Question

. Find the flux of the vector field F~ (x, y, z) = <y,-x,z> over a surface with downward orientation, whose parametric equation is given by r(s, t) = <2s, 2t, 5 − s 2 − t 2 > with s^2 + t^2 ≤ 1

Answer #1

Find the flux of the vector field F (x, y, z) =< y, x, e^xz
> outward from the z−axis and across the surface S, where S is
the portion of x^2 + y^2 = 9 with x ≥ 0, y ≥ 0 and −3 ≤ z ≤ 3.

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) = −xi − yj + z3k,
S is the part of the cone z =
x2 + y2
between the planes
z = 1
and
z = 2
with downward orientation

Find the flux of the vector field F(x, y, z) = x, y, z through
the portion of the parabaloid z = 16 - x^2-y^2 above the
plane ? = 7 with upward pointing normal.

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) = y i − x j + z2 k S is the helicoid (with
upward orientation) with vector equation r(u, v) = u cos v i + u
sin v j + v k, 0 ≤ u ≤ 5, 0...

Find the flux of the vector field F =
x i +
e6x j +
z k through the surface S given
by that portion of the plane 6x + y +
3z = 9 in the first octant, oriented upward.
PLEASE EXPLAIN STEPS. Thank you.

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

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) = 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...

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.

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 9 minutes ago

asked 13 minutes ago

asked 19 minutes ago

asked 25 minutes ago

asked 28 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago