\(\Pi =\cfrac { \partial \, }{ \partial t } \int _{ 0 }^{ r }{ xtan(\theta )e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } } } d\, r\)
Consider the differentiation wrt \(t\) first,
\( III=\left( xtan(\theta )e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } } \right) ^{ ' }=\cfrac { dx }{ dt } tan(\theta )e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } }+xsec^{ 2 }(\theta )e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } }\cfrac { d\theta }{ dt } -xtan(\theta ).e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } }.\cfrac { 1 }{ a_{ e }cos(\theta ) } \cfrac { dx }{ dt } -xtan(\theta ).e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } }.\cfrac { x }{ a_{ e } } tan(\theta )sec(\theta )\cfrac { d\theta }{ dt } \)
\( =e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } }\left\{ tan(\theta )\cfrac { dx }{ dt } +xsec^{ 2 }(\theta )\cfrac { d\theta }{ dt } -\cfrac { x }{ a_{ e } } tan(\theta )sec(\theta )\cfrac { dx }{ dt } -\cfrac { x^{ 2 } }{ a_{ e } } tan^{ 2 }(\theta )sec(\theta )\cfrac { d\theta }{ dt } \right\} \)
Since,
\( \cfrac { r }{ x } =tan(\theta )\)
\( -\cfrac { r }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } =sec^{ 2 }(\theta )\cfrac { d\, \theta }{ d\, t } \)
\( xsec^{ 2 }(\theta )\cfrac { d\, \theta }{ d\, t } =-\cfrac { r }{ x } \cfrac { d\, x }{ d\, t } =-\cfrac { d\, x }{ d\, t } tan(\theta )\)
So,
\( III=e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } }\left\{ tan(\theta )\cfrac { dx }{ dt } -\cfrac { d\, x }{ d\, t } tan(\theta )-\cfrac { x }{ a_{ e } } tan(\theta )sec(\theta )\cfrac { dx }{ dt } +\cfrac { x }{ a_{ e } } \cfrac { tan^{ 2 }(\theta ) }{ sec(\theta ) } \cfrac { dx }{ dt } \right\} \)
\( III=e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } }\cfrac { x }{ a_{ e } } \left\{ \cfrac { tan^{ 2 }(\theta ) }{ sec(\theta ) } -tan(\theta )sec(\theta ) \right\} \cfrac { dx }{ dt } \)
Substitute back into \(\Pi\),
\( \Pi =\int _{ 0 }^{ r }{ e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } }\cfrac { x }{ a_{ e } } \left\{ \cfrac { tan^{ 2 }(\theta ) }{ sec(\theta ) } -tan(\theta )sec(\theta ) \right\} \cfrac { dx }{ dt } } d\, r\)
But,
\( \cfrac { r }{ x } =tan(\theta )\)
\( \cfrac { 1 }{ x } =sec^{ 2 }(\theta )\cfrac { d\, \theta }{ dr } \)
\( dr=xsec^{ 2 }(\theta ){ d\, \theta }\)
So,
\( \Pi =\cfrac { x^{ 2 } }{ a_{ e } } \cfrac { dx }{ dt } \int _{ 0 }^{ \theta_0 }{ e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } }\left\{ \cfrac { tan^{ 2 }(\theta ) }{ sec(\theta ) } -tan(\theta )sec(\theta ) \right\} } sec^{ 2 }(\theta ){ d\, \theta }\)
\( \Pi =\cfrac { x^{ 2 } }{ a_{ e } } \cfrac { dx }{ dt } \int _{ 0 }^{ \theta_0 }{ e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } }\left\{ tan^{ 2 }(\theta )sec(\theta )-tan(\theta )sec^{ 3 }(\theta ) \right\} } { d\, \theta }\)
\( \Pi =\cfrac { x^{ 2 } }{ a_{ e } } \cfrac { dx }{ dt } \int _{ 0 }^{ \theta_0 }{ e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } }\left\{ tan^{ 2 }(\theta )sec(\theta )-\cfrac { sin(\theta ) }{ cos^{ 4 }(\theta ) } \right\} } { d\, \theta }\)
Since,
\( tan(\theta)d(sec(\theta))=tan^{ 2 }(\theta)sec(\theta)d(\theta)=\sqrt { sec^{ 2 }(\theta)-1 } d(sec(\theta))\)
and
\(d(cos(\theta))=-sin(\theta)\)
We break \(\Pi\) into two parts,
\(\Pi _{ 1 }=\cfrac { x^{ 2 } }{ a_{ e } } \cfrac { dx }{ dt } \int _{ 0 }^{ \theta_0 }{ e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } }\left\{ tan^{ 2 }(\theta )sec(\theta ) \right\} } { d\, \theta }\)
\(\Pi _{ 1 }=\cfrac { x^{ 2 } }{ a_{ e } } \cfrac { dx }{ dt } \int _{ 0 }^{ sec(\theta_0) }{ e^{ -\cfrac { x.sec(\theta ) }{ a_{ e } } }\left\{ \sqrt { sec^{ 2 }(\theta)-1 } \right\} } { d\, sec(\theta ) }\)
\(\Pi _{ 1 }=\cfrac { x^{ 2 } }{ a_{ e } } \cfrac { dx }{ dt } \int _{ 0 }^{y_0 }{ e^{ -\cfrac { x.y }{ a_{ e } } }\left\{ \sqrt { y^{ 2 }-1 } \right\} } { d\, y }\)
where \(y_0=sec(\theta_0)\)
and
\( \Pi _{ 2 }=\cfrac { x^{ 2 } }{ a_{ e } } \cfrac { dx }{ dt } \int _{ 0 }^{ \theta_0 }{ e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } }\left\{ -\cfrac { sin(\theta ) }{ cos^{ 4 }(\theta ) } \right\} } { d\, \theta }\)
\( \Pi _{ 2 }=\cfrac { x^{ 2 } }{ a_{ e } } \cfrac { dx }{ dt } \int _{ 1 }^{ cos(\theta_0) }{ e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } }\left\{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } \right\} } { d\, cos(\theta ) }\)
\( \Pi _{ 2 }=\cfrac { x^{ 2 } }{ a_{ e } } \cfrac { dx }{ dt } \int _{ 1 }^{ y_0 }{ e^{ -\cfrac { x }{ a_{ e }y } }\left\{ \cfrac { 1 }{ y^{ 4 } } \right\} } { d\,y }\)
where \(y_0=cos(\theta_0)\)
Combining, just to look at it,
$$\Pi=\cfrac { x^{ 2 } }{ a_{ e } }\cfrac { dx }{ dt } \left\{ \int _{ 1 }^{sec(\theta_0) }{ e^{ -\cfrac { x.y }{ a_{ e } } }\left\{ \sqrt { y^{ 2 }-1 } \right\} } { d\, y }+ \int _{ 1 }^{ cos(\theta_0) }{ e^{ -\cfrac { x }{ a_{ e }y } }\left\{ \cfrac { 1 }{ y^{ 4 } } \right\} } { d\,y }\right\}$$
At every point along \(x\), the integration is done from zero to \(\theta_0\). When \(x\) changes, \(\theta_0\) changes, the limit of integration changes but not \(\theta\). The variable \(\theta\) is independent of \(x\).
This corrects the post "Electron Orbit B Field II" dated 17 Oct 2014.
I apologize, I have half a brain and it is infected...
