Thread: Double Integration - Polar Co-ordinates

\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

---------------------------------------------------------------------------------

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.

doesn't become , it becomes , and CAN be integrated using a substitution.

with substitution i get -0.25e^(-2(r^2)) and evaluting between one and zero i get:

(-0.25e^(-2))+0.25 d theta

theta*((-0.25e^(-2))+0.25) then sub in pi/2

seem correct?

Yes, so now evaluate