Line integral w/ constants only only in integrand

In summary, the integrand for the line integration over the curve C is composed of constants, however, the line integrals are not substituting correctly.
  • #1
AO eye 5
20
0
Hello,

I want help in the line integration of:

Integral( 1 dy + 3 dx ), over the curve C. Where C is the union of two line segments:

Line 1 from point (0,0) to (1, -3)

Line 2 from point (1, -3) to (2,0)

The thing is I do not know what to do with the integrand being composed of constants, as I am only aware of line integral examples with functions in the integand and subsequent substitution with parameterizations.

Hints are welcome, I am stuck upon the integrand substitution with the parameterization of the lines into the constant integrand.

I have C parameterized as:

Line 1: x = t y = -3t t [0,1]
Line 2: x = t+1 y = 3t-3 t [1,2]
 
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  • #2
line integrals...

[tex]\int_{C} 1 \, dy + 3 \, dx[/tex]

where C is the union of line 1 and line 2. Now on line 1, we have x=t, y=-3t, [tex]t\in \left[ 0,1\right][/tex] so that dx = dt and dy = -3dt on line 1, the integral over line 1 is then

[tex]\int_{\mbox{Line 1}} 1 \, dy + 3 \, dx = \int_{0}^{1} 1(-3 dt)+3 (dt) = 0[/tex]

now you do the integral over Line #2.
 
  • #3
AO eye 5 said:
Hello,

I want help in the line integration of:

Integral( 1 dy + 3 dx ), over the curve C. Where C is the union of two line segments:

Line 1 from point (0,0) to (1, -3)

Line 2 from point (1, -3) to (2,0)

The thing is I do not know what to do with the integrand being composed of constants, as I am only aware of line integral examples with functions in the integand and subsequent substitution with parameterizations.

Hints are welcome, I am stuck upon the integrand substitution with the parameterization of the lines into the constant integrand.

I have C parameterized as:

Line 1: x = t y = -3t t [0,1]
Line 2: x = t+1 y = 3t-3 t [1,2]
So on line 1, dx= dt, dy= -3dt. The integral on that is:
[tex]\int_0^1 1(-3dt)+ 3(dt)= 0\int_0^1dt= 0[/tex]

On line 2, dx= dt, dy= 3dt. The integral on the second line is:
[tex]\int_0^1 1(3dt)+ 3(dt)= 6\int_0^1dt= 6[/tex]

Actually, there's an easy way to do that: [itex]\int_a^b dx= b-a[/itex] so the integral of dy as y goes from 0 to -3 is -3, the integral of 3dx as x goes from 0 to 1 is 3, the integral of dy as y goes from -3 to 0 is 3, the integral of 3dx as x goes from 1 to 2 is 3(1)= 3. The entire integral is
-3+ 3+ 3+ 3= 6.
 

Related to Line integral w/ constants only only in integrand

1. What is a line integral with constants only in the integrand?

A line integral with constants only in the integrand is a type of integral that involves integrating a function along a line or curve in a two-dimensional or three-dimensional space. The integrand only contains constants, which means that the function does not change along the path of integration.

2. How is a line integral with constants only in the integrand different from a regular line integral?

A regular line integral involves integrating a function that can vary along a line or curve, while a line integral with constants only in the integrand involves integrating a function that remains constant along the path of integration. This makes the calculation of the integral simpler and more straightforward.

3. What are some common applications of line integrals with constants only in the integrand?

Line integrals with constants only in the integrand are commonly used in physics and engineering to calculate work done by a constant force, mass flow rate, and other physical quantities. They also have applications in vector calculus, where they are used to calculate the circulation of a vector field along a closed curve.

4. Can a line integral with constants only in the integrand be evaluated using any integration method?

Yes, a line integral with constants only in the integrand can be evaluated using any integration method, such as the fundamental theorem of calculus, substitution, or integration by parts. However, since the integrand is constant, the integration becomes simpler and does not require any advanced techniques.

5. Is it possible for a line integral with constants only in the integrand to have a non-zero value?

Yes, it is possible for a line integral with constants only in the integrand to have a non-zero value. This can occur if the path of integration is not a closed curve or if the constants in the integrand have different values. In these cases, the integral will evaluate to a constant multiple of the length of the path of integration.

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