Question

The Flux of the Vector field F(1, -2, 0) through any closed surface is zero? Why?

How do you find a vector field that the flux through any closed surface is equal to the volume enclosed?

Answer #1

compute the flux of the vector field F through the parameterized
surface S. F= zk and S is oriented upward and given, for 0 ≤ s ≤ 1,
0 ≤ t ≤ 1, by x = s + t, y = s – t, z = s2 +
t2.
the answer should be 4/3.

Evaluate the surface integral
Evaluate the surface integral
S
F · dS
for the given vector field F and the oriented
surface S. In other words, find the flux of
F across S. For closed surfaces, use the
positive (outward) orientation.
F(x, y, z) = x i + y j + 9 k
S is the boundary of the region enclosed by the
cylinder
x2 + z2 = 1
and the planes
y = 0 and x + y =...

Set up a double integral to find the flux of the vector field F
= <−x, −y, z^3 > through the surface S, where S is the part
of the cone z = sqrt( x^2 + y^2) between z = 1 and z = 3. You do
not have to evaluate the double integral.

Evaluate the surface integral S F · dS for the given vector
field F and the oriented surface S. In other words, find the flux
of F across S. For closed surfaces, use the positive (outward)
orientation. F(x, y, z) = yi − xj + 4zk, S is the hemisphere x^2 +
y2^ + z^2 = 4, z ≥ 0, oriented downward

Evaluate the surface integral
S
F · dS
for the given vector field F and the oriented
surface S. In other words, find the flux of
F across S. For closed surfaces, use the
positive (outward) orientation.
F(x, y, z) = xy i + yz j + zx k
S is the part of the paraboloid
z = 2 − x2 − y2 that lies above the square
0 ≤ x ≤ 1, 0 ≤ y ≤ 1,
and...

The
electric flux through a closed surface is found to be 2.5 N*m^2/ C
.What quantity of the charge is enclosed by the surface?

Evaluate the surface integral
S
F · dS
for the given vector field F and the oriented
surface S. In other words, find the flux of
F across S. For closed surfaces, use the
positive (outward) orientation.
F(x, y, z) = yi − xj + 2zk,
S is the hemisphere
x2 + y2 + z2 = 4,
z ≥ 0,
oriented downward

Evaluate the surface integral
F · dS
for the given vector field F and the oriented
surface S. In other words, find the flux of
F across S. For closed surfaces, use the
positive (outward) orientation.
F(x, y, z) = x2 i + y2 j + z2
k
S is the boundary of the solid half-cylinder 0 ≤ z ≤
4 − y2
, 0 ≤ x ≤ 5

Evaluate the surface integral
S
F · dS
for the given vector field F and the oriented
surface S. In other words, find the flux of
F across S. For closed surfaces, use the
positive (outward) orientation.
F(x, y, z) = xy i + yz j + zx k
S is the part of the paraboloid
z = 4 − x2 − y2 that lies above the square
0 ≤ x ≤ 1, 0 ≤ y ≤ 1,
and has...

Evaluate the surface integral
S
F · dS
for the given vector field F and the oriented
surface S. In other words, find the flux of
F across S. For closed surfaces, use the
positive (outward) orientation.
F(x, y, z) = xy i + yz j + zx k
S is the part of the paraboloid
z = 6 − x2 − y2 that lies above the square
0 ≤ x ≤ 1, 0 ≤ y ≤ 1,
and has...

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