\iint(lower case B) e^-2((x^2)+(y^2)) dydx

where B is the quarter of the circle of radius 1 in the upper right-hand quadrant

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ok so e^-2((x^2)+(y^2)) becomes e^-2(r^2)

the inner limits are between 0<r<1 since r=1

and the integral goes from dydx to dr d(theta)

the outer limit is 0<theta<(pi/2) since were are dealing with the top right hand quadrant?

can the integral e^-2(r^2) dr be done? i cant see how, unless we change the order? could be way off there though.