Fundamental Theorem for Line Integrals

**saxyliz** · November 16th 2008, 02:14 PM

Suppose that

If

, find

.
Hint: As a first step, define a path from (0,0,0) to (1, 1, 5) and compute a line integral.

I have no idea how to even set this up! I've been having a lot of troubles with complex line integrals, please help!

**Jhevon** · November 16th 2008, 02:22 PM

Originally Posted by saxyliz

Suppose that

If

, find

.
Hint: As a first step, define a path from (0,0,0) to (1, 1, 5) and compute a line integral.

I have no idea how to even set this up! I've been having a lot of troubles with complex line integrals, please help!

parameterize the line curve.

$x = t,~y = t, ~z = 5t$ for $0 \le t \le 1$ works as a parameterization.

then find the required integral and then apply the fundamental theorem.

**saxyliz** · November 16th 2008, 02:30 PM

Originally Posted by Jhevon

parameterize the line curve.

$x = t,~y = t, ~z = 5t$ for $0 \le t \le 1$ works as a parameterization.

then find the required integral and then apply the fundamental theorem.

I'm having trouble integrating, I'm supposed to use the F(r(t)) [dot] r'(t) form for integration, right?

**Jhevon** · November 16th 2008, 02:42 PM

Originally Posted by saxyliz

I'm having trouble integrating, I'm supposed to use the F(r(t)) [dot] r'(t) form for integration, right?

the theorem says

$\int_C \nabla f \cdot d \bold{r} = f(\bold{r}(b)) - f(\bold{r}(a))$

but we also know that $\int_C \bold{F} \cdot d \bold{r} = \int_C \nabla f \cdot d \bold{r}$

Thus you have $\int_C \nabla f \cdot d \bold{r} = f(1,1,5) - f(0,0,0)$

where $C$ is the curve i parameterized for you. you can find the integral on the left (similar to the way you pointed out, except you do not want big F), and you know the value of $f(0,0,0)$ so you can find $f(1,1,5)$

**saxyliz** · November 16th 2008, 02:53 PM

Originally Posted by Jhevon

the theorem says

$\int_C \nabla f \cdot d \bold{r} = f(\bold{r}(b)) - f(\bold{r}(a))$

but we also know that $\int_C \bold{F} \cdot d \bold{r} = \int_C \nabla f \cdot d \bold{r}$

Thus you have $\int_C \nabla f \cdot d \bold{r} = f(1,1,5) - f(0,0,0)$

where $C$ is the curve i parameterized for you. you can find the integral on the left (similar to the way you pointed out, except you do not want big F), and you know the value of $f(0,0,0)$ so you can find $f(1,1,5)$

I know I must sound really stupid right now, but I knew all of that, I just been having a heck of a time integrating $\int_C \nabla f \cdot d \bold{r}$ . I know that I use the given function and the parameterization you graciously gave me and plug it in, but I just can't integrate the resulting functions!

**Jhevon** · November 16th 2008, 03:03 PM

Originally Posted by saxyliz

I know I must sound really stupid right now, but I knew all of that, I just been having a heck of a time integrating $\int_C \nabla f \cdot d \bold{r}$ . I know that I use the given function and the parameterization you graciously gave me and plug it in, but I just can't integrate the resulting functions!

the curve is $r(t) = \left< t, t, 5t \right>$ for $0 \le t \le 1$ . and so, $r'(t) = \left< 1,1,5 \right>$

so that $\int_C \nabla f \cdot d \bold{r} = \int_0^1 \nabla f(t,t,5t) \cdot r'(t)~dt$

can you finish now? (or did you know that already?)

tell me the integral you ended up with

**saxyliz** · November 16th 2008, 03:12 PM

Originally Posted by Jhevon

the curve is $r(t) = \left< t, t, 5t \right>$ for $0 \le t \le 1$ . and so, $r'(t) = \left< 1,1,5 \right>$

so that $\int_C \nabla f \cdot d \bold{r} = \int_0^1 \nabla f(t,t,5t) \cdot r'(t)~dt$

can you finish now? (or did you know that already?)

tell me the integral you ended up with

I did already know that... the only integral I can't solve is 10t^3e^t^2 (sorry, I don't know how to use the math function properly, hopefully you get my idea), the first integral. Is that integral wrong?

**Jhevon** · November 16th 2008, 03:24 PM

Originally Posted by saxyliz

I did already know that... the only integral I can't solve is 10t^3e^t^2 (sorry, I don't know how to use the math function properly, hopefully you get my idea), the first integral. Is that integral wrong?

(why didn't you say that from the beginnining?)

to integrate $10t^3 e^{t^2}$ , make a substitution $y = t^2$ , then the integral becomes $5 \int ye^y~dy$ . which you can do by parts (relatively) easily

**saxyliz** · November 16th 2008, 03:31 PM

Originally Posted by Jhevon

(why didn't you say that from the beginnining?)

to integrate $10t^3 e^{t^2}$ , make a substitution $y = t^2$ , then the integral becomes $5 \int ye^y$ . which you can do by parts (relatively) easily

Oh man, thank you so much! I'm in calc for engineers, so we don't do complex integrals anymore, I had forgotten. Thank you so so much for your time and patience!!

**Jhevon** · November 16th 2008, 04:06 PM

Originally Posted by saxyliz

Oh man, thank you so much! I'm in calc for engineers, so we don't do complex integrals anymore, I had forgotten. Thank you so so much for your time and patience!!

don't mention it. and say "complicated integrals", "complex integrals" actually refers to a special class of integrals--not the type we're doing now

good luck with your class

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