# Find the area of the surface. The part of the

Find the area of the surface. The part of the plane $3x+2y+z=6$ that lies in the first octant.
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Philip Williams
Step 1
Remember that: Area of the surface which is part of
is
$A\left(S\right)=\int {\int }_{D}\sqrt{1+{\left[{f}_{x}\right]}^{2}+{\left[{f}_{y}\right]}^{2}}dxdy$
Where D is the projection of the surface on the xy plane.
Step 2
In the given problem
Area of the surface is
$A\left(S\right)=\int {\int }_{D}\sqrt{1+{\left(-3\right)}^{2}+{\left(-2\right)}^{2}}dA$
$A\left(S\right)=\int {\int }_{D}\sqrt{1+9+4}dA$
$A\left(S\right)=\sqrt{14}\int {\int }_{D}dA$
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 and

Step 4
Note that: $\int {\int }_{D}dA$ is the area of the region inside D
Area of the triangle $=\frac{1}{2}×\text{Base}×\text{Height}=\frac{1}{2}×2time3=3$
Therefore
$A\left(S\right)=\sqrt{14}\int {\int }_{D}dA=3\sqrt{14}$
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GaceCoect5v
Step 1
We know that, area of surface which is part of is given as
$A\left(s\right)=\int {\int }_{D}\sqrt{1+{\left[{f}_{x}\right]}^{2}+{\left[{f}_{y}\right]}^{2}}dA$
Now, we have

Thus,
${f}_{x}=-3$ and ${f}_{y}=-2$
Area of surface is calculated as
$A\left(s\right)=\int {\int }_{D}\sqrt{1+{\left[-3\right]}^{2}+{\left[-2\right]}^{2}}dA$
$A\left(s\right)=\int {\int }_{D}\sqrt{14}dA$
$A\left(s\right)=\sqrt{14}\int {\int }_{D}dA$
Step 2
The x, y and z intercepts of the plane $3x+2y+z=6$ 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)
$\text{Area of triangle}=\frac{1}{2}\left(2\right)×\left(3\right)=3units$
Therefore,
$A\left(s\right)=\sqrt{14}\int {\int }_{D}dA$
$=\left(\sqrt{14}\right)×\left(3\right)$
$=3\sqrt{14}units$
Hence, area of surface will be $3\sqrt{14}$ units.