Question

Use polar coordinates to set up and evaluate the double integral \(\int_{R} \int f(x, y) d A\). $$ f(x, y)=e^{-\left(x^{2}+y^{2}\right) / 2}, R: x^{2}+y^{2} \leq 25, x \geq 0 $$

Short answer

The result of the double integration of the function \(f(x, y) = e^{-\left(x^{2}+y^{2}\right) / 2}\) over the region \(R: x^{2}+y^{2} \leq 25, x \geq 0\) is \(2\pi(1 - e^{-25/2})\).

Step-by-step solution

Convert to Polar Coordinates

In a polar coordinate system, we know that \(r^2 = x^2 + y^2\) and \(x = r\cos(\theta)\), \(y = r\sin(\theta)\), and the differential area element in rectangular coordinates \(dA = dx \, dy\), becomes \(r \, dr \, d\theta\) in polar coordinates. So, the function can be written as, \(f(r, θ) = e^{-r^2/2}\) and the region revolves from \( r = 0 \) to \( r = 5 \) (as \( \sqrt{25} = 5 \)) and \( \(\theta\) = 0 \) to \( \(\theta\) = \(\pi\) \) (since \( x \geq 0 \) indicates the area lies in the first and fourth quadrants).

Setup the Double Integral

The double integral for this problem can now be set up in polar coordinates as: \[\int_{0}^{5}\int_{0}^{\pi} e^{-r^2/2} \cdot r \, dr \, d\theta\]

Evaluate the Inner Integral

First, we can evaluate the radial integral. By letting \( u = r^{2}/2 \), we get that \( du = r \, dr \). This simplifies the inner integral into \(-2\int e^{-u} \, du\), which evaluates to \(-2e^{-u} = -2e^{-r^2/2}\)

Evaluate the Outer Integral

The outer integral is now: \[ -2\int_{0}^{\pi}\left(e^{-25/2} - e^{0}\right) \, d\theta = -2e^{-25/2}\theta \Bigg|_0^{\pi} + 2\theta \Bigg|_0^{\pi}.\] Evaluating at the limits then gives \( -2\pi e^{-25/2} + 2\pi \).

Final Answer

The final part of this process is to simply clean up the output of the double integration, and simplify the expression. This gives a final answer of \(2\pi(1 - e^{-25/2})\).

Key concepts

Double Integral

When we come across problems that involve calculating the volume under a surface or the area of an irregular shape, we often use a technique known as double integration. The idea is to break down the area into infinitesimally small rectangles, find the area of each rectangle, and then sum up all these tiny areas. This is what the double integral does.
In the context of our exercise, the double integral is expressed as \[\int_{R} \int f(x, y) d A\], which means we're integrating the function f(x, y) over the region R. Imagine slicing the region R into thin strips, stacking them together, and then adding up the volume of the slivers. That's the physical interpretation of the double integral.
Now, while it might sound straightforward, evaluating double integrals directly can get tricky, especially with non-rectangular regions. This is where the use of polar coordinates comes into play, simplifying the integration process over circular regions.

Polar Coordinate System

The polar coordinate system is a two-dimensional coordinate system where each point on a plane is determined by a distance from a fixed point and an angle from a fixed direction. The fixed point is called the pole, similar to the origin in the Cartesian system, and the fixed direction emanates from the pole, which is akin to the positive x-axis.
In polar coordinates, a point is represented as (r, θ), where r is the radius - the distance from the pole to the point - and θ is the angle - measured in radians from the fixed direction to the line connecting the pole to the point. This system is ideal for dealing with problems that have a circular symmetry, such as the one in the exercise, where R is given by the circular region \(x^2 + y^2 \leq 25, x \geq 0\).

Converting to Polar Coordinates

To make solving easier for regions like a semicircle, we convert the Cartesian coordinates (x, y) to polar coordinates (r, θ). Our exercise showcases this by using the relationships \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\) to rewrite the given function, making the integration process more manageable.

Change of Variables

A change of variables in integration is akin to switching languages to make communication easier. When the original variables make a problem seem insurmountable, we switch to a new set of variables that simplifies the integral. This usually involves substituting a set of variables with another set that turns the region of integration or the integrand into a more workable form.
In our problem, the switch to polar coordinates turns the complex function \(f(x, y)=e^{-\left(x^{2}+y^{2}\right) / 2}\) into a simple radial function \(f(r, \theta) = e^{-r^2/2}\) and the region of integration into a straightforward interval for r and θ. Not to be overlooked is the Jacobian determinant, which arises when changing variables. In polar coordinates, this is represented by r, accounting for the area expansion or contraction that happens when moving between coordinate systems. That's why our differential area element \(dA\) becomes \(r \, dr \, d\theta\) in polar coordinates.

Definite Integral Evaluation

Evaluating definite integrals is central to calculating areas, volumes, and other quantities represented by the integral. In the context of our exercise, the definite integral evaluation involves solving the double integral over a specified region. This includes tackling both the inner and outer integrals separately, systematically working from the innermost integral to the outermost.

Inner and Outer Integrals

The inner integral pertains to integrating with respect to r from 0 to 5 while holding θ constant, followed by integrating the result of the inner integral with respect to θ from 0 to π.

• Inner Integral: Here, we're focused on the r dependence, and by using a u-substitution, we simplify the r integral.
• Outer Integral: Post the inner integration, we move onto θ, considering the results of the inner integral while treating θ as the variable of integration.

Executing these steps results in the area under the curve over the given region, which, once the limits are applied, gives us the final answer of the problem.