = -2 \left[ \frac{x^2}{2} \right]_{0}^{1} = -1 \] \[ \int_{0}^{1} \frac{4}{3} \, dx = \frac{4}{3} \left[ x \right]_{0}^{1} = \frac{4}{3} \] Combine these results: \[ V = 1 - \frac{7}{12} - 1 + \frac{4}{3} \] Convert to common denominators: \[ V = \frac{12}{12} - \frac{7}{12} - \frac{12}{12} + \frac{16}{12} \] \[ V = \frac{12 - 7 - 12 + 16}{12} = \frac{9}{12} = \frac{3}{4} \] ### Final Answer The volume of the solid bounded by the given surfaces is \( \boxed{\frac{3}{4}} \). |
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| gpto1 | To find the volume of the solid bounded by the elliptical paraboloid \( z = 2x^2 + y^2 + 1 \), the plane \( x + y = 1 \), and the coordinate planes (\( x = 0 \) and \( y = 0 \)), we can set up and evaluate a double integral over the appropriate region in the \( xy \)-plane. **Step 1: Determine the Region of Integration** First, we need to determine the projection \( D \) of the solid onto the \( xy \)-plane. The solid lies above the coordinate planes (\( x = 0 \) and \( y = 0 \)) and below the plane \( x + y = 1 \). Thus, \( D \) is the triangular region bounded by \( x = 0 \), \( y = 0 \), and \( x + y = 1 \). **Step 2: Set Up the Double Integral** The volume \( V \) of the solid can be found by integrating the function \( z = 2x^2 + y^2 + 1 \) over the region \( D \): \[ V = \iint_D (2x^2 + y^2 + 1) \, dx \, dy \] Since \( D \) is a triangular region, it's convenient to integrate with respect to \( y \) first, from \( y = 0 \) to \( y = 1 - x \), and then with respect to \( x \), from \( x = 0 \) to \( x = 1 \): \[ V = \int_{x=0}^1 \int_{y=0}^{1-x} (2x^2 + y^2 + 1) \, dy \, dx \] **Step 3: Compute the Inner Integral** Integrate \( (2x^2 + y^2 + 1) \) with respect to \( y \): \[ \int_{y=0}^{1-x} (2x^2 + y^2 + 1) \, dy = 2x^2 (1 - x) + \frac{(1 - x)^3}{3} + (1 - x) \] Simplify the expression: \[ 2x^2 (1 - x) = 2x^2 - 2x^3 \] \[ \frac{(1 - x)^3}{3} = \frac{1 - 3x + 3x^2 - x^3}{3} \] So the result of the inner integral is: \[ I(x) = (2x^2 - 2x^3) + \frac{1 - 3x + 3x^2 - x^3}{3} + 1 - x \] Combine like terms: \[ I(x) = \left(2x^2 - 2x^3 + \frac{1 - 3x + 3x^2 - x^3}{3} + 1 - x\right) \] Simplify the expression by combining fractions and like terms: \[ I(x) = \frac{9x^2 - 7x^3 - 6x + 4}{3} \] **Step 4: Compute the Outer Integral** Now, integrate \( I(x) \) with respect to \( x \): \[ V = \int_{x=0}^1 \frac{9x^2 - 7x^3 - 6x + 4}{3} \, dx = \frac{1}{3} \int_0^1 (9x^2 - 7x^3 - 6x + 4) \, dx \] Compute the antiderivative: \[ \int (9x^2 - 7x^3 - 6x + 4) \, dx = 3x^3 - \frac{7x^4}{4} - 3x^2 + 4x + C \] Evaluate from \( x = 0 \) to \( x = 1 \): \[ F(1) = 3(1)^3 - \frac{7(1)^4}{4} - 3(1)^2 + 4(1) = 3 - \frac{7}{4} - 3 + 4 = \frac{9}{4} \] \[ F(0) = 0 \] Compute the definite integral: \[ \int_0^1 (9x^2 - 7x^3 - 6x + 4) \, dx = F(1) - F(0) = \frac{9}{4} - 0 = \frac{9}{4} \] Thus, the volume is: \[ V = \frac{1}{3} \cdot \frac{9}{4} = \frac{9}{12} = \frac{3}{4} \] **Answer:** The volume of the solid is **¾ cubic units**; that is, \( V = \dfrac{3}{4} \). |
