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. $$ \begin{aligned} &f(x, y)=4 x+4 y ; R \text { is the region enclosed by the circle }\\\ &x^{2}+y^{2}=4 \end{aligned} $$

Short answer

The integral evaluates to zero.

Step-by-step solution

Understanding the Region and Setting Up the Integral

The region \(R\) is a circle centered at the origin with a radius of 2, described by \(x^2 + y^2 = 4\). To change to polar coordinates, recall that \(x = r \cos \theta\) and \(y = r \sin \theta\), with \(r^2 = x^2 + y^2\). Thus, the region can be described in polar coordinates as \(0 \leq r \leq 2\) and \(0 \leq \theta < 2\pi\).

Applying Polar Coordinate Transformation

In polar coordinates, the function becomes \(f(x, y) = 4x + 4y = 4r(\cos \theta + \sin \theta)\). The differential area element \(dA\) becomes \(r \, dr \, d\theta\) in polar coordinates. Therefore, the integral becomes \(\iint_{R} (4r(\cos\theta + \sin\theta)) \cdot r \, dr \, d\theta\).

Setting Up the Double Integral in Polar Coordinates

The integral now becomes \(\int_{0}^{2\pi} \int_{0}^{2} 4r^2(\cos\theta + \sin\theta) \, dr \, d\theta\).

Evaluating the Inner Integral with Respect to \(r\)

First, compute the integral \(\int_{0}^{2} 4r^2 \, dr\), which evaluates to \(\left[\frac{4}{3}r^3\right]_{0}^{2} = \frac{32}{3}\). This integral will be multiplied by \(\cos\theta + \sin\theta\) when we evaluate the outer integral.

Evaluating the Outer Integral with Respect to \(\theta\)

Next, compute the integral \(\frac{32}{3} \int_{0}^{2\pi} (\cos\theta + \sin\theta) \, d\theta\). Notice that the integrals of \(\cos\theta\) and \(\sin\theta\) over \(0\) to \(2\pi\) both equal zero. Thus, the entire integral evaluates to zero.

Key concepts

Polar Coordinates

Polar coordinates offer a different way to describe points in the plane compared to the usual Cartesian coordinates. Instead of using

• two Cartesian coordinates \(x\) and \(y\),
• polar coordinates use a radius \(r\) and an angle \(\theta\).

This system is particularly useful when dealing with problems involving circles or circular symmetry. With polar coordinates, any point in the plane can be described in terms of how far it is from the origin (the radius) and the angle it makes with the positive x-axis.
The transformation from Cartesian to polar is straightforward:

• For the coordinate \(x\): \(x = r \cos \theta\)
• For the coordinate \(y\): \(y = r \sin \theta\)
• The differential area element \(dA\) in polar coordinates is given by \(r \, dr \, d\theta\)

These equations are essential for setting up integrals in polar coordinates.

Circle Equation

The equation of a circle \(x^2 + y^2 = r^2\) defines all points in the plane that are at a constant distance \(r\) from the origin. In the context of double integration in polar coordinates, this equation becomes even simpler.
In polar coordinates, a circle centered at the origin with radius 2 is simply described by \(r = 2\). There are no other variables in the equation, which reduces it from \(x^2 + y^2 = 4 \) to considering just \(r\) in the given limits: \(0 \leq r \leq 2\).
This description makes integration over circular regions much more straightforward by focusing only on the radial distance \(r\) and angle \(\theta\) ranges without involving the square terms that are necessary in Cartesian coordinates.

Inner and Outer Integrals

When performing double integration, it is crucial to understand the roles of inner and outer integrals. In the context of polar coordinates,

• the inner integral, usually with respect to \(r\), determines how the function behaves as the radius varies,
• while the outer integral, often with respect to \(\theta\), encompasses the behavior as the angle changes.

Consider our example involving the integration \(\int_{0}^{2\pi} \, \int_{0}^{2} 4r^2(\cos\theta + \sin\theta) \, dr \, d\theta\). Here, the inner integral \(\int_{0}^{2} 4r^2 \, dr\) focuses on accumulating the values along the radius of the circle, while the outer integral accounts for how these values vary as the circle sweeps through angles from \(0\) to \(2\pi\).
Each step in evaluating these integrals provides insight into the geometry and symmetry of the region being integrated.

Region Description in Polar Coordinates

Describing a region in polar coordinates can simplify complex shapes, especially circles in the plane. Here's how it's generally done:

• The radius \(r\) dictates the distance of every point in the region from the origin,
• while the angle \(\theta\) sets the direction that \(r\) points to.

For this exercise, we examine a region defined as a circle with a radius of 2. In polar terms, the region is fully captured by \(0 \leq r \leq 2\) and \(0 \leq \theta < 2\pi\).
Describing regions in this way accommodates complex boundaries with simple numerical limits, making integration more tractable. The whole region is envisioned as a sweeping radius covering all angles from the starting point back to the starting point, a full circle around the origin.