Miscellaneous surface integrals Evaluate the following integrals using the method of your choice. Assume normal vectors point either outward or upward . 50. ∬ S ∇ ln | r | ⋅ n d S , where S is the hemisphere x 2 + y 2 + z 2 = a 2 , for z ≥ 0, and where r = ( x , y , z )
Miscellaneous surface integrals Evaluate the following integrals using the method of your choice. Assume normal vectors point either outward or upward . 50. ∬ S ∇ ln | r | ⋅ n d S , where S is the hemisphere x 2 + y 2 + z 2 = a 2 , for z ≥ 0, and where r = ( x , y , z )
Miscellaneous surface integralsEvaluate the following integrals using the method of your choice. Assume normal vectors point either outward or upward.
50.
∬
S
∇
ln
|
r
|
⋅
n
d
S
, where S is the hemisphere x2 + y2 + z2 = a2, for z ≥ 0, and where r = (x, y, z)
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.
For an area A in the x-y plane, in the expression I₂ = 1x + ly, the term /₂ is the:
Minimum rectangular moment of inertia or second moment of area.
O Product of inertia.
Polar moment of inertia.
O Maximum rectangular moment of inertia or second moment of area.
Evaluate the surface integral.
J y ds, S is the helicoid with vector equation r(u, v) = (u cos(v), u sin(v), v), 0sus 6,0 SV SR.
[(10) ()-1]
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Calculate ff f(x, y, z) d.S for the given surface and function.
x² + y² = 25, 0≤ z ≤ 4; f(x, y, z) = e¯²
Consider the shown work.
To =
T, =
аф
де
=
д
(5 cos 0, 5 sin 0, z) = (-5 sin 0, 5 cos 0, 0)
do
d
-(5 cos 0, 5 sin 0, z) = (0,0,1)
дz
i
N(0, z) = T₁ × T₂ = -5 sin 0
0
||N(0, z)|| =
5 cos 0
0
2π 4
[[ f(x, y, 2) ds = [²* ["^ e
S
(5 cos 0)² + (5 sin 0)² + 0 =
e² do dz
k
0 = (5 cos 0)i + (5 sin 0)j =
1
Identify the first error in the work shown.
/25 (cos² 0 + sin²0)
The surface integral is written incorrectly.
No errors exist in the work shown.
The parametrization of the cylinder is incorrect.
The normal vector N(0, z) is incorrect.
(5 cos 0, 5 sin 0, 0)
√25 = 5
Calculus, Single Variable: Early Transcendentals (3rd Edition)
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