Question

Let S be the upper part of the boundary of W - in other words S is the paraboloid surface z = 1 − x 2 − y 2 for z ≥ −1. Use Lagrangian multipliers to find a point on S which is closest to the origin.

Answer #1

Let S be the boundary of the solid bounded by the paraboloid
z=x^2+y^2 and the plane z=16
S is the union of two surfaces. Let S1 be a portion of the plane
and S2 be a portion of the paraboloid so that S=S1∪S2
Evaluate the surface integral over S1
∬S1 z(x^2+y^2) dS=
Evaluate the surface integral over S2
∬S2 z(x^2+y^2) dS=
Therefore the surface integral over S is
∬S z(x^2+y^2) dS=

Calculus III. Please show all work and mark the
answer(s)!
1) Use Lagrange multipliers to find the maximum and minimum
values of the function f(x, y) = x^2 + y^2 subject to the
constraint xy = 1.
2) Use Lagrange Multipliers to find the point on the curve 2x +
3y = 6 that is closest to the origin. Hint: let f(x, y) be the
distance squared from the origin to the point (x, y), then find the
minimum of...

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...

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...

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...

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
≤(9-y^2)^1/2
, 0 ≤ x ≤ 3
Please provide a final answer as this is where I have an
issue.

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 =...

In the following problems, the surface S is the part of the
paraboloid z= x^2 + y^2 which lies below the plane z= 4, and
includes the circular intersection with this plane. This single
surface S could also be described as being contained inside the
cylinder x^2+y^2= 4.
(a) Iterate, but do not evaluate, the integral ∫∫S(z+x) dS in
terms of two parameters. Write the integrand in simplest form.
(b) Use Stoke’s theorem to rewrite ∫S(delta X F) · ndS...

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

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