Find the area of the part of the plane 3x 2y 6 that les in the first octant. Part1ofs The surface Sis the graph of a function zx. y) over a region Din the xy-plane. The surface area of S can be calculated by A(S) -L V-()-() dA. The plane 3r+ 2y+6 is the graph of the function z fx, y)-6- -2y. We, therefore, have - and - 2 Part 2 ofS Therefore. Since we want the area of the section of the plane that les abeve the first octant. then the region D will be the triangular region bounded by the axis, the y-axis, and the line formed by the intersection of the plane with the plane z-0. Substituting z0 into 3x + 2y 6 and solving for y, we find that this line of intersection has the equation Semt Shyocnot.come bac
Find the area of the part of the plane 3x 2y 6 that les in the first octant. Part1ofs The surface Sis the graph of a function zx. y) over a region Din the xy-plane. The surface area of S can be calculated by A(S) -L V-()-() dA. The plane 3r+ 2y+6 is the graph of the function z fx, y)-6- -2y. We, therefore, have - and - 2 Part 2 ofS Therefore. Since we want the area of the section of the plane that les abeve the first octant. then the region D will be the triangular region bounded by the axis, the y-axis, and the line formed by the intersection of the plane with the plane z-0. Substituting z0 into 3x + 2y 6 and solving for y, we find that this line of intersection has the equation Semt Shyocnot.come bac
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Tutorial Exercise
Find the area of the part of the plane 3x + 2y + z = 6 that lies in the first octant.
Part 1 of 5
2
The surface S is the graph of a function z = f(x, y) over a region D in the xy-plane. The surface area of S can be calculated by A(S) =
az
1 +
dz
dA.
ду
f(x, y) = 6 – 3x
az
2y. We, therefore, have
ax
əz
-3 and
-2
The plane 3x + 2y + z = 6 is the graph of the function z =
3.x
ду
Part 2 of 5
Therefore,
=7 V1+(-3)² + (-2)² dA =
V14
dA.
Since we want the area of the section of the plane that lies above the first octant, then the region D will be the triangular region bounded by the x-axis, the y-axis, and the line formed by the intersection of the plane with the plane z = 0. Substituting z = 0 into 3x + 2y + z = 6 and solving for y, we find that this line of intersection has the equation
y =
x +
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