Question

In Exercises \(3-10\), a function \(f(x, y)\) is given and a region \(R\) of the \(x-y\) plane is described. Set up and evaluate \(\iint_{R} f(x, y) d A\) using polar coordinates. $$ \begin{aligned} &f(x, y)=1-x^{2}-y^{2} ; R \text { is the region enclosed by the circle }\\\ &x^{2}+y^{2}=1 \end{aligned} $$

Short answer

The integral is \( \frac{\pi}{2} \).

Step-by-step solution

Identify the Given Information

The function provided is \( f(x, y) = 1 - x^2 - y^2 \) and the region \( R \) is enclosed by the circle \( x^2 + y^2 = 1 \). In polar coordinates, this circle can be described using \( r \) and \( \theta \).

Convert to Polar Coordinates

To use polar coordinates, we express \( x \) and \( y \) in terms of \( r \) and \( \theta \). Let \( x = r \cos \theta \) and \( y = r \sin \theta \). The Jacobian of this transformation is \( r \), meaning \( dA = r \, dr \, d\theta \).

Transform the Function

Substitute \( x = r \cos \theta \) and \( y = r \sin \theta \) into \( f(x, y) \). Thus, we have:\[ f(x, y) = 1 - (r \cos \theta)^2 - (r \sin \theta)^2 = 1 - r^2 \]

Define Region Limits in Polar Coordinates

The region \( R \) described by the equation \( x^2 + y^2 = 1 \) in polar coordinates is \( 0 \leq r \leq 1 \) and \( 0 \leq \theta \leq 2\pi \).

Set Up the Integral in Polar Coordinates

The double integral \( \iint_{R} f(x, y) \, dA \) becomes:\[\int_{0}^{2\pi} \int_{0}^{1} (1 - r^2) \, r \, dr \, d\theta\]

Evaluate the Integral over \( r \)

Integrate with respect to \( r \):\[\int_{0}^{1} (r - r^3) \, dr = \left[ \frac{r^2}{2} - \frac{r^4}{4} \right]_{0}^{1} = \left( \frac{1}{2} - \frac{1}{4} \right) = \frac{1}{4}\]

Evaluate the Integral over \( \theta \)

Integrate the result from the previous step with respect to \( \theta \):\[\int_{0}^{2\pi} \frac{1}{4} \, d\theta = \frac{1}{4} \left[ \theta \right]_{0}^{2\pi} = \frac{1}{4} (2\pi - 0) = \frac{\pi}{2}\]

Final Result

The value of the integral is \( \frac{\pi}{2} \).

Key concepts

Polar Coordinates

Polar coordinates provide a powerful way of working with certain types of problems, especially those involving circular or rotational symmetry. Unlike Cartesian coordinates, which specify a point by its horizontal and vertical distances from the origin (denoted as \(x\) and \(y\)), polar coordinates use a distance from the origin \(r\) and an angle \(\theta\) from the positive x-axis.
For example, a point in polar coordinates is written as \((r, \theta)\). Here, \(r\) is the radius or radial distance, and \(\theta\) is the angular coordinate. This makes polar coordinates particularly useful for expressing circular regions or objects that are naturally symmetric around a point.
By transforming a function and its region of integration into polar coordinates, integrals can become simpler to solve, especially when dealing with circles or arcs, like in our original exercise.

Region of Integration

When solving double integrals, specifying the region of integration is critical for setting up the integral accurately. In the context of our exercise, the region of integration \(R\) is enclosed by the circle with the equation \(x^2 + y^2 = 1\).
This means that all points \((x, y)\) within this circle satisfy the condition, forming a disk centered at the origin with a radius of 1.
Using polar coordinates simplifies defining this region: the radius \(r\) varies from 0 to 1, and the angle \(\theta\) sweeps from 0 to \(2\pi\).
This clear and bounded area allows us to set the limits of the double integral neatly without needing to transform complex boundaries or corners that might be present in a Cartesian system.

Jacobian Transformation

In converting from one coordinate system to another, like Cartesian to polar, it's essential to account for changes in area element. This is where the Jacobian transformation comes into play.
The Jacobian is essentially a scaling factor that adjusts the differential area element \(dA\) for the new coordinate system.
For polar coordinates, this transformation yields a Jacobian of \(r\). Thus, the differential area element in polar coordinates becomes \(dA = r \, dr \, d\theta\).
This factor \(r\), which appears due to the shape properties of circles, accounts for changes in the area slice sizes as you move radially outward. By incorporating this factor, the integral correctly computes areas and functions over the specified region.

Conversion to Polar Coordinates

Converting to polar coordinates is straightforward once you understand the basics. This involves expressing \(x\) and \(y\) in terms of \(r\) and \(\theta\). The standard transformations are:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)

These equations convey how each point \((x, y)\) on a plane corresponds to a point \((r, \theta)\) in polar coordinates.
Through substitution, the function to be integrated and region limits shift accordingly, allowing us to set up a solvable polar integral.
In our particular exercise, expressing the function \(f(x, y) = 1 - x^2 - y^2\) converts smoothly to \(f(r, \theta) = 1 - r^2\), reflecting how each point on the disk can be represented in terms of \(r\) and \(\theta\), domesticating the challenge of the original integral.