Let S be the portion of the surface x 2 + z 2 = 1 lying in the first octant and bounded by x = 0 , y = 0 , z = 0 and y = 4 − 2x . Sketch the projection of S on the xy-plane

To sketch the projection of the surface \( S \) onto the \( xy \)-plane, we need to analyze the surface, the boundary conditions, and how they interact in that plane. Here's a step-by-step breakdown:

### Step 1: Surface Equation
The surface is given by:
\[
x^2 + z^2 = 1
\]
This equation represents a cylinder with radius 1, oriented along the \( y \)-axis. Since the equation does not involve \( y \), the surface extends infinitely along the \( y \)-direction.

### Step 2: Bounded Region in the First Octant
The first octant corresponds to \( x \geq 0 \), \( y \geq 0 \), and \( z \geq 0 \). We also have an additional boundary defined by \( y = 4 - 2x \), which further constrains the region in the \( xy \)-plane.

### Step 3: Projection onto the \( xy \)-Plane
We need to project the surface onto the \( xy \)-plane. To do so, we eliminate the \( z \)-coordinate. The projection of the surface \( x^2 + z^2 = 1 \) onto the \( xy \)-plane is the curve:
\[
x^2 \leq 1 \quad \text{(since \( z^2 \geq 0 \))}.
\]
This means that \( x \) can range from 0 to 1 in the first octant. The equation \( y = 4 - 2x \) bounds \( y \) as a function of \( x \), which gives us the upper boundary for the region in the \( xy \)-plane.

### Step 4: Sketching the Projection
To summarize, the projection onto the \( xy \)-plane is the region described by:
- \( x \in [0, 1] \),
- \( y = 4 - 2x \) forms the upper boundary of the projection.

Thus, the projection is a triangular region in the \( xy \)-plane:
- The base of the triangle is along the \( x \)-axis from \( x = 0 \) to \( x = 1 \),
- The height of the triangle at \( x = 0 \) is \( y = 4 \), and the height at \( x = 1 \) is \( y = 2 \) (since \( y = 4 - 2(1) \)).

The projection will be a triangle with vertices at:
- \( (0, 0) \),
- \( (1, 2) \),
- \( (0, 4) \).

### Graph:
Here is a rough outline of what the projection would look like:

```
y
|
4 |     *
  |    /|
  |   / |
2 |  *--*
  | /  
  |/
0 *----------
   0   1    x
```

### Summary:
The projection onto the \( xy \)-plane is a triangle with the vertices at \( (0, 0) \), \( (0, 4) \), and \( (1, 2) \).

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Would you like more details on this, or do you have any questions?

Here are 5 related questions to explore further:
1. How would the projection change if the surface were defined by a different equation?
2. What would happen if the boundary condition \( y = 4 - 2x \) were different?
3. How can we calculate the area of this triangular region in the \( xy \)-plane?
4. How would the projection look in a different octant?
5. Can we parametrize the surface \( S \) to understand its 3D structure better?

**Tip:** When projecting a surface onto a plane, eliminate the variable corresponding to the plane's normal direction. For example, for the \( xy \)-plane, remove \( z \) from the surface equation.

Let S be the portion of the surface x^2 + z^2 = 1 lying in the first octant and bounded by x = 0, y = 0, z = 0 and y = 4 − 2x. Sketch the projection of S on the xy-plane.

Explore the projection of the cylindrical surface x^2 + z^2 = 1 onto the xy-plane, bounded by x = 0, y = 0, z = 0, and y = 4 - 2x in the first octant. This detailed explanation covers surface projections, cylindrical geometry, and boundary constraints in analytic geometry. Suitable for college-level calculus students.

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