Work and Force. Double Integration?

[Idoke]

Homework Statement

A particle in a path limited to the XY plane is acted upon by a force described in the following function:
{\bf F}(x,y)=9y \hat{\bf x}+7x \hat{\bf y}

(*The numerical coefficients are in kg/s^2)
What is the work that the force does along these paths, between the points A = (0,0) [m] and B = (1,2) [m]:

1. First along the X axis and then along the Y axis.
2. On the curve y=2x^2
3. On the curve y=\sqrt{2x}

Homework Equations

W_C = \int_{C} \bold{F} \cdot \mathrm{d}\bold{s}

The Attempt at a Solution

If this would be a one variable function of the force I would calculate the definite integral between the two points. But I am stumped by two things:
1. The function of the force is a two variable function, which I haven't learned to integrate, reading "Multiple Integers" on wikipedia got me more confused.
2. I know this is probably a non-conservative force and so the path does matter, but I don't know how to account for that in my calculations.

I feel like I should know this and it's getting kind of frustrating, but I'm sure someone could help because it all seems kind of basic.

Thanks.

 

[ApexOfDE]

Why don't you try to write a path in vector form, and then multiply it with force.

 

[Idoke]

Hey,
Thanks for replying.
What do you mean writing the path in vector form?
For the first question I get:
A) along the X axis:
Fd=9y (the force in the X direction) and y=0 so W=0.
B) along the Y axis:
Fd=7x. and of course the change in the x direction is again 0 (because i am moving on the y axis) so it's 0.
0+0=0 (or so I hear :) )

Is this right? and what do I do on the two other parts?

 

[ApexOfDE]

yes, that's right.

you can write ds = xdx + ydy and then plug it to work formula.

 

[ehild]

The problem asks the work when the point moves from (0,0) to (1,2). The first question refers to a rectangular path, with one piece coincident with an axis, the other parallel with the other one. The formulation is not clear, but it is certainly meant in this way. If the point moves along the axes only, it never reaches from (0.0) to (1,2).

The work along the first path is


W= \int _{(0,0)}^{(1,0)}{F_xdx}+\int_{(1,0)}^{(1,2)}{F_ydy}.

If you start the path along the y axis,

W= \int _{(0,0)}^{(0,2)}{F_ydy}+\int_{(0,2)}^{(1,2)}{F_xdx}

When you integrate along a line given with a Y(x) function, you have to use dy=dy/dx*dx in the integrand.

If y=x², dy/dx = 2x,

W= \int _{0}^{1}{(18x^2 +7x*2x)dx}

ehild

 

[Idoke]

Thank you!
Your method did work, although I still don't know exactly how, I understand that I will learn it soon so I guess I'll be OK.

Both of you have been immensely helpful,
Cheers.