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A diagram of a curve in quadrant one. Several points and segments are labeled. Starting at the left, the first points are P_0 and P_1. The segment between them is labeled delta S_1. The next points are P_i-1, P_i, and P_i+1. The segments connecting them are delta S_i and delta S_j+1. Point P_i starred and point P_i+1 starred are located on each segment, respectively. The last two points are P_n-1 and P_n, connected by segment S_n.
Curve C has been divided into n pieces, and a point inside each piece has been chosen.

You may have noticed a difference between this definition of a scalar line integral and a single-variable integral. In this definition, the arc lengths Δ s 1 , Δ s 2 ,… , Δ s n aren’t necessarily the same; in the definition of a single-variable integral, the curve in the x -axis is partitioned into pieces of equal length. This difference does not have any effect in the limit. As we shrink the arc lengths to zero, their values become close enough that any small difference becomes irrelevant.

Definition

Let f be a function with a domain that includes the smooth curve C that is parameterized by r ( t ) = x ( t ) , y ( t ) , z ( t ) , a t b . The scalar line integral    of f along C is

C f ( x , y , z ) d s = lim n i = 1 n f ( P i * ) Δ s i

if this limit exists ( t i * and Δ s i are defined as in the previous paragraphs). If C is a planar curve, then C can be represented by the parametric equations x = x ( t ) , y = y ( t ) , and a t b . If C is smooth and f ( x , y ) is a function of two variables, then the scalar line integral of f along C is defined similarly as

C f ( x , y ) d s = lim n i = 1 n f ( P i * ) Δ s i ,

if this limit exists.

If f is a continuous function on a smooth curve C , then C f d s always exists. Since C f d s is defined as a limit of Riemann sums, the continuity of f is enough to guarantee the existence of the limit, just as the integral a b g ( x ) d x exists if g is continuous over [ a , b ] .

Before looking at how to compute a line integral, we need to examine the geometry captured by these integrals. Suppose that f ( x , y ) 0 for all points ( x , y ) on a smooth planar curve C . Imagine taking curve C and projecting it “up” to the surface defined by f ( x , y ) , thereby creating a new curve C that lies in the graph of f ( x , y ) ( [link] ). Now we drop a “sheet” from C down to the xy-plane. The area of this sheet is C f ( x , y ) d s . If f ( x , y ) 0 for some points in C , then the value of C f ( x , y ) d s is the area above the xy-plane less the area below the xy-plane. (Note the similarity with integrals of the form a b g ( x ) d x . )

A diagram in three dimensions. The original curve C in the (x,y) plane looks like a parabola opening to the left with vertex in quadrant 1. The surface defined by f(x,y) is shown always above the (x,y) plane. A curve on the surface directly above the original curve C is labeled as C’. A blue sheet stretches down from C’ to C.
The area of the blue sheet is C f ( x , y ) d s .

From this geometry, we can see that line integral C f ( x , y ) d s does not depend on the parameterization r ( t ) of C . As long as the curve is traversed exactly once by the parameterization, the area of the sheet formed by the function and the curve is the same. This same kind of geometric argument can be extended to show that the line integral of a three-variable function over a curve in space does not depend on the parameterization of the curve.

Finding the value of a line integral

Find the value of integral C 2 d s , where C is the upper half of the unit circle.

The integrand is f ( x , y ) = 2 . [link] shows the graph of f ( x , y ) = 2 , curve C , and the sheet formed by them. Notice that this sheet has the same area as a rectangle with width π and length 2. Therefore, C 2 d s = 2 π .

A graph in three dimensions. There is a flat plane just above the (x,y) plane. The upper half of the unit circle in quadrants 1 and 2 of the (x,y) plane is raised up to form a semicircle sheet into the z-plane.
The sheet that is formed by the upper half of the unit circle in a plane and the graph of f ( x , y ) = 2 .

To see that C 2 d s = 2 π using the definition of line integral, we let r ( t ) be a parameterization of C . Then, f ( r ( t i ) ) = 2 for any number t i in the domain of r . Therefore,

C f d s = lim n i = 1 n f ( r ( t i * ) ) Δ s i = lim n i = 1 n 2 Δ s i = 2 lim n i = 1 n 2 Δ s i = 2 ( length of C ) = 2 π .
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Questions & Answers

what is the stm
Brian Reply
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Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
LITNING Reply
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LITNING
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write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
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what king of growth are you checking .?
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What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
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why we need to study biomolecules, molecular biology in nanotechnology?
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Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
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Kyle
biomolecules are e building blocks of every organics and inorganic materials.
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anyone know any internet site where one can find nanotechnology papers?
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research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
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Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
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s. Reply
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NANO
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are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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