   Chapter 16.6, Problem 39E

Chapter
Section
Textbook Problem

Find the area of the surface.39. The part of the plane 3x + 2y + z = 6 that lies in the first octant

To determine

To find: The area for the surface of the part of the plane 3x+2y+z=6 that lies in the first octant.

Explanation

Given data:

The equation of the part of the plane is given as follows.

3x+2y+z=6

Rearrange the equation.

z=63x2y (1)

Formula used:

Write the expression to find the surface area of the plane.

A(S)=D1+(zx)2+(zy)2dA (2)

Write the expression to find area of triangular region with base and height.

A=12bh (3)

Here,

b is the base of triangular region and

h is the height of triangular region.

Write the equation of plane as follows.

3x+2y+z=6

Consider x and y as parameters and parameterize the plane as follows.

x=x,y=y,z=63x2y

The function f(x,y)=63x2y which is z intersects the xy-plane when z=0 .

Substitute 0 for z in the equation of plane to obtain the line which intersects the function f(x,y)=63x2y .

3x+2y+0=63x+2y=6

The function f(x,y)=63x2y intersects the xy-plane in the line 3x+2y=6 .

3x+2y=6 (4)

Calculation of x-intercept:

Substitute 0 for y in equation (4),

3x+2(0)=63x=6x=2

Calculation of y-intercept:

Substitute 0 for x in equation (4),

3(0)+2y=62y=6y=3

From the given data, it is clear that, D is a triangular region with base as x-intercept and height as y-intercept.

Calculation of zx :

Take partial derivative for equation (1) with respect to x.

zx=x(63x2y)=030=3

Calculation of zy :

Take partial derivative for equation (1) with respect to y

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