- #1
kidsasd987
- 143
- 4
Exact differential of a scalar function f takes the form of
∇f⋅dr=Σ∂ifdxi (where dr is a vector)
f:R->Rnand I am not sure why this equation is valid in the sense that if we integrate the equation,
∫∇f⋅dr=∫{Σ∂ifdxi}
∫df=∫{Σ∂ifdxi}
the above equation is true because integration is a linear operator, and if we think of R.H.S only,
∫{Σ∂1fdx1}+∫{Σ∂i≠1fdxi≠1}
=f+c(x2,x3...,xn)+∫{Σ∂i≠1fdxi≠1}
and if we think of each integration of differential w.r.t variable x1,x2,x3 and so on, it should generate f+c respectively as the result of each integration.
therefore R.H.S has to be
nf+C(x1,x2,x3...,xn) but according to textbook, it says we don't add up each integration but we compare them to eradicate constants and "merge" each equation to one right answer, f.
I am a little bit confused about how to interpret the integration within the quotation mark ∫df="∫"{Σ∂ifdxi} because it seems it is linearly applied to each of partial differential, but does not spit out nf(n number of partial differentials so there must be n number of integrations on them so adding them up would give nf+C(x1,x2,x3...,xn)
Please enlighten me. I would really appreciate
∇f⋅dr=Σ∂ifdxi (where dr is a vector)
f:R->Rnand I am not sure why this equation is valid in the sense that if we integrate the equation,
∫∇f⋅dr=∫{Σ∂ifdxi}
∫df=∫{Σ∂ifdxi}
the above equation is true because integration is a linear operator, and if we think of R.H.S only,
∫{Σ∂1fdx1}+∫{Σ∂i≠1fdxi≠1}
=f+c(x2,x3...,xn)+∫{Σ∂i≠1fdxi≠1}
and if we think of each integration of differential w.r.t variable x1,x2,x3 and so on, it should generate f+c respectively as the result of each integration.
therefore R.H.S has to be
nf+C(x1,x2,x3...,xn) but according to textbook, it says we don't add up each integration but we compare them to eradicate constants and "merge" each equation to one right answer, f.
I am a little bit confused about how to interpret the integration within the quotation mark ∫df="∫"{Σ∂ifdxi} because it seems it is linearly applied to each of partial differential, but does not spit out nf(n number of partial differentials so there must be n number of integrations on them so adding them up would give nf+C(x1,x2,x3...,xn)
Please enlighten me. I would really appreciate