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Evaluate the Following Line Integral Part 2

  • Thread starter bugatti79
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  • #1
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Homework Statement



[itex]\displaystyle \int_c zdx+xdy+ydz[/itex] where C is given by [itex]t^2\vec i +t^3 \vec j +t^2 \vec k[/itex]

Can this [itex]\displaystyle \int_c zdx+xdy+ydz[/itex] be written as [itex]\displaystyle \int_c z\vec i+x \vec j+y \vec k[/itex]?

I believe I need to evalute the integral [itex]\displaystyle \int_c \vec F( \vec r (t)) d\vec r(t)[/itex]....?

How do I proceed?
 

Answers and Replies

  • #2
I like Serena
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Hi again bugatti79! :smile:

Homework Statement



[itex]\displaystyle \int_c zdx+xdy+ydz[/itex] where C is given by [itex]t^2\vec i +t^3 \vec j +t^2 \vec k[/itex]

Can this [itex]\displaystyle \int_c zdx+xdy+ydz[/itex] be written as [itex]\displaystyle \int_c z\vec i+x \vec j+y \vec k[/itex]?
It should be:
[itex]\int_c (z\vec i+x \vec j+y \vec k) \cdot d\vec r(t)[/itex]


I believe I need to evalute the integral [itex]\displaystyle \int_c \vec F( \vec r (t)) d\vec r(t)[/itex]....?

How do I proceed?
What would [itex]d\vec r(t)[/itex] be?
 
  • #3
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Hi again bugatti79! :smile:



It should be:
[itex]\int_c (z\vec i+x \vec j+y \vec k) \cdot d\vec r(t)[/itex]
How did you establish this?

What would [itex]d\vec r(t)[/itex] be
From what you have shown I rearranged to find [itex]d\vec r = \vec i dx+\vec j dy+ \vec k dz[/itex]...........?

No, that wouldnt be right as dr is a function of t.....

[itex]d\vec r = d(t^2 \vec i +t^3\vec j+ t^2\vec k)[/itex]
 
Last edited:
  • #4
I like Serena
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How did you establish this?
What you have is a dot product:
[itex]zdx+xdy+ydz=(z,x,y) \cdot (dx,dy,dz)[/itex]

The vector (z,x,y) corresponds to your function [itex]\vec F(\vec r)[/itex]

The vector (dx,dy,dz) corresponds to your [itex]d\vec r(t)[/itex]

The vector (x,y,z) corresponds to your [itex]\vec r(t)[/itex]



From what you have shown I rearranged to find [itex]d\vec r = \vec i dx+\vec j dy+ \vec k dz[/itex]...........?

No, that wouldnt be right as dr is a function of t.....

[itex]d\vec r = d(t^2 \vec i +t^3\vec j+ t^2\vec k)[/itex]
Yes, the latter.
Can you work this out as a derivative?
 
  • #5
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Hi again bugatti79! :smile:



It should be:
[itex]\int_c (z\vec i+x \vec j+y \vec k) \cdot d\vec r(t)[/itex]




What would [itex]d\vec r(t)[/itex] be?
[itex]\displaystyle \int_c (z\vec i+x \vec j+y \vec k) \cdot d\vec r(t)=\int_c (t^2 \vec i+t^2 \vec j+t^3 \vec k) \cdot (2t \vec i + 3t^2 \vec j + 2t \vec k)dt = \int_{0}^{1} (2t^3 +3t^4+2t^4) dt[/itex].......?
 
  • #6
I like Serena
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[itex]\displaystyle \int_c (z\vec i+x \vec j+y \vec k) \cdot d\vec r(t)=\int_c (t^2 \vec i+t^2 \vec j+t^3 \vec k) \cdot (2t \vec i + 3t^2 \vec j + 2t \vec k)dt = \int_{0}^{1} (2t^3 +3t^4+2t^4) dt[/itex].......?
Yep!
(That was easy, was it not? :wink:)
 
  • #7
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Cheers! There will be more trickier ones along the way. :-) Thanks
 

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