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| deepseekv3 | To find the volume of the solid bounded by the elliptical paraboloid \( z = 2x^2 + y^2 + 1 \), the plane \( x + y = 1 \), and the coordinate planes, we follow these steps: --- ### Step 1: Identify the Region of Integration The solid is bounded by: 1. The coordinate planes: \( x = 0 \), \( y = 0 \), and \( z = 0 \). 2. The plane \( x + y = 1 \). 3. The elliptical paraboloid \( z = 2x^2 + y^2 + 1 \). The region in the \( xy \)-plane is bounded by: - \( x = 0 \), - \( y = 0 \), - \( x + y = 1 \). This is the triangular region in the first quadrant with vertices at \( (0, 0) \), \( (1, 0) \), and \( (0, 1) \). --- ### Step 2: Set Up the Integral The volume \( V \) is given by the double integral of the height function \( z = 2x^2 + y^2 + 1 \) over the region in the \( xy \)-plane: \[ V = \iint_D (2x^2 + y^2 + 1) \, dA \] Since the region \( D \) is bounded by \( x = 0 \), \( y = 0 \), and \( x + y = 1 \), we integrate \( x \) from \( 0 \) to \( 1 \) and \( y \) from \( 0 \) to \( 1 - x \): \[ V = \int_{x=0}^1 \int_{y=0}^{1-x} (2x^2 + y^2 + 1) \, dy \, dx \] --- ### Step 3: Integrate with Respect to \( y \) First, integrate \( 2x^2 + y^2 + 1 \) with respect to \( y \): \[ \int_{y=0}^{1-x} (2x^2 + y^2 + 1) \, dy = \left[ 2x^2 y + \frac{1}{3} y^3 + y \right]_0^{1-x} \] Substitute the limits \( y = 1 - x \) and \( y = 0 \): \[ = 2x^2 (1 - x) + \frac{1}{3} (1 - x)^3 + (1 - x) \] Simplify: \[ = 2x^2 (1 - x) + \frac{1}{3} (1 - 3x + 3x^2 - x^3) + (1 - x) \] \[ = 2x^2 - 2x^3 + \frac{1}{3} - x + x^2 - \frac{1}{3}x^3 + 1 - x \] Combine like terms: \[ = \left( -2x^3 - \frac{1}{3}x^3 \right) + \left( 2x^2 + x^2 \right) + \left( -x - x \right) + \left( \frac{1}{3} + 1 \right) \] \[ = -\frac{7}{3}x^3 + 3x^2 - 2x + \frac{4}{3} \] --- ### Step 4: Integrate with Respect to \( x \) Now integrate the result with respect to \( x \): \[ V = \int_{x=0}^1 \left( -\frac{7}{3}x^3 + 3x^2 - 2x + \frac{4}{3} \right) \, dx \] Integrate term by term: \[ V = \left[ -\frac{7}{12}x^4 + x^3 - x^2 + \frac{4}{3}x \right]_0^1 \] Substitute the limits \( x = 1 \) and \( x = 0 \): \[ V = -\frac{7}{12}(1)^4 + (1)^3 - (1)^2 + \frac{4}{3}(1) \] \[ V = -\frac{7}{12} + 1 - 1 + \frac{4}{3} \] Simplify: \[ V = -\frac{7}{12} + \frac{16}{12} = \frac{9}{12} = \frac{3}{4} \] --- ### Final Answer: The volume of the solid is: \[ \boxed{\frac{3}{4}} \] |
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