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which is analogous to property iv.
Find the value of integral ${\int}_{C}\text{F}\xb7\text{T}}ds,$ where C is the rectangle (oriented counterclockwise) in a plane with vertices $\left(0,0\right),\left(2,0\right),\left(2,1\right),\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\left(0,1\right),$ and where $\text{F}=\u27e8x-2y,y-x\u27e9$ ( [link] ).
Note that curve C is the union of its four sides, and each side is smooth. Therefore C is piecewise smooth. Let ${C}_{1}$ represent the side from $\left(0,0\right)$ to $\left(2,0\right),$ let ${C}_{2}$ represent the side from $\left(2,0\right)$ to $\left(2,1\right),$ let ${C}_{3}$ represent the side from $\left(2,1\right)$ to $\left(0,1\right),$ and let ${C}_{4}$ represent the side from $\left(0,1\right)$ to $\left(0,0\right)$ ( [link] ). Then,
We want to compute each of the four integrals on the right-hand side using [link] . Before doing this, we need a parameterization of each side of the rectangle. Here are four parameterizations (note that they traverse C counterclockwise):
Therefore,
Notice that the value of this integral is positive, which should not be surprising. As we move along curve C _{1} from left to right, our movement flows in the general direction of the vector field itself. At any point along C _{1} , the tangent vector to the curve and the corresponding vector in the field form an angle that is less than 90°. Therefore, the tangent vector and the force vector have a positive dot product all along C _{1} , and the line integral will have positive value.
The calculations for the three other line integrals are done similarly:
and
Thus, we have ${\int}_{C}\text{F}}\xb7d\text{r}=2.$
Calculate line integral ${\int}_{C}\text{F}}\xb7d\text{r},$ where F is vector field $\u27e8{y}^{2},2xy+1\u27e9$ and C is a triangle with vertices $\left(0,0\right),$ $\left(4,0\right),$ and $\left(0,5\right),$ oriented counterclockwise.
0
Scalar line integrals have many applications. They can be used to calculate the length or mass of a wire, the surface area of a sheet of a given height, or the electric potential of a charged wire given a linear charge density. Vector line integrals are extremely useful in physics. They can be used to calculate the work done on a particle as it moves through a force field, or the flow rate of a fluid across a curve. Here, we calculate the mass of a wire using a scalar line integral and the work done by a force using a vector line integral.
Suppose that a piece of wire is modeled by curve C in space. The mass per unit length (the linear density) of the wire is a continuous function $\rho \left(x,y,z\right).$ We can calculate the total mass of the wire using the scalar line integral ${\int}_{C}\rho \left(x,y,z\right)ds.$ The reason is that mass is density multiplied by length, and therefore the density of a small piece of the wire can be approximated by $\rho \left(x*,y*,z*\right)\text{\Delta}s$ for some point $\left(x*,y*,z*\right)$ in the piece. Letting the length of the pieces shrink to zero with a limit yields the line integral ${\int}_{C}\rho \left(x,y,z\right)ds.$
Calculate the mass of a spring in the shape of a curve parameterized by $\u27e8t,2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t,2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t\u27e9,$ $0\le t\le \frac{\pi}{2},$ with a density function given by $\rho (x,y,z)={e}^{x}+yz$ kg/m ( [link] ).
To calculate the mass of the spring, we must find the value of the scalar line integral ${\int}_{C}\left({e}^{x}+yz\right)ds},$ where C is the given helix. To calculate this integral, we write it in terms of t using [link] :
Therefore, the mass is $\sqrt{5}\left({e}^{\pi \text{/}2}+1\right)$ kg.
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