How do you calculate the area of the section of the plane defined by the equation 4x + 3y + z = 12 that is located in the first octant?

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Calculating the Area of a Plane Section in the First Octant

To find the area of the part of the plane given by the equation 4x + 3y + z = 12 that lies in the first octant, we can follow a systematic approach.

Step 1: Identify the Intercepts

The first step in our calculation is to determine the intercepts of the plane with the axes:

  • x-intercept: Set y and z to 0:
    • 4x = 12 → x = 3 → (3, 0, 0)
  • y-intercept: Set x and z to 0:
    • 3y = 12 → y = 4 → (0, 4, 0)
  • z-intercept: Set x and y to 0:
    • z = 12 → (0, 0, 12)

Step 2: Set Up the Integral

The area of the triangle formed by these intercepts in the xy-plane needs to be found. The triangle is defined by the vertices:

  • (3, 0, 0)
  • (0, 4, 0)
  • (0, 0, 12)

To calculate the area in the xy-plane, we will use a double integral:


A = 
∫03 ∫0(12 - 4x)/3 dy dx

Step 3: Evaluate the Integral

First, we calculate the inner integral:

∫0(12 - 4x)/3 dy = (12 - 4x)/3

Now substitute this result into the outer integral:

A = ∫03 (12 - 4x)/3 dx

Next, evaluate this integral:

A = (1/3) ∫03 (12 - 4x) dx

The inner integral calculates as:

-4 (3/2) + 12 (3)

Finally, evaluate it to find:

A = (1/3) [12x - 2x2] |03 = 6

Step 4: Conclusion

Thus, the area of the part of the plane 4x + 3y + z = 12 that lies in the first octant is 6 square units.

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