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4. F(x, y, 2) = yĩ+ (z – y)7+ xk. S: The portion of the plane (S is the top of the tetrahedron with vertices (0,0,0), (6,0, 0), (0,6,0), and (0,0, 6).) x + y + z = 6 in the first octant (where x, Y, z > 0). z = 6 – x – Y 0 < x < 6 – y 0

4. F(x, y, 2) = yĩ+ (z – y)7+ xk. S: The portion of the plane (S is the top of the tetrahedron with vertices (0,0,0), (6,0, 0), (0,6,0), and (0,0, 6).) x + y + z = 6 in the first octant (where x, Y, z > 0). z = 6 – x – Y 0 < x < 6 – y 0

Advanced Engineering Mathematics
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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A TILTED surface: 4. F(x, y, z) = y7+(z – y)7+ xk. S: The portion of the plane (S is the top of the tetrahedron with vertices (0,0,0), (6,0, 0), (0,6, 0), and (0,0,6).) x + y + z = 6 in the first octant (where x, y, z > 0). Equations/inequalities for the surface: S: z = 6 – x – y 0 < x < 6 – y 0 <y < 6 On S, F.î = F = dS = V3 dx dy the area on this tilted surface is v3 times the horizontal area dxdy /| F . î dS = S
Transcribed Image Text:A TILTED surface: 4. F(x, y, z) = y7+(z – y)7+ xk. S: The portion of the plane (S is the top of the tetrahedron with vertices (0,0,0), (6,0, 0), (0,6, 0), and (0,0,6).) x + y + z = 6 in the first octant (where x, y, z > 0). Equations/inequalities for the surface: S: z = 6 – x – y 0 < x < 6 – y 0 <y < 6 On S, F.î = F = dS = V3 dx dy the area on this tilted surface is v3 times the horizontal area dxdy /| F . î dS = S
Expert Solution
Solution:

The vector field is F→x, y, z=yi→+z-yj→+xk→

The plane is x+y+z=6 in the first octant.

Then, we have fx, y, z=x+y+z-6

Obtain the normal vector:

n→=∇f∇f=i→+j→+k→1+1+1=13i→+j→+k→

F→x, y, 6-x-y=yi→+6-x-2yj→+xk→

 

Given: dS=3dxdy

 

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