![Types of Line Integrals:
1. Scalar line integral: ∫[C] f(x,y,z) ds, where f is a scalar
function and ds is the arc length element.
2. Vector line integral: ∫[C] F(x,y,z) · dr, where F is a vector
field and dr is the differential displacement vector.
Applications:
1.Work calculation: Line integrals can be used to calculate the
work done by a force field on an object moving along a curve.
2. Flux calculation: Line integrals can be used to calculate the flux
of a vector field across a curve.](https://image.slidesharecdn.com/lec-line-integrals-23042025-085907pm-250623174220-b3b69fc2/85/Line-Integral-Multivariable-Caclulus-great-slides-2-320.jpg)
![Suppose we have a force field
F(x,y,z) = (-y, x, z)
and an object moves along a curve C defined by
x = cos(t), y = sin(t), z = t for t in [0, 2π].
The work done by the force field can be calculated
using the line integral:](https://image.slidesharecdn.com/lec-line-integrals-23042025-085907pm-250623174220-b3b69fc2/85/Line-Integral-Multivariable-Caclulus-great-slides-8-320.jpg)
![W = ∫[C] F · dr= ∫[0,2π] (-sin(t), cos(t), t) ·
(-sin(t), cos(t), 1) dt
= ∫[0,2π] (sin^2(t) + cos^2(t) + t) dt= ∫[0,2π]
(1 + t) dt
= [t + t^2/2] from 0 to 2π= 2π + 2π^2](https://image.slidesharecdn.com/lec-line-integrals-23042025-085907pm-250623174220-b3b69fc2/85/Line-Integral-Multivariable-Caclulus-great-slides-9-320.jpg)
![Suppose we have a force field
F(x,y,z) = <2xy, 4yz, xz> and an object
moves along a curve C defined by
<t^2, t, 2t> for t in [0, 2].
The work done by the force field can be
calculated using the line integral:
W = ∫[C] F · dr](https://image.slidesharecdn.com/lec-line-integrals-23042025-085907pm-250623174220-b3b69fc2/85/Line-Integral-Multivariable-Caclulus-great-slides-10-320.jpg)
![W = ∫[C] F · dr
= ∫[0,2] <2t^3, 8t^2, 2t^3 > ·
<2t, 1 , 2> dt
= ∫[0,2] (4t^4 + 8t^2+ 4t^3) dt
=[4t^5/5 + 8t^3/3 + t^4 ]
from 0 to 2](https://image.slidesharecdn.com/lec-line-integrals-23042025-085907pm-250623174220-b3b69fc2/85/Line-Integral-Multivariable-Caclulus-great-slides-11-320.jpg)
Line Integral Multivariable Caclulus great slides
- 1. Line integrals A line integral is a mathematical concept that calculates the total value of a function along a curve or path in a multidimensional space.
- 2. Types of Line Integrals: 1. Scalar line integral: ∫[C] f(x,y,z) ds, where f is a scalar function and ds is the arc length element. 2. Vector line integral: ∫[C] F(x,y,z) · dr, where F is a vector field and dr is the differential displacement vector. Applications: 1.Work calculation: Line integrals can be used to calculate the work done by a force field on an object moving along a curve. 2. Flux calculation: Line integrals can be used to calculate the flux of a vector field across a curve.
- 3. Evaluation: Line integrals can be evaluated using various techniques, including: 1.Parameterization: Parameterize the curve C and evaluate the integral using the parameterization. 2.Vectorization: Taking vectors of function and the corresponding curve
- 8. Suppose we have a force field F(x,y,z) = (-y, x, z) and an object moves along a curve C defined by x = cos(t), y = sin(t), z = t for t in [0, 2π]. The work done by the force field can be calculated using the line integral:
- 9. W = ∫[C] F · dr= ∫[0,2π] (-sin(t), cos(t), t) · (-sin(t), cos(t), 1) dt = ∫[0,2π] (sin^2(t) + cos^2(t) + t) dt= ∫[0,2π] (1 + t) dt = [t + t^2/2] from 0 to 2π= 2π + 2π^2
- 10. Suppose we have a force field F(x,y,z) = <2xy, 4yz, xz> and an object moves along a curve C defined by <t^2, t, 2t> for t in [0, 2]. The work done by the force field can be calculated using the line integral: W = ∫[C] F · dr
- 11. W = ∫[C] F · dr = ∫[0,2] <2t^3, 8t^2, 2t^3 > · <2t, 1 , 2> dt = ∫[0,2] (4t^4 + 8t^2+ 4t^3) dt =[4t^5/5 + 8t^3/3 + t^4 ] from 0 to 2
- 12. = 25.6 +21.3+16 – 0.8- 2.66 -1 = 58.44