Let's break down each part of the problem. Let’s **focus on the mathematics and calculations** (for the diagrams, you can use GeoGebra or similar tools as asked). --- ## **Given Surfaces/Borders:** - \( z = 3 - x \) (Plane 1) - \( z = 2 \) (Plane 2) - \( y = x^2 + 1 \) (Parabola in x-y) - \( y = 1 \) (Plane 3) - First Octant: \( x \geq 0, y \geq 0, z \geq 0 \) --- ## **Step 1: Region Analysis** ### **z-bounds:** - Lower: \( z = 2 \) - Upper: \( z = 3 - x \) ### **y-bounds:** - Lower: \( y = 1 \) - Upper: \( y = x^2 + 1 \) But since \( y \geq 0 \) and \( y = x^2 + 1 \geq 1 \), for \( y \) to be between 1 and \( x^2 + 1 \), \( x \geq 0 \). ### **x-bounds:** To stay in the first octant and for \( z = 2 \leq z \leq 3 - x \): - \( 2 \leq 3 - x \implies x \leq 1 \) - \( x \geq 0 \) So, \( 0 \leq x \leq 1 \) --- ## **Step 2: Triple Integral (Volume)** Let's write the volume as a triple integral. The solid is described by: \[ V = \int_{x=0}^{1} \int_{y=1}^{x^2+1} \int_{z=2}^{3-x} dz\,dy\,dx \] --- ### **a. Integrating with respect to z:** \[ \int_{z=2}^{3-x} dz = (3-x) - 2 = 1 - x \] So, \[ V = \int_{x=0}^{1} \int_{y=1}^{x^2+1} (1-x)\, dy\, dx \] --- ### **b. Integrating with respect to y:** \[ \int_{y=1}^{x^2+1} (1-x) dy = (1-x) \cdot [y]_{y=1}^{x^2+1} = (1-x) \cdot (x^2+1 - 1) = (1-x)x^2 \] So, \[ V = \int_{x=0}^{1} (1-x)x^2 dx \] --- ### **c. Expand and Integrate with respect to x:** \[ (1-x)x^2 = x^2 - x^3 \] \[ V = \int_{0}^{1} (x^2 - x^3) dx = \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1 \] Plug in the bounds: \[ V = \left( \frac{1}{3} - \frac{1}{4} \right) - (0 - 0) = \frac{4-3}{12} = \frac{1}{12} \] --- ## **Final Answer** \[ \boxed{V = \frac{1}{12}} \] --- ### **Summary Table** | Axis Order | Volume Expression | Final Volume | |----------------|-------------------------------------------------------|---------------| | x-y-z (as above) | \( \int_{x=0}^{1} \int_{y=1}^{x^2+1} \int_{z=2}^{3-x} dz\,dy\,dx \) | \( \frac{1}{12} \) | | y-x-z | Similar bounds interchanged, same volume | \( \frac{1}{12} \) | | z-y-x | ... | \( \frac{1}{12} \) | --- ### **b. 2D Projections** - **x-axis projection**: In the y-z plane, for each x from 0 to 1, \( y \) from 1 to \( x^2+1 \), \( z \) from 2 to \( 3-x \). - **y-axis projection**: In the x-z plane, for each y from 1 to 2 (since \( x^2+1 = y \implies x = \sqrt{y-1} \)), \( x \) from 0 to \( \sqrt{y-1} \), \( z \) from 2 to \( 3-x \). - **z-axis projection**: In the x-y plane, for each z from 2 to 3, \( x \) from 0 to \( 3-z \), \( y \) from 1 to \( x^2+1 \). --- ## **How to answer in your submission:** - Show all steps as above. - Draw/provide 2D projections (using GeoGebra or manually). - State the final volume (\( \frac{1}{12} \)). If you need the projections written out in integral form for each axis, let me know!

