Find the area of the surface. The part of the

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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Expert Answer

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, y) is
A(S)=D1+[fx]2+[fy]2dxdy
Where D is the projection of the surface on the xy plane.
Step 2
In the given problem f(x, y)=63x2y
Area of the surface is
A(S)=D1+(3)2+(2)2dA
A(S)=D1+9+4dA
A(S)=14DdA
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:
xa+yb+zc=1
Write the given equation of the plane in intercept form:
x2+y3+z6=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: DdA is the area of the region inside D
Area of the triangle =12×Base×Height=12×2time3=3
Therefore
A(S)=14DdA=314
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GaceCoect5v
Answered 2021-12-20 Author has 26 answers
Step 1
We know that, area of surface which is part of z=f(x, y) is given as
A(s)=D1+[fx]2+[fy]2dA
Now, we have
z=f(x, y)=63x2y
Thus,
fx=3 and fy=2
Area of surface is calculated as
A(s)=D1+[3]2+[2]2dA
A(s)=D14dA
A(s)=14DdA
Step 2
The x, y and z intercepts of the plane 3x+2y+z=6 from the intercept form can be calculated as
x2+y3+z6=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=12(2)×(3)=3units
Therefore,
A(s)=14DdA
=(14)×(3)
=314units
Hence, area of surface will be 314 units.
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