Question

Evaluate:latex: \int_0^{\infty}\int_0^{\infty}e^{-\left(x^2+y^2\right)}dxdy

Answer 1

Double Integrals - Changing Coordinate System

Question:

Evaluate the following expression:

$\displaystyle \int_0^{\infty}\int_0^{\infty}e^{-\left(x^2+y^2\right)}\, dx \, dy$

Answer:

The given expression is in the standard Cartesian Coordinate System, with the X and Y axes.  

Here, we notice the $x^2+y^2$ part, which can be easily expressed in Polar Coordinates.  

It seems that the problem could be easier to solve if we change the coordinate system to Polar. Let's do just that!

In Polar Coordinates, we have:

$x = r\cos \theta \\ \\ y = r\sin \theta \\ \\ \\ \rightarrow r = \sqrt{x^2+y^2}$

Also, we know that:

$\boxed{dx \, dy = r \, dr \, d\theta}$

So, we can proceed to solve this integral.

For the limits, we see that both x and y range from $0$ to $\infty$. So, $r^2=x^2+y^2$ would also range from $0$ to $\infty$.

$\displaystyle I=\int\limits_0^\infty\int\limits_0^\infty e^{-\left(x^2+y^2\right)}dx\, dy \\\\\\\implies I=\int\limits_{r=0}^\infty\ \int\limits_{\theta=0}^{2\pi}e^{-r^2}\, r\, dr\, d\theta\\\\\\\implies I=2\pi\int\limits_0^\infty re^{-r^2} dr\\\\\\\text{Put }r^2=t \implies 2r\, dr=dt\\\\\\\implies I=2\pi\int\limits_0^\infty\frac{1}{2}e^{-t}\,dt\\\\\\\implies I=\pi\left[\frac{e^{-t}}{-1}\right]_0^\infty\\\\\\\implies I=-\pi (e^{-\infty}-e^0)\\\\\\\implies I=-\pi(0-1)\\\\\\\implies I=\pi$

$\implies \boxed{I=\int\limits_0^\infty\int\limits_0^\infty e^{-\left(x^2+y^2\right)}dxdy = \pi}$

Thus, the Answer of the Integral is π.

$\rule{300}{1}$

Extra Info

Proof

for $dx \, dy = r\, dr \, d\theta$

1) Geometric Approach

dx and dy represent small changes in x and y directions. So, in Polar Coordinates, we must think of what would represent small changes in the r and $\theta$ directions.

In the radial direction, i.e. the r direction, the small change would be dr.

In the angular direction, or what we call the azimuthal direction, we would consider the small arc. This arc length is represented as $r\, d\theta$. So this becomes the small change in the azimuthal direction.

Hence, we have:

$dx \, dy \equiv (dr)\ (r\, d\theta) = r\, dr \, d\theta$

2) Algebraic Approach

[Skip if you do not have any idea of Multivariable Calculus]

This goes a bit into Multivariable Calculus. We are essentially changing our Domain of the Integration. So, we need to find a mapping function which would represent the old variables in form of new variables and find the determinant of the Jacobian Matrix.

Namely, this is what we do:

$\displaystyle dx\, dy=\underbrace{\left|\left|\begin{array}{cc}\dfrac{\partial x}{\partial r} & \dfrac{\partial x}{\partial \theta} \\\\\dfrac{\partial y}{\partial r} & \dfrac{\partial y}{\partial \theta}\end{array}\right|\right|}_{\substack{\text{Jacobian}\\ \text{Matrix} \\ \text{Determinant}}}\ dr\, d\theta$

And now, since we know that

$x = r\cos\theta \\\\ y = r\sin\theta$

We can compute the required things:

$\displaystyle \implies dx\, dy=\left|\left|\begin{array}{cc}\cos\theta & -r\sin\theta\\\\\sin\theta & r\cos\theta\end{array}\right|\right|\ dr\, d\theta \\\\\\ \implies dx \, dy = (r\cos^2\theta+r\sin^2\theta)\ dr\, d\theta \\\\\\ \implies dx\, dy = r\, dr \, d\theta$