# Homework Help: Integrating along paths

1. Feb 5, 2012

### Sekonda

Hey guys,

I have :

df=c(x^2)(y^2)dx + (x^3)(y)dy

along paths (0,0) to (1,1); and also paths (0,0) to (0,1) to (1,1) (where (x,y))

where c is some constant.

I am having difficulty doing this particular integral, what type of integral is it and how do I go about solving it?

Thanks!

2. Feb 5, 2012

### Clever-Name

It's just a line integral. It's just asking you to do the same integral in two different ways (along two different paths). For the first case, can you think of a way to parametrize the path into a single variable? For the second case, think about what x, y, dx, and dy are between each point.

3. Feb 5, 2012

### Sekonda

So would I use parameters of x=sint, y=-cost for 0<t<pi for path (0,0) to (1,1)

and for the path (0,0) to (0,1) to (1,1) the fact that dx is zero for the first path and dy is zero for the second path?

4. Feb 5, 2012

### Sekonda

I managed to attain (2^0.5)(c+1) and c/3 as my answers however I am not convinced that these are correct. For the path (0,0) to (1,1) I used parameterization : x=sin(t) y=sin(t) for 0<t<pi/2

Is this correct?

5. Feb 5, 2012

### Sekonda

I just used x=y for the first path to attain a seemingly more likely answer of (c+1)/5, now I'm just stuck on the second path!

6. Feb 5, 2012

### Sekonda

For the second path (0,0) to (0,1) to (1,1) I attained 2(c+1)/5, for the first path (0,0) to (1,1) I attained (c+1)/5.

Is this right?

7. Feb 5, 2012

### vela

Staff Emeritus
This parameterization won't work because (x(t), y(t)) doesn't pass through (1,1).

Yes.

That parameterization will work, but you didn't get the right result. For the second path, the answer is indeed c/3.

That's right.