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

dedica66em
2021-12-18
Answered

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

Answered 2021-12-19
Author has **39** answers

Step 1

Remember that: Area of the surface which is part of

$z=f(x,\text{}y)$ 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$f(x,\text{}y)=6-3x-2y$

Area of the surface is

$A\left(S\right)=\int {\int}_{D}\sqrt{1+{(-3)}^{2}+{(-2)}^{2}}dA$

$A\left(S\right)=\int {\int}_{D}\sqrt{1+9+4}dA$

$A\left(S\right)=\sqrt{14}\int {\int}_{D}dA$

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$(0,\text{}0),\text{}(2,\text{}0)$ and $(0,\text{}3)$

Step 4

Note that:$\int {\int}_{D}dA$ is the area of the region inside D

Area of the triangle$=\frac{1}{2}\times \text{Base}\times \text{Height}=\frac{1}{2}\times 2time3=3$

Therefore

$A\left(S\right)=\sqrt{14}\int {\int}_{D}dA=3\sqrt{14}$

Remember that: Area of the surface which is part of

Where D is the projection of the surface on the xy plane.

Step 2

In the given problem

Area of the surface is

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:

Write the given equation of the plane in intercept form:

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

Step 4

Note that:

Area of the triangle

Therefore

GaceCoect5v

Answered 2021-12-20
Author has **26** answers

Step 1

We know that, area of surface which is part of$z=f(x,\text{}y)$ 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

$z=f(x,\text{}y)=6-3x-2y$

Thus,

${f}_{x}=-3$ and ${f}_{y}=-2$

Area of surface is calculated as

$A\left(s\right)=\int {\int}_{D}\sqrt{1+{[-3]}^{2}+{[-2]}^{2}}dA$

$A\left(s\right)=\int {\int}_{D}\sqrt{14}dA$

$A\left(s\right)=\sqrt{14}\int {\int}_{D}dA$

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)

$\text{Area of triangle}=\frac{1}{2}\left(2\right)\times \left(3\right)=3units$

Therefore,

$A\left(s\right)=\sqrt{14}\int {\int}_{D}dA$

$=\left(\sqrt{14}\right)\times \left(3\right)$

$=3\sqrt{14}units$

Hence, area of surface will be$3\sqrt{14}$ units.

We know that, area of surface which is part of

Now, we have

Thus,

Area of surface is calculated as

Step 2

The x, y and z intercepts of the plane

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)

Therefore,

Hence, area of surface will be

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