# Thread: Work field

1. ## Work field

Find the work realized by field
$F(x,y) = (2xy, x^2 + cosy)$

to move a particle from the initial point to end point of the curve $r(t) = ti + costj$ where $0 \leq t \leq \pi$

How is the integral ???

2. Can I use green's theorem?

$w= \int \int_S [\frac{ \partial (x^2+cosy)}{ \partial x} - \frac{ \partial (2xy)}{ \partial y}]dxdy$

$0 \int \int_S dxdy = 0$

3. Originally Posted by Apprentice123
Can I use green's theorem?

$w= \int \int_S [\frac{ \partial (x^2+cosy)}{ \partial x} - \frac{ \partial (2xy)}{ \partial y}]dxdy$

$0 \int \int_S dxdy = 0$

Greens theorem is only valid for piecewise simple closed curves, and this path is not closed.

$\vec r(t)=t\vec i +\cos(t) \vec j$

$\vec r'(t) = \vec i + -\sin(t) \vec j$

$F(x,y,)=(2xy,x^2+\cos(y))$

$\int_{0}^{\pi} (2t\cos(t),t^2+\cos(\cos(t)))\cdot (1,-\sin(t))dt$

$\int_{0}^{\pi}2t\cos(t)-t^2\sin(t)-\cos(\cos(t))\sin(t) dt$

You should be able to finish from here

4. Originally Posted by TheEmptySet
Greens theorem is only valid for piecewise simple closed curves, and this path is not closed.

$\vec r(t)=t\vec i +\cos(t) \vec j$

$\vec r'(t) = \vec i + -\sin(t) \vec j$

$F(x,y,)=(2xy,y+\cos(y))$

$\int_{0}^{\pi} (2t\cos(t),\cos(t)+\cos(\cos(t)))\cdot (1,-\sin(t))dt$

$\int_{0}^{\pi}2t\cos(t)-\sin(t)\cos(t)-\cos(\cos(t))\sin(t) dt$

You should be able to finish from here
Not is $\int_0^{\pi} (2tcost, t^2+cos(cost)).(1,-sint)dt$ ??

5. Originally Posted by Apprentice123
Not is $\int_0^{\pi} (2tcost, t^2+cos(cost)).(1,-sint)dt$ ??

I was a bit careless with notation

These are Vectors and the dot means the dot product
$\int_{0}^{\pi} (2t\cos(t),t^2+\cos(\cos(t)))\cdot (1,-\sin(t))dt$

$
\int_{0}^{\pi} 2t\cos(t)\vec i+ [t^2+\cos(\cos(t)]\vec j)\cdot (\vec i,-\sin(t)\vec j)dt
$

Then this gives the bottome integral in the post before

6. Originally Posted by TheEmptySet
I was a bit careless with notation

These are Vectors and the dot means the dot product
$\int_{0}^{\pi} (2t\cos(t),\cos(t)+\cos(\cos(t)))\cdot (1,-\sin(t))dt$

$
\int_{0}^{\pi} 2t\cos(t)\vec i+ [\cos(t)+\cos(\cos(t)]\vec j)\cdot (\vec i,-\sin(t)\vec j)dt
$

Then this gives the bottome integral in the post before

What I said was: $F(x,y) = (2xy)i + (x^2 + cosy)j => (2tcost)i + (t^2 + cos(cost))j$ What did you do with $x^2$ ?

7. Originally Posted by Apprentice123
What I said was: $F(x,y) = (2xy)i + (x^2 + cosy)j => (2tcost)i + (t^2 + cos(cost))j$ What did you do with $x^2$ ?
Ahhh. I see why you are confused. Sorry

I have edited the above posts, I wrote down the wrong equation and had y instead of x^2.

8. Originally Posted by TheEmptySet
Ahhh. I see why you are confused. Sorry

I have edited the above posts, I wrote down the wrong equation and had y instead of x^2.
Ok. Thank you