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Question Number 203349 by Mathspace last updated on 17/Jan/24

c a l c u l a t e \int \int_{{[0, a]}^{2}} e^{- x^{2} - y^{2}} d x d y

c a n y o u f i n d \int_{0}^{a} e^{- x^{2}} d x ?

a > 0

Answered by witcher3 last updated on 17/Jan/24

the idee \int \int_{{[0, \infty]}^{2}} f (x, y) dxdy = \int_{0}^{\frac{π}{2}} \int_{0}^{\infty} f (rcos (a), rsin (a)) rdrda

{can}^{'} t hellp ther

for a > 0 {[0, a]}^{2} is a square if you try to fill it withe disc

is not easy

0 ⩽ rcos (z) ⩽ a

0 ⩽ rsin (z) ⩽ a

0 < r < \frac{a}{\cos (z)}, 0 ⩽ r < \frac{a}{\sin (z)}

\Rightarrow r < \min (\frac{a}{\cos (z)}, \frac{a}{\sin (z)})

= \int_{0}^{\frac{π}{4}} \int_{0}^{\frac{a}{\cos (z)}} {re}^{- r^{2}} drdz + \int_{\frac{π}{4}}^{\frac{π}{2}} \int_{0}^{\frac{a}{\sin (z)}} {re}^{- r^{2}} drdz

= \int_{0}^{\frac{π}{4}} \int_{0}^{\frac{a}{\cos (z)}} {2 re}^{- r^{2}} drdz

= \int_{0}^{\frac{π}{4}} {(- e^{- r^{2}}]}_{0}^{\frac{a}{\cos (z)}} dz

= \int_{0}^{\frac{π}{4}} 1 - e^{- \frac{a^{2}}{\cos^{2} (z)}} dz

tg (z) = t

= \int_{0}^{1} (1 - e^{- a^{2} (1 + t^{2})}) \frac{dt}{1 + t^{2}}, not easy from heree

we can use feyneman and givre answer withe “ \erf^{″}

\int_{0}^{a} e^{- x^{2}} dx \dots . should bee first Quation = \frac{\sqrt{π}}{2} \erf (a)

\erf (z) = \frac{2}{\sqrt{π}} \int_{0}^{z} e^{- t^{2}} dt

Than \int \int e^{- x^{2} - y^{2}} dxdy = \int_{0}^{a} e^{- x^{2}} dx . \int_{0}^{a} e^{- y^{2}} dy

= {(\int_{0}^{a} e^{- x^{2}})}^{2} dx = \frac{π}{4} . \erf^{2} (a)

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