Question Number 218195 by david2409 last updated on 01/Apr/25
$${P}=\frac{\Sigma{Fz}}{{Npil}} \\ $$
Answered by MrGaster last updated on 02/Apr/25
![(1):ΣFz=∫_∂Σ FzdS Npil=∫_∂Σ dS P=(1/(Npil))∫_∂Σ FzdS ∫_∂Σ Fzds=∫^R _0 ∫_∂Σ FzdS dr ∫_∂Σ dS=∫_0 ^R ∫_∂Σ dS dr P=((∫_0 ^R ∫_∂Σ Fz dS dr)/(∫_0 ^R ∫_∂Σ dS dr)) ∫_∂Σ dS=∫_∂Σ Fz(∂Σ/∂x_i )dx_i P=((∫_0 ^R (∫_∂Σ Fz(∂Σ/∂x_i )dx_i )dr)/(∫_0 ^R (∫_∂Σ (∂x/∂x_i ))dr)) ∫_∂Σ (∂Σ/∂x_i )=∫_∂Σ ∂S P=((∫_0 ^R (∫_∂Σ FzdS)dr)/(∫_0 ^R (∫_∂Σ dS)dr)) ∫_∂Σ FzdS=∫_∂Σ Fz(√(Σ_(i=1) ^n ((∂Σ/∂x_i ))^2 ))dx_1 ∧dx_2 ∧…∧dx_(n−1) P=((∫_0 ^R (∫_∂Σ Fz(√(Σ_(i=1) ^n ((∂Σ/∂x_i ))^2 ))dx_1 ∧dx_2 ∧…∧dx_(n−1) )dr)/(∫_0 ^R (∫_∂Σ (√(Σ_(i=1) ^n ((∂Σ/∂x_i ))^2 ))∂x_1 ∧∂x_2 ∧…∧dx_(n−1) )dr)) P=((∫_0 ^R ∫_∂Σ Fz dS dr)/(∫_0 ^R ∫_∂Σ dS dr)) (2):Let F(z)=Σ_(k=0) ^∞ (((z−z_0 )^k F^((k,0)) (z_0 ∣m))/(k!)) ΣFz=∫_(z_0 −ε) ^(z_0 +ε) F(z)dz=∫_(z_0 −ε) ^(z_0 +ε) Σ_(k=0) ^∞ (((z−z_0 )^k F^((k,0)) (z_0 ∣m))/(k!))dz ΣFz=Σ_(k=0) ^∞ ((F^((k,0)) (z_0 ∣m))/(k!))∫_(z_0 −ε) ^(z_0 +ε) (z−z_0 )dz ∫_(z_0 −ε) ^(z_0 +ε) (z−z_0 )^k dz= [(((z−z_0 )^(k+1) )/(k+1))]_(z_0 −ε) ^(z_0 +ε) =((2ε^(k+1) )/(k+1)) ΣFz=Σ_(k=0) ^∞ ((F^((k,0)) (z_0 ∣m))/(k!)) ((2ε^(k+1) )/(k+1)) ΣFz=2εΣ_(k=0) ^∞ ((F^((k,0)) (z_0 ∣m))/((k+1)!))ε^k Now,divide dy Npil: P=((2ε)/(Npil))Σ_(k=0) ^∞ ((F^((z,0)) (z_0 ∣m))/((k+1)!))ε^k (3):=Σ_(n=3) ^(10) ((n^4 +^n +1)/(n^7 −n))⊕(1/(Npil)) =Σ_(n=3) ^(10) (((n^2 −n+1)(n^2 +n+1))/(n(n−1)(n+1)(n^2 +n+1)(n^2 −n+1)))⊕(1/(Npil)) =Σ_(n=3) ^(10) (1/(n(n−1)(n+1)))⊕(1/(Npil)) =(1/2)Σ_(n=3) ^(10) ((1/((n−1)n))−(1/(n(n+1))))⊕(1/(Npil)) =(1/2)((1/(2×3))−(1/(10×11)))⊕(1/(Npil)) =((1/(12))−(1/(120)))⊕(1/(Npil)) =((13)/(165))⊕(1/(Npil)) determinant (((P=((13)/(165Npil)))))](https://www.tinkutara.com/question/Q218228.png)
$$\left(\mathrm{1}\right):\Sigma{Fz}=\int_{\partial\Sigma} {FzdS} \\ $$$${Npil}=\int_{\partial\Sigma} {dS} \\ $$$${P}=\frac{\mathrm{1}}{{Npil}}\int_{\partial\Sigma} {FzdS} \\ $$$$\int_{\partial\Sigma} {Fzds}=\underset{\mathrm{0}} {\int}^{{R}} \int_{\partial\Sigma} {FzdS}\:\:{dr} \\ $$$$\int_{\partial\Sigma} {dS}=\int_{\mathrm{0}} ^{{R}} \int_{\partial\Sigma} {dS}\:{dr} \\ $$$${P}=\frac{\int_{\mathrm{0}} ^{{R}} \int_{\partial\Sigma} {Fz}\:{dS}\:{dr}}{\int_{\mathrm{0}} ^{{R}} \int_{\partial\Sigma} {dS}\:{dr}} \\ $$$$\int_{\partial\Sigma} {dS}=\int_{\partial\Sigma} {Fz}\frac{\partial\Sigma}{\partial{x}_{{i}} }{dx}_{{i}} \\ $$$${P}=\frac{\int_{\mathrm{0}} ^{{R}} \left(\int_{\partial\Sigma} {Fz}\frac{\partial\Sigma}{\partial{x}_{{i}} }{dx}_{{i}} \right){dr}}{\int_{\mathrm{0}} ^{{R}} \left(\int_{\partial\Sigma} \frac{\partial{x}}{\partial{x}_{{i}} }\right){dr}} \\ $$$$\int_{\partial\Sigma} \frac{\partial\Sigma}{\partial{x}_{{i}} }=\int_{\partial\Sigma} \partial{S} \\ $$$${P}=\frac{\int_{\mathrm{0}} ^{{R}} \left(\int_{\partial\Sigma} {FzdS}\right){dr}}{\int_{\mathrm{0}} ^{{R}} \left(\int_{\partial\Sigma} {dS}\right){dr}} \\ $$$$\int_{\partial\Sigma} {FzdS}=\int_{\partial\Sigma} {Fz}\sqrt{\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\partial\Sigma}{\partial{x}_{{i}} }\right)^{\mathrm{2}} }{dx}_{\mathrm{1}} \wedge{dx}_{\mathrm{2}} \wedge\ldots\wedge{dx}_{{n}−\mathrm{1}} \\ $$$${P}=\frac{\int_{\mathrm{0}} ^{{R}} \left(\int_{\partial\Sigma} {Fz}\sqrt{\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\partial\Sigma}{\partial{x}_{{i}} }\right)^{\mathrm{2}} }{dx}_{\mathrm{1}} \wedge{dx}_{\mathrm{2}} \wedge\ldots\wedge{dx}_{{n}−\mathrm{1}} \right){dr}}{\int_{\mathrm{0}} ^{{R}} \left(\int_{\partial\Sigma} \sqrt{\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\partial\Sigma}{\partial{x}_{{i}} }\right)^{\mathrm{2}} }\partial{x}_{\mathrm{1}} \wedge\partial{x}_{\mathrm{2}} \wedge\ldots\wedge{dx}_{{n}−\mathrm{1}} \:\right){dr}} \\ $$$${P}=\frac{\int_{\mathrm{0}} ^{{R}} \int_{\partial\Sigma} {Fz}\:{dS}\:{dr}}{\int_{\mathrm{0}} ^{{R}} \int_{\partial\Sigma} {dS}\:{dr}} \\ $$$$\left(\mathrm{2}\right):\mathrm{Let}\:{F}\left({z}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({z}−{z}_{\mathrm{0}} \right)^{{k}} {F}^{\left({k},\mathrm{0}\right)} \left({z}_{\mathrm{0}} \mid{m}\right)}{{k}!} \\ $$$$\Sigma{Fz}=\int_{{z}_{\mathrm{0}} −\epsilon} ^{{z}_{\mathrm{0}} +\epsilon} {F}\left({z}\right){dz}=\int_{{z}_{\mathrm{0}} −\epsilon} ^{{z}_{\mathrm{0}} +\epsilon} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({z}−{z}_{\mathrm{0}} \right)^{{k}} {F}^{\left({k},\mathrm{0}\right)} \left({z}_{\mathrm{0}} \mid{m}\right)}{{k}!}{dz} \\ $$$$\Sigma{Fz}=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{F}^{\left({k},\mathrm{0}\right)} \left({z}_{\mathrm{0}} \mid{m}\right)}{{k}!