Problem 25 As needed, use a computer to plo... [FREE SOLUTION]

Chapter 5: Problem 25

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

Expert verified

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, $\textstyle 2\pi \int_{0}^{\infty} r e^{-r} d r = 2\pi \cdot 1 = 2\pi$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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