What type of integral is this and how do I solve it?

[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!

 

[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.

 

[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?

 

[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?

 

[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!

 

[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?

 

[vela]

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

This parameterization won't work because (x(t), y(t)) doesn't pass through (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?

Yes.

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?

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

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!

That's right.

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?

Show us your work.

 

[Sekonda]

Thanks for neatly reviewing all my random progressions through this questions; I realized where I made an error or two and now have the paths as c/3 and (c+1)/5.

Thanks Vela & Clever-Name!