A definite Double Integral!

Calculus Level 3

Find

e ( x 2 + y 2 ) d x d y \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy


The answer is 3.14159.

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2 solutions

Anish Puthuraya
Apr 3, 2014

I = e ( x 2 + y 2 ) d x d y I = \iint_{-\infty}^{\infty} e^{-(x^2+y^2)} dx dy

I = e x 2 e y 2 d x d y I = \iint_{-\infty}^{\infty} e^{-x^2} e^{-y^2} dx dy

I = e y 2 d y e x 2 d x I = \int_{-\infty}^{\infty}e^{-y^2} dy\int_{-\infty}^{\infty} e^{-x^2}dx

I = e y 2 ( π ) d y I = \int_{-\infty}^{\infty} e^{-y^2} (\sqrt{\pi}) dy

I = π e y 2 d y I = \sqrt{\pi} \int_{-\infty}^{\infty} e^{-y^2} dy

I = π π = π = 3.14159 I = \sqrt{\pi} \sqrt{\pi} = \pi = \boxed{3.14159}

Cool! You could have also used the concept of Jacobians, by using polar coordinates substitutions for x and y. (I actually expected that as one of the solutions)

Chandramouli Chowdhury - 7 years, 2 months ago

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Your wish is granted, Chandramouli…..6 years later! See my cylindrical coordinates solution above.

tom engelsman - 10 months, 2 weeks ago

You can actually see its the modification of the pdf of a bivariate normal with means 0, sds 1/sqrt(2) each; and they are independent.

Baidehi Chattopadhyay - 7 years, 1 month ago
Tom Engelsman
Jul 31, 2020

Let x 2 + y 2 = r 2 x^2 + y^2 = r^2 in polar form. Since we are integrating over the entire x y xy- plane, we require r [ 0 , ) , θ [ 0 , 2 π ] . r \in [0, \infty), \theta \in [0 , 2\pi]. Thus, the double integration transforms into:

0 2 π 0 r e r 2 d r d θ = π . \int_{0}^{2\pi} \int_{0}^{\infty} re^{-r^2} dr d\theta = \boxed{\pi}.

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