Lateral Surface Area In Exercises 65-72, find the area of the lateral surface (see figure) over the curve C in the xy-plane and under the surface z = f ( x , y ) where Lateral surface area = ∫ C f ( x , y ) d s . f ( x , y ) = x 2 − y 2 + 4 , C : x 2 + y 2 = 4
Lateral Surface Area In Exercises 65-72, find the area of the lateral surface (see figure) over the curve C in the xy-plane and under the surface z = f ( x , y ) where Lateral surface area = ∫ C f ( x , y ) d s . f ( x , y ) = x 2 − y 2 + 4 , C : x 2 + y 2 = 4
Solution Summary: The author calculates the value of the lateral surface over the curve C in xy -plane. The parametrization form is r(t)=2mathrmco
Lateral Surface Area In Exercises 65-72, find the area of the lateral surface (see figure) over the curve C in the xy-plane and under the surface
z
=
f
(
x
,
y
)
where Lateral surface
area
=
∫
C
f
(
x
,
y
)
d
s
.
Evaluating a Surface Integral. Evaluate ∫∫ f(x, y, z)dS, where S
f(x,y,z)=√(x2+y2+z2), S:x2+y2 =9, 0⩽x⩽3, 0⩽y⩽3, 0⩽z⩽9.
Stokes’ Theorem for evaluating surface integrals Evaluate the line integral in Stokes’ Theorem to determine the value of the surface integral ∫∫S (∇ x F) ⋅ n dS. Assume n points in an upward direction.
F = ⟨4x, -8z, 4y⟩; S is the part of the paraboloidz = 1 - 2x2 - 3y2 that lies within the paraboloid z = 2x2 + y2 .
Stokes’ Theorem for evaluating surface integrals Evaluate the line integral in Stokes’ Theorem to determine the value of the surface integral ∫∫S (∇ x F) ⋅ n dS. Assume n points in an upward direction.
F = ⟨ex, 1/z, y⟩; S is the part of the surface z = 4 - 3y2 thatlies within the paraboloid z = x2 + y2.
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