Question

Find the area of the surface.

The part of the plane\(3x + 2y + z = 6\)that lies in the first octant.

Short answer

The required surface area is \(3\sqrt {14} \) .

Step-by-step solution

Given

Consider the equation of a plane.

\(3x + 2y + z = 6\)

The part of the plane lies in the first octant. That is \(x \ge 0,y \ge 0,z \ge 0\) .

The objective is to find the area of the surface of the part of the plane lies in the first octant.

Figure

The surface of the plane in the first octant is shown below:

From the above graph observe that the projection of the plane on the \(xy\) -plane is a triangle bounded by the coordinate axis and \(3x + 2y = 6\) , described by

\(D = \left\{ {\left( {x,y} \right)|0 \le x \le 2,0 \le y \le 3 - \frac{3}{2}x} \right\}\)

Graph

The region \(D\) is shown below:

Finding the domain

To find the surface area of the required plane use the following surface area formula.

Surface area formula with surface \(S\) with equation \(z = g\left( {x,y} \right)\) is,

where the \(D\) is the domain of integration.

Here, \(3x + 2y + z = 6\) .

Then, \(z = 6 - 3x - 2y\) .

Determine partial derivatives \(\frac{{\partial z}}{{\partial x}}\) and \(\frac{{\partial z}}{{\partial y}}\) .

\(\begin{array}{c}\frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {6 - 3x - 2y} \right)\\ = - 3\end{array}\)

\(\begin{array}{c}\frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {6 - 3x - 2y} \right)\\ = - 2\end{array}\)

Calculating surface area

The surface area is

Substitute \(\frac{{\partial z}}{{\partial x}} = - 3,\frac{{\partial z}}{{\partial y}} = - 2\)

\(\begin{array}{c}A\left( s \right) = \int_0^2 {\int_0^{\frac{{6 - 3x}}{2}} {\sqrt {{{\left( { - 3} \right)}^2} + {{\left( { - 2} \right)}^2} + 1} dydx} } \\ = \int_0^2 {\int_0^{3 - \frac{3}{2}x} {\sqrt {14} dydx} } \\ = \sqrt {14} \int_0^2 {\left[ y \right]_0^{3 - \frac{3}{2}x}} dx\\ = \sqrt {14} \int_0^2 {\left[ {3 - \frac{3}{2}x} \right]dx} \end{array}\)

\(\begin{array}{c}A\left( s \right) = \sqrt {14} \left[ {3x - \frac{3}{2} \cdot \frac{{{x^2}}}{2}} \right]_{x = 0}^{x = 2}\\ = \sqrt {14} \left[ {3x - \frac{3}{4}{x^2}} \right]_{x = 0}^{x = 2}\\ = \sqrt {14} \left[ {3\left( 2 \right) - \frac{3}{4}{{\left( 2 \right)}^2} - 0} \right]\\ = \sqrt {14} \left[ {6 - 3} \right]\end{array}\)

\(A\left( s \right) = 3\sqrt {14} \)

Therefore, the required surface area of the plane is \(3\sqrt {14} \) .