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E&M line integral setup

My problem has force decreasing with F=(1/r^2)r, where F is a vecotr and r is unit vector. i need to find a).work done in moving from a point at r=sqrt(2) to a point at r=2*sqrt(2) by a direct radial path and (b) by a path from (1,1)-->(2,1)-->(2,2). Compare my answers.

a)I did direct radial path using Work=Integral[1/r^2] from sqrt(2) to 2*sqrt(2). I got sqrt(2)/2.

b) This where problem is: from (1,1)-->(2,1) x:1-->2, y=1, dl=dx x, so i get F (dot) dl = (x^2+y^2)^2, however, I am not sure I am setting up the x and y components correctly. from (2,1)-->(2,2) y:1-->2, x=2 dl=dy y and again same issues of x and y components.

i did read that r vector=sin(theta)cos(phi)x+sin(theta)sin(phi)y, but was not sure how to incorparate this into line integral part of problem.


any help would be appreciated, thanks
 
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First of all this is a vector-integral. This means that you are gonna have to calculate this integral for each component. In your case we have two components x and y so r²=x² + y².

Now (1/r²)rdr = (1/r²)dx + (1/r²)dy. Because the unitvector multiplied by the vector dr gives you dx and dy (via the dot-product)

Now when you move from (1,1) to (2,1) you will need a parameter-equation for the integration. y = 1 and x=l with l the parameter and l : 1-->2. So in the integral replace x by l and dx=dl and y = 1 and dy =0

The x-component of the integration is integral(dl/(l²+1²)) and l starts in 1 and stops in 2. this integral is easy because you know that generally
integral(1/(1 + x²)dx) yields arctan(x).

Since along this path dy = 0 the second term of the integral vanishes...
So the only thing you need to do now is calculate the integral correctly. But the solution of this integral will be an arcus-tangens...


The other path is calculated in the exact same way.


regards
marlon
 
thanks a bunch, I was sort of close, but never looked at the vector part of it. My first integral where dy=0 I was using (x^2+1)dx from 1-->2. I should be able to complete it now, again thank you.
 

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