# Line integrals

## 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:

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

## 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?

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HallsofIvy
Homework Helper

## 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:

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

## 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 $e^6 e^{5t}$.

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