To find the area of the part of the plane defined by the equation 4x + 3y + z = 12 that lies in the first octant, we first need to understand a few key concepts.
The first octant is the portion of three-dimensional space where all coordinates (x, y, z) are non-negative. This means that x ≥ 0, y ≥ 0, and z ≥ 0. We start by determining where the plane intersects the axes:
- For the x-intercept: Set y = 0 and z = 0 in the equation. This gives us:
- For the y-intercept: Set x = 0 and z = 0. This gives us:
- For the z-intercept: Set x = 0 and y = 0. This gives us:
4x = 12 → x = 3
3y = 12 → y = 4
z = 12
These intercepts form a triangular region in the first octant with vertices at points (3, 0, 0), (0, 4, 0), and (0, 0, 12).
Now, to find the area of this triangular region projected onto the xy-plane, we can use the formula for the area of a triangle:
Area = 0.5 * base * height
In this case, the base of the triangle is along the x-axis from (0,0) to (3,0), which has a length of 3, and the height is along the y-axis from (0,0) to (0,4), which has a height of 4.
Thus, we calculate:
Area = 0.5 * 3 * 4 = 6
However, since this area is projected onto the xy-plane, and we need the area of the actual region formed by the triangle on the plane, we need to consider the slope of the plane. The formula for the area of a surface is given by:
Area = ∬_D sqrt(1 + (dz/dx)^2 + (dz/dy)^2) dA
Where D is the region in the xy-plane. We first find dz/dx and dz/dy:
From the equation of the plane, we can express z as:
z = 12 – 4x – 3y
This gives us:
- dz/dx = -4
- dz/dy = -3
Now we plug these into the area formula:
Area = ∬_D sqrt(1 + (-4)^2 + (-3)^2) dA
= ∬_D sqrt(1 + 16 + 9) dA
= ∬_D sqrt(26) dA
The area of the triangle projected onto the xy-plane is 6, hence:
Area on surface = 6 * sqrt(26)
Thus, the area of the part of the plane 4x + 3y + z = 12 that lies in the first octant is 6√26 square units.