How Do You Calculate the Line Integral of a Vector Field Along a Segment?

[-EquinoX-]

Homework Statement

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

\int\limits_C \vec{F} d\vec{r}

Homework Equations

The Attempt at a Solution

First I parametrize C as:

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

Then I set up the integral as:

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

is this correct so far?

 

[HallsofIvy]

Homework Statement

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


I need to find:

\int\limits_C \vec{F} d\vec{r}

Homework Equations

The Attempt at a Solution

First I parametrize C as:

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

I assume you mean \vec{r}= (6+ 5t)\vec{i}+ (7+ 3t)\vec{j}- 2\vec{k}.
When t= 0 this is \vec{r}= 6\vec{i}+ 7\vec{j}- 2\vec{k} which is one of the given points. When t= -1 this is \vec{r}= \vec{i}+ 4\vec{j}- 2\vec{k} 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:

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

is this correct so far?

Yes, that is correct. You will probably wnat to "separate" e^{6+ 5t} as e^6 e^{5t}.

 

[-EquinoX-]

and so after doing all the computation I got:

\frac{12(y^2-x^2)}{(x^2+y^2)^2} being evaluated from -1 to 0. Is this correct? Reason I ask is because the integral was kind of complex

 

[HallsofIvy]

Why did you change from t to x and y?

 

[-EquinoX-]

oops.. I guess you're right

 

[-EquinoX-]

I got \frac{9}{5} - 4e