# Evaluating Line Integral F over C: a Step-by-Step Guide

• boneill3
In summary, the homework statement is to evaluate the integral \int_{C}F . dr where C is the line segment from (0,0) to (1,1).

## Homework Statement

Let F=(3x+2y)i+(2x-y)j Evaluate $\int_{C}F . dr$ where C is the line segment from (0,0) to (1,1)

## Homework Equations

$\int_{a}^{b}[f(x(t),y(t))x'(t) + g(x(t),y(t))y'(t)]dt$

## The Attempt at a Solution

How do I choose the correct values of x and y as a function of t?
And do I just integrate over a = 0 to b = 1 ?

regards

What parametrization have you found for x(t) and y(t)?

i.e. what is a parameterization for the line from (0,0) to (1,1)?

I'm using the formula to find
$r(t) = (1-t)(x_{0},y_{0})+t(x_{1},y_{1}) =(1-t)(0,0)+t(1,1)\\ \text{ so } x = t y = t$ for 0<t<1

So is the integral to evaluate

$\int_{0}^{1} [(3t+2t)3+(2t-t)]dt$

regards

boneill3 said:
I'm using the formula to find
$r(t) = (1-t)(x_{0},y_{0})+t(x_{1},y_{1}) =(1-t)(0,0)+t(1,1)\\ \text{ so } x = t y = t$ for 0<t<1
So x= t, y= t. Yes, that goes from (0,0) to (1,1)

boneill3 said:
So is the integral to evaluate

$\int_{0}^{1} [(3t+2t)3+(2t-t)]dt$

regards
where did you get the "3" multiplying (3t+ 2t)?

Let F=(3x+2y)i+(2x-y)j

$\int_{a}^{b}[f(x(t),y(t))x'(t) + g(x(t),y(t))y'(t)]dt$

I got the three from the partial derivative dx of 3x+2y = 3

But I think I may have stuffed up because the formula
says x'(t) which would be 1.

So the integral would be

$\int_{0}^{1} [(3t+2t)+(2t-t)]dt$

Is that better?

boneill3 said:
So is the integral to evaluate

$\int_{0}^{1} [(3t+2t)3+(2t-t)]dt$

regards

yes your limits would be from 0 to 1

Hi Guys

If C instead of a line segment is y= x^2 can I substitute x = t so that C is now just
[latex}
fy(t)= t^2
[/itex]

regards

Sorry,

$fy(t)= t^2$

I've used the new parameterization of x = t^2 y = t^2 ( for C y=x^2)

On the original function

So is the integral to evaluate now

$\int_{0}^{1} [(3t+2t)2t+(2t-t)2t]dt$

Where x'(t) = x'(t^2) = 2t

and

Where y'(t) = y'(t^2) = 2t

Is this right so far?

regards
Brendan

You need to put x=t^2 and y=t^2 into the definition of the vector F as well, right? Or if you are doing y=x^2, x=t, y=t^2, dx=dt, dy=2tdt. You seem to be mixing up your parametrizations.

So I need to put it as
$\int_{0}^{1} [(3t^2+2t^2)2t+(2t^2-t^2)2t]dt$

regards

Yes, that's right for parametrizing along the line between (0,0) and (1,1) using x=t^2 and y=t^2. And you should get the same answer as using x=t and y=t.

Thanks for all your help guys

## 1. What is a line integral?

A line integral is a type of integral used in mathematics to calculate the area under a curve or the work done along a path in a vector field. It is represented by the symbol ∫ and is used to evaluate a function along a given curve.

## 2. What is the purpose of evaluating line integrals?

Evaluating line integrals allows us to calculate the amount of work done along a path in a vector field, which has many practical applications in physics and engineering. It also helps us understand the behavior of functions along a given curve.

## 3. How is a line integral over a curve represented mathematically?

A line integral over a curve C is represented by the expression ∫C F(x,y) ds, where F(x,y) is the function being evaluated and ds is the infinitesimal length along the curve C.

## 4. What are the steps involved in evaluating a line integral over a curve?

The steps involved in evaluating a line integral over a curve C are:
1. Parametrize the curve C to represent it as a function of a single variable t.
2. Find the derivative of the parametric function to calculate ds.
3. Substitute the parametric function and ds into the line integral expression.
4. Evaluate the resulting integral using appropriate techniques, such as substitution or integration by parts.

## 5. What are some common mistakes to avoid when evaluating line integrals?

Common mistakes to avoid when evaluating line integrals include:
1. Forgetting to parametrize the curve C.
2. Using the wrong limits of integration.
3. Forgetting to include the ds term in the integral.
4. Making errors in the calculation of ds.
5. Not simplifying the resulting integral before attempting to evaluate it.