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Line integral is an integral that represents the integration of a function along a curve. A route integral, curvilinear integral, or curve integral is another name for a line integral. Line integrals are very useful for calculating work done by a force on an item traveling in a vector field, the mass of a wire, calculation of reaction rates, the position of a celestial body the center of mass of a wire, moments of inertia of a wire, the magnetic field that surrounds a conductor (Ampere’s law) and voltage created in a loop, Faraday's Law of Magnetic Induction.
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Key Terms: Line Integral, Green’s Theorem, Stoke’s Theorem, Vector Field, Scalar Field, Work done, Moment of Inertia, Divergence Theorem
Line Integral Definition
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An integral in which a function is integrated along a curve in the coordinate system is known as a line integral. A scalar field or a vector field can be used to represent the function to be integrated. Along a curve, we can integrate both scalar-valued and vector-valued functions. Summing all of the values of the points on the vector field yields the value of the vector line integral.
Definition of Line Integral
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Line Integral of the Vector Field
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The integral of a function along a curve is called a line integral. A scalar-valued function can be integrated along a curve to determine, for example, the mass of a wire from its density. A certain type of vector-valued function can also be integrated along a curve. These are vector-valued functions, which have the same input and output dimensions and are typically represented as vector fields.
Integrals Detailed Video Explanation
Line Integral Formula
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The formulas of line integral for the scalar field and vector field are mentioned below.
Line Integral Formula for Scalar Field
For a scalar field with function f: U ⊆ Rn → R, a line integral along with a smooth curve, C ⊂ U is defined as,
| \(\int C f(r) ds = \int_a^b f [r(t)] \mid r’(t) \mid dt \) |
r(a) and r(b) give the endpoints of C and a < b.
Line Integral Formula for Vector Field
For a vector field with function, F: U ⊆ Rn → Rn, a line integral along with a smooth curve C ⊂ U, in the direction “r” is defined as,
| \(\int C F(r). dr = \int_a^b F[r(t)] . r’(t)dt\) |
Applications of Line Integral
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There are a variety of uses for line integrals. The following are some examples of line integral applications in vector calculus.
Applications of Line Integral
A line integral can be used to compute the mass of a wire, as well as its moment of inertia and center of mass. Assume that a three-dimensional curve C describes a piece of wire. The mass of the wire per unit length is a continuous function ρ (x,y,z).
The total mass of the wire is then given as the scalar function's line integral as m= \(\int_C\) ρ(x,y,z)ds.
When applying Ampere's law, the line integral of a magnetic field B around a closed channel 'C' equals the total current flowing through the area limited by boundary 'C'.
The line integral describes the work done W by a force F on an object traveling along a curve C, W=\(\int_C\)F.dr
- where F is the vector force field acting on the object and dr is the unit tangent vector.
- Line integrals are used in classical mechanics to compute the work done by an object of mass m traveling in a gravitational field.
- The three-dimensional surface areas can be calculated using line integrals. It's an extension of simple integrals that's best used on curved surfaces.
If the currents in all places can be precisely anticipated, one may calculate how many calories a swimmer will burn swimming along a specific route. The overall amount of effort he needs to complete depends on the current intensity and direction. As a result, a line integral over his route will aid in calculating the overall labor done or calories burned while swimming along the intended path.
- Line integrals are used in electrical engineering to determine the precise length of power cable required to connect two substations that may be thousands of miles apart.
- Line integrals are frequently used by space flight engineers for long trips. When launching exploratory satellites, the path of the various circling velocities of Earth and the planet the probe is aimed at is taken into account.
- Line integrals are also used to calculate an object's velocity and trajectory, estimate the positions of the planets, and gain a better understanding of electromagnetism.
- It is also utilized in statistics to assess survey data and devise effective methods.
- Line integrals are used in chemistry to calculate reaction rates and to gather information on radioactive decay reactions.
Facts about Line Integral
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- Area and volume calculation methods extend back to ancient Greek mathematics.
- The ideas of integration were separately developed by Isaac Newton and Gottfried Wilhelm Leibniz in the late 17th century.
- A line integral is used in Faraday's Law of Magnetic Induction to calculate the voltage created in a loop.
- The line integral is useful for calculating the work done by a force on a moving object in a vector field.
- A line integral might be used to calculate how much radiation a pirate would be exposed to from a radiation source near his treasure.
Some Important Formulas
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Line Integral formula for scalar field
| \(\int C f(r) ds = \int_a^b f[r(t)] \mid r’(t) \mid dt \) |
Line Integral formula for vector field
| \(\int C F(r). dr = \int_a^b F[r(t)] . r’(t)dt\) |
Formula for Work Done
| W= \(\int_C\) F.dr |
Formula to find the mass of wire
| m = \(\int_C\) ρ (x,y,z) ds. |
Things to Remember
- Line integrals are classified into two types: scalar line integrals and vector line integrals.
- Line Integral is used to calculate the magnetic field surrounding a conductor in Ampere's Law.
- Line integrals around closed curves are related to double integrals or surface integrals by Green's theorem and Stokes' theorem.
- They can be used in electromagnetics to compute the work done on a charged particle traveling along a curve in a force field represented by a vector field.
- They can be used to calculate the work done on a mass mm traveling in a gravitational field in classical mechanics.
