Find the area of the part of the plane 3x + 2y + z = 6 that lies in the first octant.

Solution:

The diagram representing the problem statement is shown below:

Figure 2 is the triangle generated by the shaded area in Fig 1 on the x-y plane. For a surface z = f(x, y) , the surface area formula is of the form:

A = \( \int \int_{R_{xy}}^{}\sqrt{f_{x}^{2}+f_{y}^{2}+1}dxdy \) (1)

Where fₓ and fᵧ are the partial derivatives of the function:

Z = f(x) = 6 - 3x - 2y

= \( \frac{\partial f(x)}{\partial x} \) = -3

= \( \frac{\partial f(y)}{\partial y} \) = -2

Substituting these values in equation (1) and giving limits to the integrals we get:

= \( \int_{0}^{2}\int_{0}^{3 - \frac{3}{2}x}\sqrt{(-3)^{2}+(-2)^{2}+1}dxdy \)

= \( \int_{0}^{2}\int_{0}^{3 - \frac{3}{2}x}\sqrt{14}dxdy \)

= \( \sqrt{14}\int_{0}^{2}[y]_{0}^{3-\frac{3}{2}x}dxdy\)

=\( \sqrt{14}\int_{0}^{2}[3 - \frac{3}{2}x]dx\)

= \( \sqrt{14}[3x - \frac{3}{2}.\frac{1}{2}.x^{2}]_{0}^{2} \)

= \( \sqrt{14}[3 - \frac{3}{2}.\frac{1}{2}.x^{2}]_{0}^{2} \)

= \( \sqrt{14}[3.2 - \frac{3}{2}.\frac{1}{2}.3^{2}] \)

= 3√14 units²

Find the area of the part of the plane 3x + 2y + z = 6 that lies in the first octant.

Summary:

The area of the part of the plane 3x + 2y + z = 6 is 3√14 units²