##\int_0^1 F_x(x,0)dx + \int_0^1 F_y(1,y)dy##

[happyparticle]

Homework Statement

Integral

Relevant Equations

##\int_0^1 F_x(x,0)dx + \int_0^1 F_y(1,y)dy##

Hi,
This is the first time I see this kind of integral. I'm not sure how to resolve it.

##
\int_0^1 F \cdot dr
##

##
\int_0^1 F_x(x,0)dx + \int_0^1 F_y(1,y)dy
##

##F = (y,2x)##
I don't know the values of ## F_x(x,0) ## and ## F_y(1,y)##

 

[haruspex]

Homework Statement:: Integral
Relevant Equations:: ##\int_0^1 F_x(x,0)dx + \int_0^1 F_y(1,y)dy##

Hi,
This is the first time I see this kind of integral. I'm not sure how to resolve it.

##
\int_0^1 F \cdot dr
##

##
\int_0^1 F_x(x,0)dx + \int_0^1 F_y(1,y)dy
##

##F = (y,2x)##
I don't know the values of ## F_x(x,0) ## and ## F_y(1,y)##

It is unclear what of the above you are given and what is your own work. Where did ##F = (y,2x)## come from?
You should be able to perform the integral ##\int_0^1 F_x(x,0)dx##. (I assume you realize it means ##\int_0^1 \frac{\partial F(x,0)}{\partial x}dx##.)

 

[happyparticle]

I don't have anymore information. F is a force.
This is a scalar product, but I don't know the value of ##F_x##

 

[pasmith]

It is unclear what of the above you are given and what is your own work. Where did ##F = (y,2x)## come from?
You should be able to perform the integral ##\int_0^1 F_x(x,0)dx##. (I assume you realize it means ##\int_0^1 \frac{\partial F(x,0)}{\partial x}dx##.)

Unless [itex]F_x[/itex] indicates the horizontal component of [itex]\mathbf{F}[/itex], which is given as [itex]y[/itex].

 

[happyparticle]

That's what I thought, but the answer should be 0 + 2, but I don't get that.

##(y,0)\cdot(x,0) \neq 0##

##(0,2x)\cdot(1,y) \neq 2##

 

[pasmith]

That's what I thought, but the answer should be 0 + 2, but I don't get that.

(y,0)⋅(x,0)!=0

(0,2x)⋅(1,y)!=2

The question is asking you to sett [itex]x = 0[/itex] in the integral with respect to [itex]x[/itex]: [tex]
\int_0^1 F_x(x,0)\,dx = \int_0^1 0\,dx.[/tex] Similarly, the question is asking you to set [itex]x = 1[/itex] in the integral with respect to [itex]y[/itex].

 

[happyparticle]

I'm still confuse. Why not y = 0 in the first integral?

In the second integral, I understand x = 1 so 2x = 2 * 1, but what about x? I'm not sure to understand what respect to means.

For me, ##F_y(1,y)## means ## x = 1## and ##y = y## so ## y = y ## and ## x = 2##

At this point, I'm so confuse I'm not even sure what I'm typing.

Basically, the first integral is on the x-axis and the second on the y axis. ##\int_0^1 F \cdot dr + \int_0^1 F \cdot dr## = ## \int_0^1 Fx \cdot (x,0) dx + \int_0^1 Fy \cdot (1,y) dy##

I thought ##Fx## = 1 so ##1(x,0) = x## and ##2x(1,y) = 2x + 2xy##

 

[haruspex]

Why not y = 0 in the first integral?

I think that's what @pasmith meant to write.