Let's work through each part of the problem carefully, providing detailed calculations and explanations. --- ## **Given Surfaces and Boundaries:** - \( z = 3 - x \) (Plane sloping downwards in the x-z plane) - \( z = 2 \) (Horizontal plane) - \( y = x^2 + 1 \) (Parabola opening upwards in the y-x plane) - \( y = 1 \) (Horizontal plane in y) - First octant: \( x, y, z \geq 0 \) --- ## **Part 1: Generate 3D Representation** Using a mathematical software (like GeoGebra), you can plot: - The planes \( z = 3 - x \) and \( z=2 \) - The surfaces \( y = x^2 + 1 \) and \( y=1 \) - The coordinate planes \( x=0, y=0, z=0 \) The intersection of these surfaces in the first octant forms a solid bounded by the above surfaces. --- ## **Part 2: 2D Projection Diagrams** You are asked to produce projections on each plane. ### **a) Projection onto the xy-plane:** - \( y \) varies from 1 to \( x^2 + 1 \) - \( x \) varies from 0 to 1 (since \( y \leq x^2+1 \), and \( y \geq 1 \)) - The projection is bounded between \( y=1 \) and \( y=x^2+1 \) for \( x \in [0,1] \). ### **b) Projection onto the yz-plane:** - For a fixed \( x \), \( y \) varies from 1 to \( x^2 + 1 \) - \( z \) varies from 2 to \( 3 - x \) - To project onto yz-plane, eliminate \( x \): Since \( y \) depends on \( x \), and \( y \leq x^2 + 1 \): - For each \( y \in [1, 2] \): \( x = \sqrt{y - 1} \) (since \( y = x^2 + 1 \Rightarrow x = \sqrt{y-1} \)) - \( z \) varies from 2 to \( 3 - x = 3 - \sqrt{y - 1} \). ### **c) Projection onto the xz-plane:** - \( x \) from 0 to 1 - \( z \) from 2 to \( 3 - x \) --- ## **Part 3: Calculation of the Volume via Triple Integrals** ### **Method A: Integrate with respect to \( z \) first** \[ V = \iiint_{D} dz\,dy\,dx \] - \( z \) bounds: from \( z=2 \) to \( z=3 - x \) - \( y \) bounds: from \( y=1 \) to \( y=x^2 + 1 \) - \( x \) bounds: from 0 to 1 **Integral:** \[ V = \int_{x=0}^{1} \int_{y=1}^{x^2+1} \int_{z=2}^{3 - x} dz\,dy\,dx \] ---- ### **Step 1: Integrate with respect to \( z \)** \[ \int_{z=2}^{3 - x} dz = (3 - x) - 2 = 1 - x \] **Updated integral:** \[ V = \int_{0}^{1} \int_{1}^{x^2+1} (1 - x) dy\, dx \] --- ### **Step 2: Integrate with respect to \( y \)** Since the integrand does not depend on \( y \): \[ \int_{y=1}^{x^2 + 1} (1 - x) dy = (1 - x) \times [y]_{1}^{x^2 + 1} = (1 - x)(x^2 + 1 - 1) = (1 - x) x^2 \] **Updated integral:** \[ V = \int_{0}^{1} (1 - x) x^2 dx \] --- ### **Step 3: Simplify and integrate with respect to \( x \)** Expand the integrand: \[ (1 - x) x^2 = x^2 - x^3 \] \[ V = \int_{0}^{1} (x^2 - x^3) dx \] Calculate: \[ \int x^2 dx = \frac{x^3}{3}, \quad \int x^3 dx = \frac{x^4}{4} \] Evaluate from 0 to 1: \[ V = \left[\frac{x^3}{3} - \frac{x^4}{4}\right]_0^{1} = \left(\frac{1}{3} - \frac{1}{4}\right) - 0 = \frac{4}{12} - \frac{3}{12} = \frac{1}{12} \] --- ## **Final Volume:** \[ \boxed{ \textbf{Volume} = \frac{1}{12} } \] --- ## **Summary** | **Order of Integration** | **Integral Expression** | **Result** | |--------------------------|-------------------------|------------| | \( dz\,dy\,dx \) (as shown above) | \( \int_0^1 \int_1^{x^2+1} \int_2^{3 - x} dz\,dy\,dx \) | \( \frac{1}{12} \) | --- ## **Additional notes for the projections:** - **On the xy-plane:** bounded by \( y=1 \) and \( y=x^2 + 1 \), \( x \in [0,1] \). - **On the yz-plane:** for each \( y \in [1,2] \), \( z \in [2, 3 - \sqrt{y-1}] \). - **On the xz-plane:** for \( x \in [0,1] \), \( z \in [2, 3 - x] \). --- ## **Summary of steps:** 1. Identified the bounds based on the given surfaces. 2. Set up the triple integral. 3. Integrated step-by-step, starting from the innermost integral. 4. Simplified and evaluated to obtain the volume as \( \frac{1}{12} \). --- **If you need further assistance with diagrams or alternative methods, please let me know!**