Question 1041431
The region of integration is the right half of the parabolic region (or to the right of the y-axis.)
{{{graph( 300, 200, -2, 2, 0, 4, 4-x^2 )}}}


{{{int( int ((xe^(2y))/(4-y), dy,0,4-x^2),dx,0,2  )}}}

={{{int( int ((xe^(2y))/(4-y), dx,0,sqrt(4-y)),dy,0,4  )}}}

(In the inner integral, I integrated wrt x only from 0 to {{{sqrt(4-y)}}}, as per the region of integration.)

={{{int( (e^(2y)/(4-y))*int(x, dx,0,sqrt(4-y)),dy,0,4 )}}}

={{{int( (e^(2y)/(4-y))*(x^2/2)0^sqrt(4-y),dy,0,4 )}}}

={{{(1/2)*int( (e^(2y)/(4-y))*(4-y),dy,0,4 )}}}
={{{(1/2)*int( e^(2y),dy,0,4 )}}}
={{{(1/4)(e^(2y))0^4}}}

={{{(1/4)(e^8 - 1)}}}