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Decide which integral of the Divergence Theorem to use and compute the outward flux of the vector fieldF=<2xcosy, siny, zcosy>across the surface​S, where S is the boundary of the region bounded by the planesx=2​, y=​0, y=π​/2, z=​0, and z=x

Question

Decide which integral of the Divergence Theorem to use and compute the outward flux of the vector field

F=<2xcosy, siny, zcosy>

across the surface​S, where S is the boundary of the region bounded by the planes

x=2​, y=​0, y=π​/2, z=​0, and z=x
check_circleAnswer
Step 1

From the Divergence theorem, for a positively oriented boundary surface (outward Normal) of region E the flux is equal to triple integral of divF over the region E.

E
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E

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Step 2

Find the range of x,y and z as follows.

The range of z varies from value of z 0 to value ofz on plane z = x which implies
0zx
The range of y varies from value of y 0 to value of z on plane y which implies
2
0s ys
2
The range ofx varies from value of x = 0 to value ofz on plane x = 2 which implies
0xs2
The divF can be calculated as follows
divF = div( (2xcos y, sin y, z cos y))
(sin y) +(zcos y)
ôy
(2.xcos y)
x
ôz
=2 cos y cos y + cos y
= 4cos y
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The range of z varies from value of z 0 to value ofz on plane z = x which implies 0zx The range of y varies from value of y 0 to value of z on plane y which implies 2 0s ys 2 The range ofx varies from value of x = 0 to value ofz on plane x = 2 which implies 0xs2 The divF can be calculated as follows divF = div( (2xcos y, sin y, z cos y)) (sin y) +(zcos y) ôy (2.xcos y) x ôz =2 cos y cos y + cos y = 4cos y

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Step 3

Outward flux can be calcu...

Fds-
divFdV
=I1[4cos ydzdydx
0 0 0
2 2
4[f cos y[, dvdx
0 0
4 xdxcos ydy
x
= 4
2
sin y
0
= 4(2-0)(1-0)
= 8
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Fds- divFdV =I1[4cos ydzdydx 0 0 0 2 2 4[f cos y[, dvdx 0 0 4 xdxcos ydy x = 4 2 sin y 0 = 4(2-0)(1-0) = 8

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Math

Calculus

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