Green's Theorem in 3 dimensions problem

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Homework Help Overview

The problem involves evaluating a line integral in three dimensions, specifically the integral of the form \(\int_C{xydx - yzdy + xzdz}\) along a given parametric curve defined by \(\vec{r}(t) = t\vec{i} + t^2\vec{j} + t^4\vec{k}\) for \(0 \leq t \leq 1\). The original poster notes the inapplicability of Green's Theorem in this context.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the standard definition of a line integral and suggest expressing the variables in terms of the parameter \(t\). There is also mention of the fundamental theorem of line integrals and how to derive the differentials \(dx\), \(dy\), and \(dz\) from the parametric equations.

Discussion Status

The discussion is progressing with participants providing guidance on how to approach the integral using parametric equations. There is an acknowledgment of the original poster's understanding of the limitations of Green's Theorem, and participants are exploring the integration process without reaching a consensus on the final evaluation.

Contextual Notes

Participants are working under the constraint that Green's Theorem does not apply in this three-dimensional scenario, prompting a shift to line integral techniques. The original poster's initial confusion about the applicability of the theorem is noted.

duki
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Homework Statement



Evaluate: \int _C{xydx - yzdy + xzdz}
C: \vec{r}(t) = t\vec{i} + t^2\vec{j} + t^4\vec{k}
o <= t <= 1

Homework Equations



The Attempt at a Solution



I understand that you cannot use Green's Theorem in 3 dimensions. How else can I go about solving this?
 
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Just use the standard definition of a "line integral".

On C, x= t, y= t2, and z= t4
dx= dt, dy= 2tdt, and dz= 4t3.

xydx- yzdy+ xzdz= (t)(t2)dt- (t2)(t4)(2tdt)+ (t)(t4)(4t3dt.

Integrate that from t= 0 to t= 1.
 


Ohhh snap. I remember talking about the fundamental theorem of line integrals now. Thanks!
 


x, y and z will all be functions of t along the curve...You should be able to simply look at the parametric equation for your curve \vec{r}(t) = t\hat{i} + t^2\hat{j} + t^4\hat{k} and read off what those functions are.:wink:...Then you can easily express dx, dy and dz in terms of dt and you integral will simply be a single variable integration...

EDIT: Halls beat me to it.
 


Ok so I got:

EDIT:
\int _0^1{4t^4 dt} = \frac{4t^5}{5} =&gt; \frac{4}{5} ?
 
Last edited:

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