Evaluate the surface integral ∬ σ f x , y , z d S f x , y , z = x − y − z ; σ is the portion of the plane x + y = 1 in the first octant between z = 0 and z = 1.
Evaluate the surface integral ∬ σ f x , y , z d S f x , y , z = x − y − z ; σ is the portion of the plane x + y = 1 in the first octant between z = 0 and z = 1.
f
x
,
y
,
z
=
x
−
y
−
z
;
σ
is the portion of the plane
x
+
y
=
1
in the first octant between
z
=
0
and
z
=
1.
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
Evaluate
F.ndS for the given F and ơ.
(b) F(x, y, z) = (x² + y) i+ xyj – (2xz + y) k,
o : the surface of the plane x + y + z = 1 in the first octant
Let S be the portion of the plane 2x + 3y + z = 2 lying between the points (−1, 1, 1), (2, 1, −5), (2, 3, −11), and (-1, 3, -5). Find
parameterizations for both the surface S and its boundary S. Be sure that their respective orientations are compatible with Stokes'
theorem.
from (-1, 1, 1) to (2, 1,-5)
from (2, 1, 5) to (2, 3, -11)
from (2, 3, -11) to (-1, 3, -5)
from (-1, 3, 5) to (-1, 1, 1)
boundary
S₁ (t) =
S₂(t) =
S3(t) =
S4(t) =
Φ(u, v) =
te [0, 1)
te [1, 2)
te [2, 3)
te [3, 4)
UE [-1, 2], VE [1, 3]
Thomas' Calculus: Early Transcendentals (14th Edition)
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.