Find the area of the part of the plane 5x + 4y + z = 20 that lies in the first octant.

Solution:

Since the plane lies in the first octant it is implied that x ≥ 0; y ≥ 0; z ≥ 0

The diagram representative of the given problem is given 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) = 20 - 5x - 4y

= \( \frac{\partial f(x)}{\partial x}\) = -5

= \( \frac{\partial f(y)}{\partial y}\) = -4

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

= \( \int_{0}^{4}\int_{0}^{5 - \frac{5}{4}x}\sqrt{(-5)^{2}+ (-4)^{2}+1}dxdy\)

= \( \int_{0}^{4}\int_{0}^{5 - \frac{5}{4}x}\sqrt{42}dxdy\)

= \( \sqrt{42}\int_{0}^{4}[y]_{0}^{5-\frac{5}{4}x}dxdy\)

= \( \sqrt{42}\int_{0}^{4}[5 - \frac{5}{4}x]dx\)

= \(\sqrt{42}[5 - \frac{5}{4}x]_{0}^{4} \)

= \( \sqrt{42}[5x - \frac{5}{4}.\frac{1}{2}.x^{2}]_{0}^{4} \)

= \( \sqrt{42}[5.4 - \frac{5}{4}.\frac{1}{2}.4^{2}] \)

= 10√42 units ²

Find the area of the part of the plane 5x + 4y + z = 20 that lies in the first octant.

Summary:

The area the part of the plane 5x + 4y + z = 20 is 10√42 units²