transform the cartesian inyegral ∫_0 ^1 ∫_0 ^(√(1−x^2 )) e^(−(x^2 +y^2 )) dy dx into polar integral and evaluate it.


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Question Number 146067 by gsk2684 last updated on 10/Jul/21

$t r a n s f o r m t h e c a r t e s i a n i n y e g r a l$ $\underset{0}{\int^{1}} \underset{0}{\int^{\sqrt{1 - x^{2}}}} e^{- (x^{2} + y^{2})} d y d x i n t o p o l a r i n t e g r a l$ $a n d e v a l u a t e i t .$
	Answered by Olaf_Thorendsen last updated on 10/Jul/21

	$A = \int_{0}^{1} \int_{0}^{\sqrt{1 - x^{2}}} e^{- (x^{2} + y^{2})} d y d x$ $x = r \cos θ, y = r \sin θ$ $d S = d y d x = r d r d θ$ $A = \int_{0}^{\frac{π}{2}} \int_{0}^{1} e^{- r^{2}} r d r d θ$ $A = \frac{π}{2} {[- e^{- r^{2}}]}_{0}^{1} = \frac{π}{2} (1 - \frac{1}{e})$
	Commented by gsk2684 last updated on 11/Jul/21

	$t h a n k y o u s i r$ $\underset{0}{\int^{\frac{Π}{2}}} {[\frac{- e^{- r^{2}}}{2}]}_{0}^{1} d θ = \frac{- 1}{2} \underset{0}{\int^{\frac{Π}{2}}} (e^{- 1} - 1) d θ$ $= \frac{- 1}{2} (\frac{1}{e} - 1) (\frac{Π}{2}) = \frac{Π}{4} (1 - \frac{1}{e})$
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