# Line Integral Help Requested

• Jun 16th 2009, 02:49 PM
affslick
Line Integral Help Requested
I'm taking a 5-week Calc III summer course, and I'm working on a practice problem in preparation for the exam tomorrow morning. I've been doing fine up until we started on Line Integrals. I've read a number of tutorials, my book, etc., but this one keeps slipping by... Trying various methods, I've gotten the following numeric answers:

On Paper (with the assistance of a TI89): 270.899
Maple (using the VectorCalculus LineInt command): 15.739
Maple (working out each step): 147.927

The problem is as follows: (with some of my likely incorrect work):

Find the work done by the force:
$\vec{F}=(x+y)\vec{i}+(2x-y)\vec{j}$

acting on a body as it moves along the curve between the values:

$t$ between $0$ and $\pi$

with

$\vec{r}=2t\vec{i}+sin(t)\vec{j}$.

I'm computing:
$x=2t$
$y=sin(t)$
$r'(t)=[2, cos(t)]$

So...
$\vec{F}=(2t+sin(t))\vec{i}+(2(2t)-sin(y))\vec{j}$

$\int_0^\pi \! \vec{F} \cdot \left|\left|r \right|\right| \, dt=\int_0^\pi \! ((2t+sin(t))\vec{i}+(2(2t)-sin(y))\vec{j} \cdot \sqrt{(2t)^2 + (sin(t))^2}) \,dt$

I'll stop there, as my mistake is probably well before this point... If anyone could help me figure this out, with the simplest language possible, it would be greatly appreciated.

Thanks,
Chris
• Jun 16th 2009, 03:00 PM
Jester
Quote:

Originally Posted by affslick
I'm taking a 5-week Calc III summer course, and I'm working on a practice problem in preparation for the exam tomorrow morning. I've been doing fine up until we started on Line Integrals. I've read a number of tutorials, my book, etc., but this one keeps slipping by... Trying various methods, I've gotten the following numeric answers:

On Paper (with the assistance of a TI89): 270.899
Maple (using the VectorCalculus LineInt command): 15.739
Maple (working out each step): 147.927

The problem is as follows: (with some of my likely incorrect work):

Find the work done by the force:
$\vec{F}=(x+y)\vec{i}+(2x-y)\vec{j}$

acting on a body as it moves along the curve between the values:

$t$ between $0$ and $\pi$

with

$\vec{r}=2t\vec{i}+sin(t)\vec{j}$.

I'm computing:
$x=2t$
$y=sin(t)$
$r'(t)=[2, cos(t)]$

So...
$\vec{F}=(2t+sin(t))\vec{i}+(2(2t)-sin(y))\vec{j}$

$\int_0^\pi \! \vec{F} \cdot \left|\left|r \right|\right| \, dt=\int_0^\pi \! ((2t+sin(t))\vec{i}+(2(2t)-sin(y))\vec{j} \cdot \sqrt{(2t)^2 + (sin(t))^2}) \,dt$

I'll stop there, as my mistake is probably well before this point... If anyone could help me figure this out, with the simplest language possible, it would be greatly appreciated.

Thanks,
Chris

I believe what you want is

$
\int_c {\bf F} \cdot {\bf r}' dt
$
• Jun 16th 2009, 03:14 PM
affslick
Oh, the dangers of reading online notes... the prime mark ran into the arrow head, and so it was very hard to read. That would probably explain the conflict I was having...

So now I have:
$\int_0^pi ((4t-sin(t)) \cdot cos(t)+2 \cdot (sin(t)+2t))) \, dt$

The result of this is $2\pi^2 - 4$ which equals $15.739$, which appears to match the answer provided by Maple.

Thanks you so much for clearing that up. I spent about 5 hours banging my head against the wall wondering why it kept getting so strange...