Question

As needed, use a computer to plot graphs and to check values of integrals. By changing to polar coordinates, evaluate $$\int_{0}^{\infty} \int_{0}^{\infty} e^{-\sqrt{x^{2}+y^{2}}} d x d y.$$

Short answer

The value of the integral is \(\textstyle 2\pi\).

Step-by-step solution

Understand the Integral

Given the double integral \(\textstyle \int_{0}^{\infty} \int_{0}^{\infty} e^{-\sqrt{x^{2}+y^{2}}} d x d y\), the goal is to evaluate it by changing to polar coordinates.

Set Up the Polar Coordinates Transformation

In polar coordinates, \((x, y) = (r \cos \theta, r \sin \theta)\) and the Jacobian of the transformation is \(|J| = r\). Moreover, \(\textstyle x^2 + y^2 = r^2\). Hence, the integral becomes \(\textstyle \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r} r d\theta d r\).

Evaluate the \( \theta \) Integral

Integrate with respect to \theta, keeping \r as a constant: \(\textstyle \int_{0}^{2\pi} e^{-r} r d\theta = r e^{-r} \int_{0}^{2\pi} d\theta = r e^{-r} \cdot 2\pi\).

Simplify the Integral

The integral now simplifies to \(\textstyle 2\pi \int_{0}^{\infty} r e^{-r} d r\).

Evaluate the Remaining Integral

Consider the integral \(\textstyle \int_{0}^{\infty} r e^{-r} d r\). Use integration by parts where \(\textstyle u = r\) and \(\textstyle dv = e^{-r} dr\), hence \(\textstyle du = dr\) and \(\textstyle v = -e^{-r} \). Applying integration by parts, we get \(\textstyle \int_{0}^{\infty} r e^{-r} d r = \left[ -r e^{-r} \right]_{0}^{\infty} + \int_{0}^{\infty} e^{-r} d r\).

Solve the Evaluated Expression

The boundary term \(\textstyle \left[ -r e^{-r} \right]_{0}^{\infty} = 0\) since \(\textstyle \lim_{r \to \infty} -r e^{-r} = 0\) and \(\textstyle -0 \cdot e^{-0} = 0\). Thus, we have \(\textstyle \int_{0}^{\infty} r e^{-r} d r = \int_{0}^{\infty} e^{-r} d r = \left[ -e^{-r} \right]_{0}^{\infty} = 1\).

Combine All Results

Combining the results, \(\textstyle 2\pi \int_{0}^{\infty} r e^{-r} d r = 2\pi \cdot 1 = 2\pi\).

Key concepts

Double Integral

A double integral is a way to integrate over a two-dimensional area. It's basically adding up little pieces of area (much like how a single integral adds up line segments). When dealing with functions of two variables, a double integral gives you a way to find the volume under a surface defined by those variables.

For example, the double integral \textstyle \int_{0}^{\infty} \int_{0}^{\infty} e^{-\sqrt{x^{2}+y^{2}}} \, dx \, dy}\, tells us to sum up all the tiny areas under the surface described by \textstyle e^{-\sqrt{x^{2}+y^{2}}}\, across all positive \( x \) and \( y \) values.

We typically write double integrals like this:

• Inner integral (\(dy\)) integrates respect to \(y\), treating \(x\) as a constant.
• Outer integral (\(dx\)) integrates the result of the inner integral, this time with respect to \(x\).

To simplify the evaluation, we often use coordinate transformations like polar coordinates.

Polar Coordinates

Polar coordinates, represented as \((r, \theta)\), describe points in the plane using a distance from the origin (\(r\)) and an angle from the positive x-axis (\(\theta\)).

For conversion:

• \(x = r \, \cos \theta\)
• \(y = r \, \sin \theta\)
• \(dx \, dy = r \, dr \, d\theta\)

So, the given problem can be converted with these transformations.

Using these transformations, the region of integration becomes simpler, and the integral \textstyle \int_{0}^{\infty} \int_{0}^{\infty} e^{-\sqrt{x^{2}+y^{2}}} \, dx \, dy\ changes to \textstyle \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r} \, r \, d\theta \, dr}\.

Notice that the limits simplify as well: \(r\) ranges from \(0\) to \(\infty\), and \(\theta\) ranges from \(0\) to \(2\pi\). This greatly simplifies the integral.

Integration by Parts

Integration by parts is a technique used to solve integrals where the standard methods aren't straightforward. It is based on the product rule for differentiation and is given by:

(uv - \int v \, du).

In the problem's context, to integrate \textstyle \int_{0}^{\infty} r e^{-r} \, dr\}, we set it up as:

• \(u = r\)
• \(dv = e^{-r} \, dr\)
• \(du = dr\)
• \(v = -e^{-r}\)

Applying the integration by parts formula, we get:
\textstyle \int_{0}^{\infty} r e^{-r} \, dr = [ -r e^{-r} ]_{0}^{\infty} + \int_{0}^{\infty} e^{-r} \, dr\. Both boundary terms resolve to 0 due to limits. Simplifying further, we integrate \(e^{-r}\) from 0 to \(\infty\), resulting in 1.

This allows us to evaluate our original integral fully.