# Question about line integral of F dot dr

1. Dec 15, 2013

### mesa

1. The problem statement, all variables and given/known data
Evaluate the line integral of F dot dr where f(x,y)=<3x^2,2x+y> and C is a straight line segment from (1,2) to (5,4)

2. Relevant equations
Unfortunately I was out with family obligations when we covered line integrals and surface integrals so am stuck with the textbook for these so I will just jump to my 'attempt at a solution' since I have limited knowledge on these subjects.

3. The attempt at a solution

First I figured the equation of our line to be y=1/2x-1/2
so if x=t then,
y=1/2t-1/2
so r(t)= ti+(1/2t+3/2)j
and r'(t)= i+1/2j

∫<3t^2i+(5/2t+3/2j>dot<i+1/2j>dt
∫(3t^2+5/4t+6/2)dt

Then we evaluate the integral from 1 to 5. How bad is it?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 15, 2013

### LCKurtz

That should work; I didn't check your numbers. A nicer, and more standard, parameterizaton of the line would be $\vec r(t) = \langle 1,2\rangle + t\langle 4,2\rangle$ for $0\le t\le 1$. I think you will find it simpler. Try it.

3. Dec 15, 2013

### mesa

Ah hah, a surprising result!

If I have an integral of F dot T ds with F(x,y)=<4x,y^2> and C is the arc of a circle of radius 2 centered at the origin then do we just take,
x=2cos t
y=2sin t
so we have the integral of <8cos t,4sin^2 t>dot<-2sin t,2cos t> with the bounds of 0 to pi/2?

4. Dec 15, 2013

### mesa

I thought I could figure this out, but...
How did you get that parameterization?

5. Dec 15, 2013

### LCKurtz

Given points $P_0,~P_1$ the displacement vector from $P_0$ to$P_1$ is $D=P_1-P_0$. Then $R= P_0 +tD$ is at $P_0$ when $t=0$ and $P_1$ when $t=1$.

6. Dec 15, 2013

### LCKurtz

Sure, if that's the part of the arc in which you are interested. You can use any parametrization and the idea is to use "nice" ones, like that one, for a circle. Note that going from one point to another over different curves may (but might not) give different results.

7. Dec 15, 2013

### mesa

Okay, so would r(t)=<5,4>+t<-4,-2> also be correct?

8. Dec 15, 2013

### mesa

Sorry, I forgot to mention 'in the first quadrant'...
Nice to see the other part is correct.

9. Dec 15, 2013

### mesa

Another couple questions, how do we take the derivative of r(t) in this form? Is it just 4i+2j? and then what do we do with our f(x,y)=<3x^2,2x+y> to set it up for the dot product in the integral?

10. Dec 15, 2013

### LCKurtz

Yes and no. Yes, it's the same line segment, but no, it is in the other direction.

Yes. The derivative of $t$ is $1$ so you just get the direction vector.

Remember $r(t) = \langle x,y\rangle$. Your parameterization gives $x$ and $y$ in terms of $t$.

11. Dec 15, 2013

### mesa

Okay so we would have the integral of <48t^2,10t>dot<4,2> and the new bounds will be from 0 to 1 and then just calculus as usual. Is that right?

Last edited: Dec 15, 2013
12. Dec 15, 2013

### LCKurtz

I haven't worked it out. But you could check by comparing the answer against your original calculation of the integral.

13. Dec 15, 2013

### mesa

Very good, although I will rejoice when the semester is finished and this material can be covered properly (hopefully my copy of Anton will resurface soon as well, Stewart is a pain).

The only other question I have is if we have multiple 'segments' then we simply evaluate each one with their respective bounds and add the up, correct?

14. Dec 15, 2013

### LCKurtz

Yes.

15. Dec 15, 2013

### mesa

Thank you LCKurtz, that should do it.