##\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 F_x indicates the horizontal component of \mathbf{F}, which is given as y.

 

[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 x = 0 in the integral with respect to x: <br /> \int_0^1 F_x(x,0)\,dx = \int_0^1 0\,dx. Similarly, the question is asking you to set x = 1 in the integral with respect to y.

 

[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.