You are Here: Home >< Maths

# Finding whether line integrals are independent of path. Watch

1. I have a question;

"Determine whether of not the line integral along the given curve C is independent of the path and evaluate the integral.

(a) along the parabola from (-1,1) to (2,4)"

I know you can work out if it's independent if but I'm not exactly sure how to do it. My lecture notes don't explain this very well, what is ? It also says I should try to parametrise C, i.e so that I can use instead.

It's all in another language, this section has been crammed in to a one page lecture note and I have no idea how I'm meant to approach this question. Can anyone help?
2. (Original post by Sasukekun)
I know you can work out if it's independent if
I think you might mean 'if the integral of every closed path is zero'.

There's a couple of equivalent conditions for a line integral being independent of its path (and only depending on the start and end points):

The curl of the integrand is zero.

The integrand is a gradient (i.e. for some scalar function ).

The line integral along any closed path (start and end points are the same) is zero.

what is ?
Talking in a very non-rigourous way, is analogous to in the calculus you are used to: we can think of the integral as taking the scalar product of the integrand f with the line element at every point along the path and summing along the path, in the same way that a "normal" integral can be thought of as multiplying f by dx and summing along the range of integration.

It also says I should try to parametrise C, i.e so that I can use instead.
The standard way to evaluate a line integral like this is to parametrise the curve as your notes suggest, which turns it into a standard scalar integral. One way of parametrising the curve C in your question would be , with t ranging form -1 to 2.
3. (Original post by Mark13)
Stuff
Thanks, I think I've done that question now due to your help. Although I'm now stuck on another. It's asking for the same thing but for for any closed circle.

Now I realise for any closed circle it essentially means where C and D are just random constants. Also and where t is the angle ranging from 0 to 2pi.

However how do I get in terms of the parameter t? From instance the in becomes which surely can't be right?

Am I doing the correct thing so far, what can I do?
4. (Original post by Sasukekun)
Am I doing the correct thing so far, what can I do?
I haven't checked your working thoroughly, but it all looks correct. However, you've probably noticed the algebra is pretty horrific.

But from my first post, note that if the integrand is a gradient, then the integral along a closed path is zero. Can you show that the integrand is a gradient in this case?
5. (Original post by Mark13)
Can you show that the integrand is a gradient in this case?
Sorry man, I'm not exactly sure. What is the integrand in this case, and how would I know if it's a gradient?
6. (Original post by Sasukekun)
Sorry man, I'm not exactly sure. What is the integrand in this case, and how would I know if it's a gradient?
No worries. The integrand just means the bit inside the integral sign, which in this case is (sin(y),xcos(y)). If it is a gradient, then we can write it as:

for some function .

If you haven't come across the notation before, all it means in this context is that . So basically, if there's a function that satisfies:

and

then (sin(y),xcos(y)) is a gradient, and therefore any line integral of this function along a closed path is zero. So, you need to find . If this all seems completely unfamiliar, then your lecturer might have wanted you to use some other method.
7. (Original post by Mark13)
No worries. The integrand just means the bit inside the integral sign, which in this case is (sin(y),xcos(y)). If it is a gradient, then we can write it as:

for some function .

If you haven't come across the notation before, all it means in this context is that . So basically, if there's a function that satisfies:

and

then (sin(y),xcos(y)) is a gradient, and therefore any line integral of this function along a closed path is zero. So, you need to find . If this all seems completely unfamiliar, then your lecturer might have wanted you to use some other method.
I wish you were the one writing my lecture notes instead. Clear and concise explanation, thanks a lot for your help man.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: December 11, 2010
Today on TSR

### Anxious about my Oxford offer

What should I do?

### Am I doomed because I messed up my mocks?

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE