transform-the-cartesian-inyegral-0-1-0-1-x-2-e-x-2-y-2-dy-dx-into-polar-integral-and-evaluate-it-

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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