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| Related Topics | ||
|---|---|---|
| Unit Vector | Calculus Formula | Scalar matrix |
| Real-Valued Functions | Operations on Rational Numbers | Vector Product |
Previous Years’ Questions
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Sample Questions
Ques: Define Line integrals? (2 Marks)
Ans: A-line integral (also known as a route integral) is the integral of a function that runs along a curve. A scalar-valued function can be integrated along a curve to determine, for example, the mass of a wire from its density. A specific sort of vector-valued function can also be integrated along a curve.
Ques: What are the applications of Line integrals? (3 Marks)
Ans: Line Integral has a wide range of uses. It's used to figure out how much space a three-dimensional shape takes up on the surface. The sole application of the line integral in vector calculus is described below.
- The magnitude of the wire is calculated using a line integral.
- Line integrals are used to calculate the work done on a mass mm traveling in a gravitational field in classical mechanics.
- The inertia moment and center of the magnitude of the wire are calculated using a line integral.
- It is employed in Ampere's law to determine the magnetic field surrounding a conductor.
- The voltage created in a loop in Faraday's law of magnetic induction can be examined using a line integral.
It's used to figure out how much work a force does on moving objects in a vector field.
Ques: Why do we employ Stokes' Theorem? (2 Marks)
Ans: Stokes' theorem establishes a connection between line and surface integrals. One integral can be computed in terms of the other for our convenience. Stokes' theorem is also used to calculate the curvature of a vector field.
Ques: Define Green’s Theorem. (1 Mark)
Ans: According to Green's theorem, the line integral equals the double integral of this quantity across the contained region.
Ques: When is Green's theorem applicable? (2 Marks)
Ans: Green's theorem transfers the line integral to the microscopic circulation's double integral. The double integral is computed over the path's region D. Only closed pathways contain a region D. Only for closed pathways can the concept of circulation make sense.
Ques: Define Green’s Theorem. (2 Marks)
Ans: According to the divergence theorem, adding up all the little amounts of outward flow in a volume using a triple integral of divergence yields the overall outward flow from that volume, as measured by the flux through its surface.
Ques: What is the formula of Line integrals for the scalar field? (3 Marks)
Ans: For a scalar field with function f: U ⊆ Rn → R, a line integral along with a smooth curve, C ⊂ U is defined as,
\(\int C f(r) ds = \int_a^b f[r(t)] \mid r’(t) \mid dt \)
where r: [a, b]→C is an arbitrary bijective parametrization of the curve.
r(a) and r(b) give the endpoints of C and a < b.
Ques: What is the formula of Line integrals for the vector field? (2 Marks)
Ans: For a vector field with function, F: U ⊆ Rn → Rn, a line integral along with a smooth curve C ⊂ U, in the direction “r” is defined as,
\(\int C F(r). dr = \int_a^b F[r(t)] . r’(t)dt\)
where “.” represents dot product.
Ques: Find the line integral ∫C F. dr where F(x, y, z) = [P(x, y, z), Q(x, y, z), R(x, y, z)] = (z, x, y), and Cis defined by the parametric equations, x = t2, y = t3 and z = t2 , 0 ≤ t ≤ 1. (5 Marks)
Ans: Given that, the function, F(x, y, z) = [P(x, y, z), Q(x, y, z), R(x, y, z)] = (z, x, y)
Parametric equations are x = t2, y = t3 and z = t2 , 0 ≤ t ≤ 1.
Now we know,
\(\int_C\) F. dr = ∫C P dx + Q dy + R dz
\(\int_C\) F. dr = \(\int_0^1\) z(t) x’(t)dt + x(t) y’(t)dt + y(t) z’(t)dt
= \(\int_0^1\) t2 (2t)dt + t2 (3t2)dt + t3 (2t) dt
= \(\int_0^1\) 2t3 dt + 3t4 dt + 2t4 dt
= \(\int_0^1\)(5t4 + 2t3) dt
=[5t5/5 + 2t4/4]\(_0^1\)
Substitute the limits, we get,
∫C F. dr = 3/2
Therefore, the line integral for the given function is ‘1.5’.
Ques: Evaluate \(\int_C\)F⋅dr where F(x,y,z) = 8x2yzi + 5zj − 4xykF(x,y,z) = 8x2yzi + 5zj − 4xyk and C is the curve given by r(t) = ti + t2j + t3k, 0 ≤ t ≤1. (5 Marks)
Ans: Okay, we first evaluate the vector field along the curve.
F(r(t)) = 8t2 (t2) (t3) i + 5t3j − 4t(t2)k = 8t7i + 5t3j − 4t3k
Next, the derivative of the parameterization are,
r′(t)= i + 2tj + 3t2k
Finally, we take dot product
F(r(t))⋅r′(t)= 8t7 + 10t4 − 12t5
Hence, The line integral is,
\(\int_C\)F⋅dr = \(\int_0^1\) 8t7 + 10t4 − 12t5dt
=[(t8 + 2t5 − 2t6)]\(_0^1\)
=1
Ques: Find the line integral of \(\int_C\) (1+x2y) ds Where C is considered as an ellipse r(t) = (2cos t) + (3sin t) for 0 ≤ t ≤ 2π (3 Marks)
Ans: ds =√(−2sint)2+(3cost)2dt =√4sin2t + 9cost2t
Now, we have the integral \(\int_0^{2 \pi}\)(1+(2cost)2(3sint))√4sin2t + 9cost2t
Hence, we get line integral = 15.87.
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