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

Question

Philip Williams · Accepted Answer

Step 1
Remember that: Area of the surface which is part of

z = f (x, y)

is

A (S) = \int \int_{D} \sqrt{1 + {[f_{x}]}^{2} + {[f_{y}]}^{2}} d x d y

Where D is the projection of the surface on the xy plane.
Step 2
In the given problem

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

Area of the surface is

A (S) = \int \int_{D} \sqrt{1 + {(- 3)}^{2} + {(- 2)}^{2}} d A

A (S) = \int \int_{D} \sqrt{1 + 9 + 4} d A

A (S) = \sqrt{14} \int \int_{D} d A

Note that D is the projection of the surface on the xy-plane
Step 3
Intercept form of plane
The x, y, z intercepts of the following plane are a, b, c respectively:

\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1

Write the given equation of the plane in intercept form:

\frac{x}{2} + \frac{y}{3} + \frac{z}{6} = 1

The x-intercept is 2 and y-intercept is 3
Projection of part of plane in 1st octant on the xy-plane is
A triangle with vertices

(0, 0), (2, 0)

and

(0, 3)

Step 4
Note that:

\int \int_{D} d A

is the area of the region inside D
Area of the triangle

= \frac{1}{2} \times Base \times Height = \frac{1}{2} \times 2 t i m e 3 = 3

Therefore

A (S) = \sqrt{14} \int \int_{D} d A = 3 \sqrt{14}

GaceCoect5v · Accepted Answer

Step 1
We know that, area of surface which is part of

z = f (x, y)

is given as

A (s) = \int \int_{D} \sqrt{1 + {[f_{x}]}^{2} + {[f_{y}]}^{2}} d A

Now, we have

z = f (x, y) = 6 - 3 x - 2 y

Thus,

f_{x} = - 3

and

f_{y} = - 2

Area of surface is calculated as

A (s) = \int \int_{D} \sqrt{1 + {[- 3]}^{2} + {[- 2]}^{2}} d A

A (s) = \int \int_{D} \sqrt{14} d A

A (s) = \sqrt{14} \int \int_{D} d A

Step 2
The x, y and z intercepts of the plane

3 x + 2 y + z = 6

from the intercept form can be calculated as

\frac{x}{2} + \frac{y}{3} + \frac{z}{6} = 1

Thus, x, y and z intercepts will be 2, 3 and 6 units respectively.
The domain D is the triangle in the first quadrant of the xy-plane bounded by (0, 0),(2, 0) and (0, 3)

Area of triangle = \frac{1}{2} (2) \times (3) = 3 u n i t s

Therefore,

A (s) = \sqrt{14} \int \int_{D} d A

= (\sqrt{14}) \times (3)

= 3 \sqrt{14} u n i t s

Hence, area of surface will be

3 \sqrt{14}

units.

Find the area of the surface. The part of the

Answered question

Answer & Explanation

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