# Double Integration - Polar Co-ordinates

• May 8th 2011, 06:13 AM
keelejody
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

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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.
• May 8th 2011, 06:22 AM
Prove It
$\displaystyle \displaystyle dy\, dx$ doesn't become $\displaystyle \displaystyle dr\,d\theta$, it becomes $\displaystyle \displaystyle r\,dr\,d\theta$, and $\displaystyle \displaystyle r\,e^{-2r^2}$ CAN be integrated using a substitution.
• May 8th 2011, 06:32 AM
keelejody
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?
• May 8th 2011, 06:38 AM
Prove It
Yes, so now evaluate $\displaystyle \displaystyle \left \left(1-e^{-2}\right)\frac{\theta}{4}\right|_0^{\frac{\pi}{2}}$