I have the folllwoing integral <msubsup> &#x222B;<!-- ∫ --> 0 <mrow class="MJX-TeXAt

reryfaikear 2022-05-20 Answered
I have the folllwoing integral
0 π / 4 0 2 π sin 2 ( ρ ) sin ( θ ) cos ( ρ ) sin 2 ( ρ ) sin ( θ )   d θ d ρ
and I would like to know why exaclty one can rewrite it as follows
( 0 π / 4 sin 2 ( ρ ) cos ( ρ ) sin 2 ( ρ ) d ρ ) 0 2 π sin θ d θ .
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Answers (1)

Julien Carrillo
Answered 2022-05-21 Author has 13 answers
First, factor out the sinθ:
0 π / 4 0 2 π ( sin 2 ( ρ ) cos ( ρ ) sin 2 ( ρ ) ) sin θ d θ d ρ .
Since ρ does not depend on θ, it can be factored out of the inner θ integral (just as a constant can be factored out of an integral):
0 π / 4 ( sin 2 ( ρ ) cos ( ρ ) sin 2 ( ρ ) ) 0 2 π sin θ d θ d ρ .
The quantity 0 2 π sin θ d θ is just a number (in this case, it is zero), so it can be factored out of the outer ρ integral just as any number can:
( 0 π / 4 ( sin 2 ( ρ ) cos ( ρ ) sin 2 ( ρ ) ) d ρ ) 0 2 π sin θ d θ .
In general, the following holds via the same logic:
a b c d f ( x ) g ( y ) d y d x = ( a b f ( x ) d x ) ( c d g ( y ) d y ) .
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