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Evaluate the line integral by two methods

  • Thread starter tony873004
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  • #1
tony873004
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Homework Statement


Evaluate the line integral by two methods: (a) directly and (b) using Green's Theorem. [tex]\oint_C {xy^2 \,dx\, + \,x^3 \,dy} [/tex]. C is the rectangle with vertices (0,0), (2,0), (2,3) and (0,3).

The Attempt at a Solution


I get the correct answer using Green's theorem. But I don't know how to do the "directly" method. I came up with 4 parametric equations, one for each line of the rectangle. Let's look at the first one for the line (0,0) - (2,0): x=t, y=0, t goes between 0 and 2.

[tex]\oint_C {xy^2 \,dx\, + \,x^3 \,dy} = \int\limits_0^2 {t\left( 0 \right)^2 \,\, + \,t^3 \,dt} = \int\limits_0^2 {t^3 \,dt} = \left[ {\frac{{t^4 }}{4}} \right]_0^2 = \left( {\frac{{2^4 }}{4}} \right) - \left( {\frac{{0^4 }}{4}} \right) = \frac{{16}}{4} = 4[/tex].

I'm pretty sure I did this step wrong. How do I go from the C integral to the 2-0 integral?

Thanks


Homework Equations


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
dx
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Think of [tex] {xy^2 \,dx\, + \,x^3 \,dy} [/tex] as the dot product of the vectors [tex](dx,dy)[/tex] and [tex](xy^2, x^3) [/tex]. On the x axis, the x component is zero, so the integral from (0,0) to (0,2) should be zero.
 
  • #3
tony873004
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sorry, I still don't understand. How do I get from dx+dy to dt?
 
  • #4
nrqed
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Homework Statement


Evaluate the line integral by two methods: (a) directly and (b) using Green's Theorem. [tex]\oint_C {xy^2 \,dx\, + \,x^3 \,dy} [/tex]. C is the rectangle with vertices (0,0), (2,0), (2,3) and (0,3).

The Attempt at a Solution


I get the correct answer using Green's theorem. But I don't know how to do the "directly" method. I came up with 4 parametric equations, one for each line of the rectangle. Let's look at the first one for the line (0,0) - (2,0): x=t, y=0, t goes between 0 and 2.

[tex]\oint_C {xy^2 \,dx\, + \,x^3 \,dy} = \int\limits_0^2 {t\left( 0 \right)^2 \,\, + \,t^3 \,dt} = \int\limits_0^2 {t^3 \,dt} = \left[ {\frac{{t^4 }}{4}} \right]_0^2 = \left( {\frac{{2^4 }}{4}} \right) - \left( {\frac{{0^4 }}{4}} \right) = \frac{{16}}{4} = 4[/tex].

I'm pretty sure I did this step wrong. How do I go from the C integral to the 2-0 integral?

Thanks


Homework Equations


Homework Statement





Homework Equations





The Attempt at a Solution


Your line is along the x axis so dy is zero for that section of the rectangle!
 
  • #5
Hootenanny
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sorry, I still don't understand. How do I get from dx+dy to dt?
So you have paremerised the line,

[tex] x = t \Rightarrow \frac{dx}{dt} = 1 \Rightarrow dx = dx[/tex]

[tex]y = 0 \Rightarrow dy = 0[/tex]

Do you follow?
 
  • #6
tony873004
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Thanks, I think I'm getting it now. So in the next line segment (the right vertical one), dx is zero, so

[tex]\oint_C {xy^2 \,dx\, + \,x^3 \,dy} = \int\limits_0^3 {2t^2 \left( 0 \right)\, + \,2^3 \,dt} = \int\limits_0^3 {8\,dt} = \left[ {8t} \right]_0^3 = 24[/tex]
 
  • #7
Hootenanny
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Thanks, I think I'm getting it now. So in the next line segment (the right vertical one), dx is zero, so

[tex]\oint_C {xy^2 \,dx\, + \,x^3 \,dy} = \int\limits_0^3 {2t^2 \left( 0 \right)\, + \,2^3 \,dt} = \int\limits_0^3 {8\,dt} = \left[ {8t} \right]_0^3 = 24[/tex]
Looks good to me :approve:.

Although from a formalism point of view, I will point out that it isn't technically correct to say that your path integral is equal to the integral over this segment, but I know what you mean.
 

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