Question

Evaluate the surface integral ∫∫_{S}**F** · *d***S** 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** - *z* **j** +
*y* **k**

*S* is the part of the sphere *x*^{2} +
*y*^{2} + *z*^{2} = 81 in the first
octant, with orientation toward the origin.

Answer #1

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 − z j + y k
S is the part of the sphere
x2 + y2 + z2 = 4
in the first octant, with orientation toward the origin

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 − z j + y k
S is the part of the sphere
x2 + y2 + z2 = 25
in the first octant, with orientation toward the origin

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 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) = x2 i + y2 j + z2 k S is the boundary of
the solid half-cylinder 0 ≤ z ≤ 25 − y2 , 0 ≤ x ≤ 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 =...

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) = xz i + x j + y k S is the hemisphere x2 +
y2 + z2 = 4, y ≥ 0, oriented in the direction of the positive
y-axis. Incorrect: Your answer is incorrect.

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

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 = 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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