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Chapter 13: Problem 24

Evaluate the double integral. $$ \int_{0}^{\infty} \int_{0}^{\infty} x y e^{-\left(x^{2}+y^{2}\right)} d x d y $$

Short Answer

Expert verified
The solution of the double integral \(\int_{0}^{\infty} \int_{0}^{\infty} x y e^{-\left(x^{2}+y^{2}\right)} d x d y\) is \(1/32\)

Step by step solution

01

Transform from Cartesian to Polar Coordinates

Since the integral is symmetric in x and y, and due to the presence of \(x^{2} + y^{2}\) in the integrand, it can be simplified by converting to polar coordinates with \(r^{2} = x^{2} + y^{2}\) and \(x = r cos(\theta)\), \(y = r sin(\theta)\), and accordingly differential area in polar coordinates will be \(dxdy = rdrd\theta\). With these transformations, the integral becomes \(\int_{0}^{\infty} \int_{0}^{\infty} r^{2} cos(\theta) sin(\theta) e^{-r^{2}} r dr d\theta\)
02

Calculate the integral

Now, the integral can be separated and factored. We get \(r^{3} cos(\theta) sin(\theta) e^{-r^{2}}\), and this bounds to 0 to \(\infty\) for r and 0 to \(\pi/2\) for \(\theta\), This integral further simplifies to equal \(1/8\)
03

Evaluate the integral over theta

Now let us calculate the remaining integral. We have already factored out the r-dependence, so we are left with \(\int_{0}^{\pi/2} cos(\theta) sin(\theta) d\theta\). This evaluates to \(1/4\)

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