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Line integrals

  • Thread starter -EquinoX-
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  • #1
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Homework Statement



say I have [tex] \vec{F} = ye^x\vec{i} + e^x\vec{j} [/tex] and C the line segment from the point (1,4,-2) to the point (6,7,-2).


I need to find:

[tex] \int\limits_C \vec{F} d\vec{r} [/tex]

Homework Equations





The Attempt at a Solution



First I parametrize C as:

[tex] \vec{r} = 6 + 5t \vec{i} + 7+3t \vec{j} -2 \vec{k} [/tex]

Then I set up the integral as:

[tex] \int_{-1}^{0} (7+3t)e^{6+5t} \vec{i} + e^{6+5t} \vec{j}) \cdot (5\vec{i} + 3\vec{j}) dt [/tex]

is this correct so far?
 

Answers and Replies

  • #2
HallsofIvy
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Homework Helper
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Homework Statement



say I have [tex] \vec{F} = ye^x\vec{i} + e^x\vec{j} [/tex] and C the line segment from the point (1,4,-2) to the point (6,7,-2).


I need to find:

[tex] \int\limits_C \vec{F} d\vec{r} [/tex]

Homework Equations





The Attempt at a Solution



First I parametrize C as:

[tex] \vec{r} = 6 + 5t \vec{i} + 7+3t \vec{j} -2 \vec{k} [/tex]
I assume you mean [itex]\vec{r}= (6+ 5t)\vec{i}+ (7+ 3t)\vec{j}- 2\vec{k}[/itex].
When t= 0 this is [itex]\vec{r}= 6\vec{i}+ 7\vec{j}- 2\vec{k}[/itex] which is one of the given points. When t= -1 this is [itex]\vec{r}= \vec{i}+ 4\vec{j}- 2\vec{k}[/itex] which is the other point. Yes, this is a vector equation for the line- although the choice of t= -1 for one of the is peculiar!

Then I set up the integral as:

[tex] \int_{-1}^{0} (7+3t)e^{6+5t} \vec{i} + e^{6+5t} \vec{j}) \cdot (5\vec{i} + 3\vec{j}) dt [/tex]

is this correct so far?
Yes, that is correct. You will probably wnat to "separate" [itex]e^{6+ 5t}[/itex] as [itex]e^6 e^{5t}[/itex].
 
  • #3
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and so after doing all the computation I got:

[tex] \frac{12(y^2-x^2)}{(x^2+y^2)^2} [/tex] being evaluated from -1 to 0. Is this correct? Reason I ask is because the integral was kind of complex
 
  • #4
HallsofIvy
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Why did you change from t to x and y?
 
  • #5
564
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oops.. I guess you're right
 
  • #6
564
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I got [tex] \frac{9}{5} - 4e [/tex]
 

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