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Finding potencial of forces - answer differs from solutions

Homework Statement
3.1 (Kibble's Classical Mechanics) Find which of the following forces are conservative, and for those that
are find the corresponding potential energy function (a and b are constants)
Homework Equations
Fx = ax + by², Fy = az + 2bxy, Fz = ay + bz²

Fr = 2ar sin θ sin ϕ, Fθ = ar cosθ sin ϕ, Fϕ = ar cosϕ
The first force components:
Fx = ax + by², Fy = az + 2bxy, Fz = ay + bz²
I calculated the integral V=-∫Fdr, using dr=(dx,dy,dz)
The result I found was
-(1/2(ax²)+2azy+2bxy²+1/3bz³)
The answer in the book (Kibble's Classical Mechanics): -(1/2(ax²)+azy+bxy²+1/3bz³)


The second force:
Fr = 2ar sin θ sin ϕ, Fθ = ar cosθ sin ϕ, Fϕ = ar cosϕ
I calculated the integral V=-∫Fdr, using dr=(dr,rdθ,rsinθdϕ)
The result I found was
-arsinθsinϕ(r+1)
The answer in the book (Kibble's Classical Mechanics): -ar²sinθsinϕ

Me and my colleagues can't find where we have mistaken. Can you help us? Our solutions are incorrect?
 

TSny

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I calculated the integral V=-∫Fdr, using dr=(dx,dy,dz)
You probably mean V = -∫F⋅dr where the integrand is the scalar product of the vectors F and dr.
Kibble's answers are correct since they agree with F = - V. You didn't show any details of your calculation, so we can't tell you where you made your mistakes. Your approach should yield the correct answer. But this is not the only way to go.

For your method, note that the integral is along some path that connects your point of zero potential energy to the point where you are calculating V. It won't matter what particular path you choose, but you need to choose a path. (Pick a path that makes things easy!) When integrating along your path, be sure to think about what the values of x,y,z [or r, θ, φ] are along your particular path.

Edit: I assumed that you had already shown that the forces are conservative. If not, then you need to do that first as @kuruman points out in the next post.
 
Last edited:

kuruman

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Before you find the potential function, you need to establish that the forces can be derived from one. How did you do that? You can calculate a line integral for each force, but it doesn't mean anything if it is not path independent.
 

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