Evaluating a Line Integral In Exercises 19-22, evaluate the line integral along the given path. ∫ C 3 ( x − y ) d s C : r ( t ) = t i + ( 2 − t ) j 0 ≤ t ≤ 1
Evaluating a Line Integral In Exercises 19-22, evaluate the line integral along the given path. ∫ C 3 ( x − y ) d s C : r ( t ) = t i + ( 2 − t ) j 0 ≤ t ≤ 1
Solution Summary: The author explains how to calculate the line integral displaystyle undersetCint3(x-y)ds along the path.
Evaluating a Line Integral In Exercises 19-22, evaluate the line integral along the given path.
∫
C
3
(
x
−
y
)
d
s
C
:
r
(
t
)
=
t
i
+
(
2
−
t
)
j
0
≤
t
≤
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.
Displacement d→1 is in the yz plane 62.8 o from the positive direction of the y axis, has a positive z component, and has a magnitude of 5.10 m. Displacement d→2 is in the xz plane 37.0 o from the positive direction of the x axis, has a positive z component, and has magnitude 0.900 m. What are (a) d→1⋅d→2 , (b) the x component of d→1×d→2 , (c) the y component of d→1×d→2 , (d) the z component of d→1×d→2 , and (e) the angle between d→1 and d→2 ?
Nonuniform straight-line motion Consider the motion of an object given by the position function r(t) = ƒ(t)⟨a, b, c⟩ + ⟨x0, y0, z0⟩, for t ≥ 0,where a, b, c, x0, y0, and z0 are constants, and ƒ is a differentiable scalar function, for t ≥ 0.a. Explain why r describes motion along a line.b. Find the velocity function. In general, is the velocity constant in magnitude or direction along the path?
Application of Green's theorem
Assume that u and u are continuously differentiable functions. Using Green's theorem,
prove that
JS
D
Ur
Vy
dA=
u dv,
where D is some domain enclosed by a simple closed curve C with positive orientation.
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