Calcularing area vector using line integral

  • #1

Homework Statement


A closed curve C is described by the following equations in a Cartesian coordinate system:
gif.gif

gif.gif

gif.gif

where the parameter t runs monotonically from 0 to 2π, thus defining the direction of C. Calculate the area vector of the planar region enclosed by C, using the formula:
gif.gif


2. The attempt at a solution
I'm mostly having trouble defining what
gif.gif
is as a physical quantity. I think it is the distance to an incremented point along the curve such that
gif.gif
it the area of the equilateral shape formed by the vectors and that half the integral of that gives the area but i'm suck here:
gif.gif

Than I can evaluate it as:
gif.gif

and this would give a result that is only in the z direction which dimensionally makes sense
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


A closed curve C is described by the following equations in a Cartesian coordinate system:
gif.gif

gif.gif

gif.gif

where the parameter t runs monotonically from 0 to 2π, thus defining the direction of C. Calculate the area vector of the planar region enclosed by C, using the formula:
gif.gif


2. The attempt at a solution
I'm mostly having trouble defining what
gif.gif
is as a physical quantity. I think it is the distance to an incremented point along the curve such that
gif.gif
it the area of the equilateral shape formed by the vectors and that half the integral of that gives the area but i'm suck here:
gif.gif

Than I can evaluate it as:
gif.gif

and this would give a result that is only in the z direction which dimensionally makes sense

If
[tex] \vec{r} = x(t) \, {\bf i} + y(t) \, {\bf j} + z(t)\, {\bf k} [/tex]
how would you compute ##d \vec{r}## in terms of ##dt##?
 
  • #3
If
[tex] \vec{r} = x(t) \, {\bf i} + y(t) \, {\bf j} + z(t)\, {\bf k} [/tex]
how would you compute ##d \vec{r}## in terms of ##dt##?
My first instinct was just to derive each of them with respect to t like such
20y%28t%29%5Chat%7By%7D+%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20t%7D%20z%28t%29%5Chat%7Bz%7D.gif

gif.gif

Is that right?
 
  • #5
SteamKing
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My first instinct was just to derive each of them with respect to t like such
20y%28t%29%5Chat%7By%7D+%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20t%7D%20z%28t%29%5Chat%7Bz%7D.gif

gif.gif

Is that right?
Yes, go ahead and finish the evaluation of the integral ...
 
  • #6
Yes, go ahead and finish the evaluation of the integral ...
So http://www.sciweavers.org/upload/Tex2Img_1442441349/eqn.png [Broken] (by trig identities)
and so the integral is
http://www.sciweavers.org/upload/Tex2Img_1442441225/eqn.png [Broken]
 
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  • #7
SteamKing
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So http://www.sciweavers.org/upload/Tex2Img_1442441349/eqn.png [Broken] (by trig identities)
and so the integral is
http://www.sciweavers.org/upload/Tex2Img_1442441225/eqn.png [Broken]

There's a small mistake in your integration. You seem to have omitted integrating the constant (3/8) in your trig identity expression
 
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  • #8
There's a small mistake in your integration. You seem to have omitted integrating the constant (3/8) in your trig identity expression
I was hoping you wouldn't notice that, I was just too lazy to retype it into LaTex, but yes, it was included in my calculations (I appreciate your thoroughness though!)
 

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