- then ∮ C ( L d x + M d y ) = ∬ D ( ∂ M ∂ x − ∂ L ∂ y ) d A {\displaystyle \
oint _{C}(L\,dx+M\,dy)=\iint _{D}\left({\frac {\partial M}{\partial x}}-{\frac...
- _{\Sigma }(\nabla \times \mathbf {F} )\cdot \mathrm {d} \mathbf {\Sigma } =\
oint _{\partial \Sigma }\mathbf {F} \cdot \mathrm {d} \mathbf {\Gamma } .} More...
- ( v l × B ) ⋅ d l {\textstyle \
oint \left(\mathbf {v} \times \mathbf {B} \right)\cdot \mathrm {d} \mathbf {l} =\
oint \left(\mathbf {v} _{l}\times \mathbf...
- {3}{4z}}}}\\&=-i\
oint _{C}{\frac {4}{3z^{3}+10z+{\frac {3}{z}}}}\,dz\\&=-4i\
oint _{C}{\frac {dz}{3z^{3}+10z+{\frac {3}{z}}}}\\&=-4i\
oint _{C}{\frac...
- x + u d y ) {\displaystyle \
oint _{\gamma }f(z)\,dz=\
oint _{\gamma }(u+iv)(dx+i\,dy)=\
oint _{\gamma }(u\,dx-v\,dy)+i\
oint _{\gamma }(v\,dx+u\,dy)} By...
- 1 2 π i ∮ γ f ( z ) z − a d z . {\displaystyle f(a)={\frac {1}{2\pi i}}\
oint _{\gamma }{\frac {f(z)}{z-a}}\,dz.\,} The
proof of this
statement uses the...
- \operatorname {Res} (f,c)={1 \over 2\pi i}\
oint _{\gamma }f(z)\,dz={1 \over 2\pi i}\sum _{n=-\infty }^{\infty }\
oint _{\gamma }a_{n}(z-c)^{n}\,dz=a_{-1}} using...
- {\displaystyle \
oint _{\Gamma _{N}}g(z)dz=\
oint _{\Gamma _{N}}\left({\frac {1}{z}}+{\frac {1}{w-z}}\right)\pi \cot(\pi z)dz=-w\
oint _{\Gamma _{N}}{\frac...
- _{\Omega }} is a
volume integral over the
volume Ω, ∮ ∂ Σ {\displaystyle \
oint _{\partial \Sigma }} is a line
integral around the
boundary curve ∂Σ, with...
- R ⋅ d S {\displaystyle \Phi (R)=\
oint _{\partial B_{R}(\mathbf {r} _{0})}\mathbf {E} _{0}\cdot d\mathbf {S} =\
oint _{\partial B_{R}(\mathbf {r} _{0})}\mathbf...
