# Find the area of the part of the plane 4x + 3y + z = 12 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) = 12 - 4x - 3y

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

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

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

= \( \int_{0}^{3}\int_{0}^{4 - \frac{4}{3}x}\sqrt{(-4)^{2}+(-3)^{2}+1}dxdy \)

= \( \int_{0}^{3}\int_{0}^{4 - \frac{4}{3}x}\sqrt{26}dxdy \)

= \( \sqrt{26}\int_{0}^{3}[y]_{0}^{4-\frac{4}{3}x}dxdy \)

=\( \sqrt{26}\int_{0}^{3}[4 - \{\frac{4}{3}x]dx \)

= \( \sqrt{26}[4x - \frac{4}{3}.\frac{1}{2}.x^{2}]_{0}^{3} \)

= \( \sqrt{26}[4x - \frac{4}{3}.\frac{1}{2}.x^{2}]_{0}^{3} \)

= \( \sqrt{26}[4(3) - \frac{4}{3}.\frac{1}{2}.3^{2}] \)

= 6√26 units²

## Find the area of the part of the plane 4x + 3y + z = 12 that lies in the first octant.

**Summary: **

The area of the part of the plane 4x + 3y + z = 12 is 6√26 units²

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