}\int_{{z}_{\mathrm{0}} −\epsilon} ^{{z}_{\mathrm{0}} +\epsilon} \left({z}−{z}_{\mathrm{0}} \right){dz} \\ $$$$\int_{{z}_{\mathrm{0}} −\epsilon} ^{{z}_{\mathrm{0}} +\epsilon} \left({z}−{z}_{\mathrm{0}} \right)^{{k}} {dz}=\:\left[\frac{\left({z}−{z}_{\mathrm{0}} \right)^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}\right]_{{z}_{\mathrm{0}} −\epsilon} ^{{z}_{\mathrm{0}} +\epsilon} =\frac{\mathrm{2}\epsilon^{{k}+\mathrm{1}} }{{k}+\mathrm{1}} \\ $$$$\Sigma{Fz}=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{F}^{\left({k},\mathrm{0}\right)} \left({z}_{\mathrm{0}} \mid{m}\right)}{{k}!}\:\frac{\mathrm{2}\epsilon^{{k}+\mathrm{1}} }{{k}+\mathrm{1}} \\ $$$$\Sigma{Fz}=\mathrm{2}\epsilon\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{F}^{\left({k},\mathrm{0}\right)} \left({z}_{\mathrm{0}} \mid{m}\right)}{\left({k}+\mathrm{1}\right)!}\epsilon^{{k}} \\ $$$$\mathrm{Now},\mathrm{divide}\:\mathrm{dy}\:{Npil}: \\ $$$${P}=\frac{\mathrm{2}\epsilon}{{Npil}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{F}^{\left({z},\mathrm{0}\right)} \left({z}_{\mathrm{0}} \mid{m}\right)}{\left({k}+\mathrm{1}\right)!}\epsilon^{{k}} \\ $$$$\left(\mathrm{3}\right):=\underset{{n}=\mathrm{3}} {\overset{\mathrm{10}} {\sum}}\frac{{n}^{\mathrm{4}} +^{{n}} +\mathrm{1}}{{n}^{\mathrm{7}} −{n}}\oplus\frac{\mathrm{1}}{{N}\mathrm{p}{il}} \\ $$$$=\underset{{n}=\mathrm{3}} {\overset{\mathrm{10}} {\sum}}\frac{\left({n}^{\mathrm{2}} −{n}+\mathrm{1}\right)\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)}{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)\left({n}^{\mathrm{2}} −{n}+\mathrm{1}\right)}\oplus\frac{\mathrm{1}}{{Npil}} \\ $$$$=\underset{{n}=\mathrm{3}} {\overset{\mathrm{10}} {\sum}}\frac{\mathrm{1}}{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)}\oplus\frac{\mathrm{1}}{{Npil}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{3}} {\overset{\mathrm{10}} {\sum}}\left(\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right){n}}−\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}\right)\oplus\frac{\mathrm{1}}{{Npil}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{10}×\mathrm{11}}\right)\oplus\frac{\mathrm{1}}{{Npil}} \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{120}}\right)\oplus\frac{\mathrm{1}}{{Npil}} \\ $$$$=\frac{\mathrm{13}}{\mathrm{165}}\oplus\frac{\mathrm{1}}{{Npil}} \\ $$$$\begin{array}{|c|}{{P}=\frac{\mathrm{13}}{\mathrm{165}{Npil}}}\\\hline\end{array} \\ $$