# Line integral question

1. Dec 3, 2011

### maogden

1. The problem statement, all variables and given/known data
Find the line integral of ∫ x+yz dx + 2x dy + xyz dz

C consists of line segments from (1,0,1) to (2,3,1) and from (2,3,1) to (2,5,2).

2. Relevant equations

r=(1-t)<r0> + t<r1> 0<t<1

3. The attempt at a solution

I split up the two line segments into C1 and C2.

For C1 I got the parametric equation of x(t)=1+t, y(t)=3t, z(t)=1
I plugged this into the original equation also using dx=dt, dy=3dt, dz=0
This gave me the answer C1=13.

I then found the parametric equations of C2 to be x(t)=2, y(t)=3-2t, z(t)=1-t.
I plugged these into the equations again with their derivatives and got the answer to be -22/3.

The correct answer should be 97/3.

2. Dec 4, 2011

### jackmell

On the first, if I let:

x=1+t
y=3t
z=1

I get 12.

On the second, how about let

x=2
y=3+2t
z=1+t

3. Dec 4, 2011

### HallsofIvy

Staff Emeritus
If you had shown how you got this, we might have been able to point out a mistake. As it is, we, like jackmell, can only say that your first integral is wrong.

Okay, when t= 0, (x, y, z)= (2, 3, 1) and when t= -1, (x, y, z)= (2, 5, 2). Did you then integrate from 0 to -1?

4. Dec 4, 2011

### maogden

Sorry I didn't explain my work. I integrated both C1 and C2 from 0 to 1.. why are you integrating it from -1 to 0?

And here's the rest of my work,

C1:
r=(1-t)<1,0,1> + t<2,3,1>
r=<1+t,3t,1>

(from 0 to 1)∫ (1+4t) (1)dt +
(from 0 to 1)∫ (2) (1+t) (3) dt +
(from 0 to 1)∫ (1+t)(3t)(1)(0) dt

[2t^2+t](0 to 1) +
6[t+(t2/2)](0 to 1) +
[t](0 to 1)

(2 + 1) + (6 + 3) + 1 = 13

C2:
r=(1-t)<2,3,1> + t<2,5,2>
r=<2,3-2t,1-t>

(from 0 to 1)∫ 2+(3+2t)(1-t)(0)dt +
(from 0 to 1)∫ 4(-2)dt +
(from 0 to 1)∫ (2)(3-2t)(1-t)(-1) dt

2[t](0 to 1) +
-8[t](0 to 1) +
-2[3t - (5/2)*t2 + (2/3)*t3](0 to 1)

adding these up i got (-19/3) when I redid the problem again.

so C1 + C2 = (20/3)

5. Dec 5, 2011

### jackmell

Ok, that third one should be zero because z=1 which means dz=0 right? There you go. Got 36/3 of them.

[/quote]

Now that second one you have:

$$\mathop\int\limits_{(2,3,2)}^{(2,5,2)} (x+yz)dx+2xdy+xyzdz$$

Ok, if I let:

x=2
y=3+2t
z=1+t

then I need to let t go from 0 to 1 so it's

$$\int_0^1 (x+yz)dx+2xdy+xyzdz,\quad x(t)=2, y(t)=3+2t, z(t)=1+t$$

you can do that I bet.

6. Dec 5, 2011

### HallsofIvy

Staff Emeritus
I'm not. I asked if you did because your parametric equations, x(t)=2, y(t)=3-2t, z(t)=1-t, give x(0)= 2, y(0)= 3, z(0)= 1, the first point, and x(-1)= 2, y(-1)= 5, z(-1)= 2, the second point.

If you integrate from 0 to 1 then, because x(1)= 2, y(1)= 3- 2= 1, z(1)= 1- 1= 0, you are integrating from (2, 3, 1) to (2, 1, 0) which is not what you wanted to do.

7. Dec 5, 2011

### maogden

Ok I think I understand..

So what you're saying is...

$0,-1\int$

8. Dec 5, 2011

### maogden

Ok I think I understand..

So what you're saying is...

(from 0 to -1)∫ (2 + (3-2t)(1+2t))(0)dt +
(from 0 to -1)∫ (2(2t))(3)dt +
(from 0 to -1)∫ (2)(3-2t)(2)dt
simplified:
4 * (from 0 to -1)∫ -4t^2 + 4t + 1 dt

4[(-4t^3)/3 + 2t^2 + t] dt <== from 0 to -1

4(4/3 + 2 - 1 - (0 + 0 + 0)) = 28/3

but,
28/3 + 13 ≠ 97/3