# Line Integrals

1. Dec 9, 2008

### hils0005

1. The problem statement, all variables and given/known data
Calculate $$\int$$F dr if C = C1 + C2 where C1 is the line segment from P1(-1,pi,-1) to P2(0,0,0) and C2 is the line segment from P2(0,0,0) to P3(2,0,4)

vectorF= yz i + (xz - e^(z)siny) j + (e^(z)cosy + xy) k

3. The attempt at a solution
Im having problems setting up the C1 integral:
I have it parametrized as x=t, y=-pi t z=t, as t goes 0 to 1

r(t)= t i - pit j + t k

r'(t)= i - pi j + k

F= -pi t^2 i + (t^2-e^(t)sin(pit)) j + (e^(t)cos(pit) + pit^2) k

F dot r'(t) = -pi t^2 + pi e^(t)sin(pit) + e^tcos(pit)

I would then take integral dt as t goes from 0 to 1

I think I have answered C2 correctly, I'm not sure if my parameters are correct

2. Dec 9, 2008

### Defennder

Big time-saving hint: What is curl F ?

3. Dec 9, 2008

### hils0005

dF2/dx - dF1/dy =0 z-z=0
dF3/dy - dF2/dz =0 (-e^zsiny + x) + (-x + e^zsiny) = 0
dF1/dz - dF3/dx = 0 y-y=0

Curl test?? how does that help?

4. Dec 9, 2008

### Defennder

So what does that tell you about F?

5. Dec 9, 2008

### hils0005

that it is path independent or conservative?

Last edited: Dec 9, 2008
6. Dec 9, 2008

### gabbagabbahey

Yes, which also tells you that you can write F as the gradient of some scalar $\varphi$...So if you can find such a scalar, you can easily compute the path integral using the fundamental theorem of gradients.

However, the way your original question is worded leads e to believe that they want you to evaluate the path integral along that specific path, so I would use the gradient method only as a check.

7. Dec 9, 2008

### gabbagabbahey

That doesn't seem right; $r(0)=(0,0,0)$ which is not your starting point, and $r(1)=(1,-\pi,1)$ which is neither your starting point or your finish point!

8. Dec 9, 2008

### hils0005

x=t-1
y= pi t + t
z=t-1

t goes from -2 to -1,
would that work?

9. Dec 9, 2008

### gabbagabbahey

Nope; now your starting point is r(-2)=(-3,-2pi-2,-3) and your finish point is just as incorrect.

Let's think this through logically instead of stabbing blindly in the dark....You know that c1 is the direct line from P1 to P2; so you can write down a general parameterized line for r: $\mathbf{r}(t)=(at+b,ct+d,ft+g)$....say you let t go from 0 to 1; what must a,b,c,d,f and g be for $\mathbf{r}(0)$ to be P1 and $\mathbf{r}(1)$ to be P2?

10. Dec 10, 2008

### hils0005

x=t-1
y=pit^2
z=t-1

11. Dec 10, 2008

### gabbagabbahey

x and z look good, but y isn't even linear in t!...how did you come up with that? Is y(1) really 0?

12. Dec 10, 2008

### hils0005

its late.....what else can i say
I don't know what to use for t going from 0 to 1
y=pi t + pi
y(0)=pi
y(1)=2pi

y=pit - pi
y(0)=-pi
y(1)=pi t - pi=0

Last edited: Dec 10, 2008
13. Dec 10, 2008

### gabbagabbahey

well now y is linear, and y(1)=0 , but y(0)=-pi not +pi

Perhaps you should get some rest and leave this until morning?

14. Dec 10, 2008

### hils0005

Great idea, thanks for your help

15. Dec 10, 2008

### kathrynag

Here's a big hint for finding x, y and z,
P1(-1,pi,-1) to P2(0,0,0)
So, we have (-1, pi, -1) + some t
(-1, pi,-1)+(1,a, b)t
So x= -1+t
I got this by having our original point and taking 0 from our second point and and -1 from the first point.
So, we have -1+[0-(-1)]t=-1+t
Follow this in the same manner for finding y and z.

16. Dec 10, 2008

### Defennder

Actually my intention wasn't to find the scalar potential which itself could be rather tedious, but to show that one could evaluate the line integral from P1 to P3 directly bypassing P2 to save time.