How do I find the area of the portion of the plane defined by the equation 3x + 2y + z = 6 that exists in the first octant?

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To find the area of the portion of the plane defined by the equation 3x + 2y + z = 6 that lies in the first octant, we will follow these steps:

  1. Identify intercepts: The first octant is where all coordinates (x, y, z) are non-negative. To find where this plane intercepts the axes, we can set two variables to zero at a time:
    • Set y = 0 and z = 0: 
                  3x + 2(0) + 0 = 6  
            => 3x = 6  
            => x = 2

      so we have the x-intercept at (2, 0, 0).

    • Set x = 0 and z = 0: 
                  3(0) + 2y + 0 = 6  
            => 2y = 6  
            => y = 3

      thus the y-intercept is at (0, 3, 0).

    • Set x = 0 and y = 0: 
                  3(0) + 2(0) + z = 6  
            => z = 6

      and the z-intercept is at (0, 0, 6).

  2. Determine vertices of the triangular region: The intercepts give us vertices of the triangular portion of the plane in the first octant, which are (2, 0, 0), (0, 3, 0), and (0, 0, 6).

Now, we will use these vertices to determine the area of the triangle lying in the first octant.

Using the formula for the area of a triangle:

The area (A) can be calculated using the formula:

A = 0.5 * base * height

To use this formula, we need to determine the base and height of our triangle:

  • Base (along x-axis): The distance between the points (0, 0, 0) and (2, 0, 0) is 2.
  • Height (along y-axis): The maximum y-coordinate at (0, 3, 0) is 3.

Now, substituting these values into the area formula:

A = 0.5 * 2 * 3 = 3

Thus, the area of the triangular region of the plane in the first octant is 3 square units.

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