Question

The population density of a city is approximated by the model \(f(x, y)=4000 e^{-0.01\left(x^{2}+y^{2}\right)}, x^{2}+y^{2} \leq 49,\) where \(x\) and \(y\) are measured in miles. Integrate the density function over the indicated circular region to approximate the population of the city.

Short answer

The population of the city is approximately \(2 \pi * (2000 - 2000e^{-0.49})\) people.

Step-by-step solution

Change to Polar Coordinates

We must first transform the function to polar coordinates as it simplifies the integration process. In polar coordinates, \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\), and \(x^{2} + y^{2} = r^{2}\). So, the function \(f(x, y)\) becomes \(f(r, \theta) = 4000e^{-0.01r^{2}}\).

Setting Bounds for Integration

Next, we need to specify the bounds for \(r\) and \(\theta\). As we're dealing with a full circle with radius 7, the bounds for \(r\) will be from 0 to 7 and for \(\theta\) it will be from 0 to \(2 \pi\).

Double Integral Calculation

Now, set up and calculate the double integral. We can split the double integral into the product of two simpler integrals. First, we integrate over \(\theta\) from 0 to \(2 \pi\), which yields \(2\pi\). Then, we integrate over \(r\) from 0 to 7. The integral is \(\int_{0}^{7} 4000r e^{-0.01 r^{2}} dr\), where the extra \(r\) is due to the change of variables to polar coordinates. Use integration by substitution to solve this integral, let \(u = r^2\), then \(du = 2r dr\), \(dr = du /(2r)\). The integral becomes \(\int_{0}^{49} 2000 e^{-0.01 u} du\). The antiderivative of \(e^{-0.01 u}\) is \(-100 e^{-0.01 u}\), so the result of this integral is \([-100*2000 [e^{-0.01(49)} - e^{0}]\) = 2000 - 2000e^{-0.49}.

Result Calculation

Finally, multiply the results of the integrals over \(r\) and \(\theta\). Therefore, the total population is \(2 \pi * (2000 - 2000e^{-0.49})\).

Key concepts

Polar Coordinates

Understanding polar coordinates is essential when dealing with problems that have a natural circular or spherical symmetry, such as integrating functions over a circular region. Instead of using the traditional Cartesian coordinates (x, y), which are a distance measured along two perpendicular axes, polar coordinates represent points on a plane through a distance from a reference point (the origin) and an angle from a reference direction.

In polar coordinates, the position of a point P is given by a pair \( (r, \theta) \), where \( r \) is the radial distance from the origin to the point, and \( \theta \) is the angular coordinate, i.e., the angle measured from the positive x-axis to the line segment connecting the origin to P. One of the main benefits of using polar coordinates, especially in the context of integration, is that it simplifies calculations involving circular or spherical regions, as seen in our population model example.

Double Integral

The concept of a double integral extends the notion of an integral to functions of two variables, allowing the computation of volumes under surfaces or, as in our case, total quantities across areas. In Cartesian coordinates, the double integral of a function f(x, y) over a region R is denoted by \( \iint_R f(x, y) \,dx\,dy \). However, when the region R is circular or annular, using polar coordinates significantly streamlines the process.

For polar coordinates, the double integral includes an additional factor of \( r \) in the integrand, which accounts for the 'expansion' of space as one moves further from the origin. Thus, the double integral becomes \( \iint_R f(r, \theta) r \,dr\,d\theta \). It's also important to properly define the limits of integration, which in polar coordinates are radii (from the center to the boundary of the region) and angles (typically 0 to \( 2\pi \) for a full circle).

Integration by Substitution

When facing a challenging integral, integration by substitution, often referred to as the reverse chain rule, is a powerful tool for simplifying the integrand. By treating one part of the integrand as a new variable \( u \)—where \( u \) is a function of \( r \) in the context of polar coordinates—we can often make complex expressions more tractable.

In our example of population density integration, the substitution \( u = r^2 \) simplifies the exponent in the integrand, transforming a difficult integral into one that is much easier to solve. Remember that when substituting, it's crucial to convert all parts of the integral, including the differentials. Thus, \( dr \) becomes \( \frac{du}{2r} \), reflecting that \( du \) is derivative of \( u \) with respect to \( r \). This substitution helps streamline computations and arrives at a solvable integral that yields the desired population figure when evaluated.