Question Number 47737 by ajfour last updated on 14/Nov/18
$$\:\:{I}\:=\:\int_{\mathrm{0}} ^{\:\:{L}/\mathrm{2}} \frac{{Rz}^{\mathrm{2}} }{\left({d}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\sqrt{{d}^{\mathrm{2}} +{z}^{\mathrm{2}} −{R}^{\mathrm{2}} }}\:{dz} \\ $$$$\:{Find}\:{I}\:. \\ $$
Answered by MJS last updated on 14/Nov/18
![R∫(z^2 /((z^2 +d^2 )(√(z^2 +d^2 −R^2 ))))dz= =R∫(dz/(√(z^2 +d^2 −R^(2 ) )))−d^2 R∫(dz/((z^2 +d^2 )(√(z^2 +d^2 −R^2 )))) R∫(dz/(√(z^2 +d^2 −R^(2 ) )))= [t=(z/(√(d^2 −R^2 ))) → dz=(√(d^2 −R^2 ))dt] =R∫(dt/(√(t^2 +1)))=Rln (t+(√(t^2 +1))) =Rln ((z+(√(z^2 +d^2 −R^2 )))/(√(d^2 −R^2 ))) −d^2 R∫(dz/((z^2 +d^2 )(√(z^2 +d^2 −R^2 ))))= [z=(√(d^2 −R^2 ))tan u → u=arctan (z/(√(d^2 −R^2 ))); dz=(√(d^2 −R^2 ))sec^2 u du] =−d^2 R∫(((√(d^2 −R^2 ))sec^2 u)/((d^2 +(d^2 −R^2 )tan^2 u)(√(d^2 −R^2 +(d^2 −R^2 )tan^2 u))))du= =−d^2 R∫((sec u)/(d^2 +(d^2 −R^2 )tan^2 u))du=−d^2 R∫((cos u)/(d^2 −R^2 sin^2 u))du= [v=sin u → du=(dv/(cos u))] =−d^2 R∫(dv/(d^2 −R^2 v^2 ))=−d^2 R∫(dv/((d+Rv)(d−Rv)))= =(dR/2)∫(dv/(d−Rv))−(dR/2)∫(dv/(d+Rv))= =(d/2)ln ((d−Rv)/(d+Rv)) =(d/2)ln ((d−Rsin u)/(d+Rsin u))= =(d/2)ln ((d−Rsin arctan (z/(√(d^2 −R^2 ))))/(d+Rsin arctan (z/(√(d^2 −R^2 )))))= =(d/2)ln ((d−R(z/(√(z^2 +d^2 −R^2 ))))/(d+R(z/(√(z^2 +d^2 −R^2 ))))) =(d/2)ln ((d(√(z^2 +d^2 −R^2 ))−Rz)/(d(√(z^2 +d^2 −R^2 ))+Rz)) ∫((Rz^2 )/((z^2 +d^2 )(√(z^2 +d^2 −R^2 ))))dz= Rln ∣((z+(√(z^2 +d^2 −R^2 )))/(√(d^2 −R^2 )))∣ +(d/2)ln ∣((Rz−d(√(z^2 +d^2 −R^2 )))/(Rz+d(√(z^2 +d^2 −R^2 ))))∣ +C](Q47751.png)
$${R}\int\frac{{z}^{\mathrm{2}} }{\left({z}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\sqrt{{z}^{\mathrm{2}} +{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }}{dz}= \\ $$$$={R}\int\frac{{dz}}{\sqrt{{z}^{\mathrm{2}} +{d}^{\mathrm{2}} −{R}^{\mathrm{2}\:} }}−{d}^{\mathrm{2}} {R}\int\frac{{dz}}{\left({z}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\sqrt{{z}^{\mathrm{2}} +{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }} \\ $$$$ \\ $$$${R}\int\frac{{dz}}{\sqrt{{z}^{\mathrm{2}} +{d}^{\mathrm{2}} −{R}^{\mathrm{2}\:} }}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{{z}}{\sqrt{{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }}\:\rightarrow\:{dz}=\sqrt{{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }{dt}\right] \\ $$$$={R}\int\frac{{dt}}{\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}={R}\mathrm{ln}\:\left({t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\right)\:={R}\mathrm{ln}\:\frac{{z}+\sqrt{{z}^{\mathrm{2}} +{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }}{\sqrt{{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }} \\ $$$$ \\ $$$$−{d}^{\mathrm{2}} {R}\int\frac{{dz}}{\left({z}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\sqrt{{z}^{\mathrm{2}} +{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }}= \\ $$$$\:\:\:\:\:\left[{z}=\sqrt{{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }\mathrm{tan}\:{u}\:\rightarrow\:{u}=\mathrm{arctan}\:\frac{{z}}{\sqrt{{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }};\:{dz}=\sqrt{{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }\mathrm{sec}^{\mathrm{2}} \:{u}\:{du}\right] \\ $$$$=−{d}^{\mathrm{2}} {R}\int\frac{\sqrt{{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }\mathrm{sec}^{\mathrm{2}} \:{u}}{\left({d}^{\mathrm{2}} +\left({d}^{\mathrm{2}} −{R}^{\mathrm{2}} \right)\mathrm{tan}^{\mathrm{2}} \:{u}\right)\sqrt{{d}^{\mathrm{2}} −{R}^{\mathrm{2}} +\left({d}^{\mathrm{2}} −{R}^{\mathrm{2}} \right)\mathrm{tan}^{\mathrm{2}} \:{u}}}{du}= \\ $$$$=−{d}^{\mathrm{2}} {R}\int\frac{\mathrm{sec}\:{u}}{{d}^{\mathrm{2}} +\left({d}^{\mathrm{2}} −{R}^{\mathrm{2}} \right)\mathrm{tan}^{\mathrm{2}} \:{u}}{du}=−{d}^{\mathrm{2}} {R}\int\frac{\mathrm{cos}\:{u}}{{d}^{\mathrm{2}} −{R}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:{u}}{du}= \\ $$$$\:\:\:\:\:\left[{v}=\mathrm{sin}\:{u}\:\rightarrow\:{du}=\frac{{dv}}{\mathrm{cos}\:{u}}\right] \\ $$$$=−{d}^{\mathrm{2}} {R}\int\frac{{dv}}{{d}^{\mathrm{2}} −{R}^{\mathrm{2}} {v}^{\mathrm{2}} }=−{d}^{\mathrm{2}} {R}\int\frac{{dv}}{\left({d}+{Rv}\right)\left({d}−{Rv}\right)}= \\ $$$$=\frac{{dR}}{\mathrm{2}}\int\frac{{dv}}{{d}−{Rv}}−\frac{{dR}}{\mathrm{2}}\int\frac{{dv}}{{d}+{Rv}}= \\ $$$$=\frac{{d}}{\mathrm{2}}\mathrm{ln}\:\frac{{d}−{Rv}}{{d}+{Rv}}\:=\frac{{d}}{\mathrm{2}}\mathrm{ln}\:\frac{{d}−{R}\mathrm{sin}\:{u}}{{d}+{R}\mathrm{sin}\:{u}}= \\ $$$$=\frac{{d}}{\mathrm{2}}\mathrm{ln}\:\frac{{d}−{R}\mathrm{sin}\:\mathrm{arctan}\:\frac{{z}}{\sqrt{{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }}}{{d}+{R}\mathrm{sin}\:\mathrm{arctan}\:\frac{{z}}{\sqrt{{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }}}= \\ $$$$=\frac{{d}}{\mathrm{2}}\mathrm{ln}\:\frac{{d}−{R}\frac{{z}}{\sqrt{{z}^{\mathrm{2}} +{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }}}{{d}+{R}\frac{{z}}{\sqrt{{z}^{\mathrm{2}} +{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }}}\:=\frac{{d}}{\mathrm{2}}\mathrm{ln}\:\frac{{d}\sqrt{{z}^{\mathrm{2}} +{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }−{Rz}}{{d}\sqrt{{z}^{\mathrm{2}} +{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }+{Rz}} \\ $$$$ \\ $$$$\int\frac{{Rz}^{\mathrm{2}} }{\left({z}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\sqrt{{z}^{\mathrm{2}} +{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }}{dz}= \\ $$$${R}\mathrm{ln}\:\mid\frac{{z}+\sqrt{{z}^{\mathrm{2}} +{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }}{\sqrt{{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }}\mid\:+\frac{{d}}{\mathrm{2}}\mathrm{ln}\:\mid\frac{{Rz}−{d}\sqrt{{z}^{\mathrm{2}} +{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }}{{Rz}+{d}\sqrt{{z}^{\mathrm{2}} +{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }}\mid\:+{C} \\ $$
Commented by MJS last updated on 14/Nov/18
$$\mathrm{it}\:\mathrm{only}\:\mathrm{works}\:\mathrm{for}\:{d}>{R} \\ $$
Commented by MJS last updated on 14/Nov/18
$$\mathrm{I}\:\mathrm{have}\:\mathrm{no}\:\mathrm{time}\:\mathrm{to}\:\mathrm{check}\:\mathrm{it}\:\mathrm{for}\:\mathrm{typos}\:\mathrm{and}\:\mathrm{other} \\ $$$$\mathrm{minor}\:\mathrm{mistakes}\:\mathrm{but}\:\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{path}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it} \\ $$
Commented by ajfour last updated on 14/Nov/18
$${Thank}\:{you}\:{Sir},\:{understood}\:{the} \\ $$$${method},\:{quite}\:{natural}. \\ $$
Commented by ajfour last updated on 14/Nov/18
$${of}\:{course},\:{it}\:{should}\:{be}\:{so},\:{please} \\ $$$${resolve}\:{our}\:{issue};\:{if}\:{at}\:{all}\:{this} \\ $$$${integral}\:{arises}\:{or}\:{not}\:? \\ $$$${Please}\:{view}\:{Q}.\mathrm{47728}\: \\ $$