Question

Sketch each region and use a double integral to find its area. The region bounded by the spiral \(r=2 \theta,\) for \(0 \leq \theta \leq \pi,\) and the \(x\) -axis

Short answer

Answer: The area of the region is \(\frac{1}{12}\pi^4\).

Step-by-step solution

Sketching the region

To start, let's recall the conversion from polar coordinates \((r, \theta)\) to Cartesian coordinates \((x, y)\): \(x = r\cos(\theta)\) \(y = r\sin(\theta)\) The given curve is \(r=2\theta\), and we are given the interval for \(\theta\), which is \(0\leq \theta\leq\pi\). We can plug in different values for \(\theta\) in this range to sketch the curve in polar coordinates. Note that as \(\theta\) increases, \(r\) also increases, and the curve spirals outward.

Setting up the double integral

With the region bounded by the spiral \(r=2\theta\) and x-axis, we can set up a double integral to find the area. To find the area of the region in polar coordinates, we can use the general formula: \(A = \frac{1}{2}\int_{\alpha}^{\beta}\int_{r_1(\theta)}^{r_2(\theta)}r^2 dr d\theta\) In our case, \(\alpha=0\), \(\beta=\pi\), \(r_1(\theta)=0\), and \(r_2(\theta)=2\theta\). So the area of the region can be expressed as a double integral: \(A = \frac{1}{2}\int_{0}^{\pi}\int_{0}^{2\theta}r^2 dr d\theta\)

Evaluating the double integral

Now, we can evaluate the double integral that we have set up. Start by integrating with respect to \(r\): \(A=\frac{1}{2}\int_{0}^{\pi}\left[\frac{1}{3}r^3\right]_{0}^{2\theta} d\theta\) \(A=\frac{1}{2}\int_{0}^{\pi}\frac{8}{3}\theta^3 d\theta\) Next, integrate with respect to \(\theta\): \(A=\frac{1}{2}\left[\frac{2}{3}\cdot\frac{1}{4}\theta^4\right]_{0}^{\pi}\) \(A=\frac{1}{12}\left[\pi^4-0^4\right]\)

Final Answer

After evaluating the double integral, we find that the area of the region bounded by the spiral \(r=2\theta\), the interval \(0\leq\theta\leq\pi\), and the \(x\)-axis is: \(A=\frac{1}{12}\pi^4\)

Key concepts

Polar Coordinates

Polar coordinates are a way of representing points in a plane using a radius and an angle. Instead of describing a point using x and y coordinates, as with Cartesian coordinates, a point in polar coordinates is described using a distance from the origin, called the radius (r), and an angle (\( \theta \)) from the positive x-axis.
This system is particularly useful for problems with circular symmetry, such as spirals or circles, where traditional Cartesian coordinates may involve more complex expressions.

• The radius, \( r \), tells us how far the point is from the origin.
• The angle, \( \theta \), indicates the direction of the point from the positive x-axis.

In the given problem, the spiral \( r=2\theta \) means that as \( \theta \) increases, the distance from the origin increases proportionally, forming a spiral shape in the plane.
This description gives us an outline of the region to analyze during integration.

Area Calculation

The area calculation of a region described in polar coordinates can be efficiently performed using a double integral. The specific setup for calculating area in polar coordinates differs slightly from Cartesian coordinates.
Rather than integrating over x and y, we integrate over \( r \) and \( \theta \). This is due to the nature of polar coordinates, where the element of area also requires considering the radius.

• In polar coordinates, the differential area element is \( r \, dr \, d\theta \).
• We must include `r` in our integral as the radius varies over the region.
• The range for \( \theta \) and \( r \) often comes directly from the problem statement or the bounds of the curve.

In our exercise, the spiral bounds the region from \( \theta = 0 \) to \( \theta = \pi \) and from \( r = 0 \) to \( r = 2\theta \), setting up our integral for the area as \( \frac{1}{2}\int_{0}^{\pi}\int_{0}^{2\theta}r^2 \, dr \, d\theta \).
This formula is derived from the geometry of the sector in polar coordinates, simplifying the processing of our area integration.

Integration

Integration in polar coordinates involves setting up a double integral over the ranges of \( r \) and \( \theta \). The unique aspect of this setup is ensuring that each variable is properly integrated within its respective bounds.
Evaluating the integral involves two steps:
1. **Integrate with respect to \( r \):** Start by evaluating the inner integral, managing the \( r^2 \) component and applying the bounds from the curve.2. **Integrate with respect to \( \theta \):** Once the inner integration is complete, tackle the outer integral with respect to \( \theta \).In our case, the step of evaluating the double integral begins with:

• First, calculate the integral \( \int_{0}^{2\theta} r^2 \, dr = \frac{1}{3} r^3 \) from 0 to \( 2\theta \).
• Then substitute the bounds to find \( \frac{8}{3} \theta^3 \).
• Finally, integrate \( \frac{8}{3} \theta^3 \) over \( \theta \) from 0 to \( \pi \).

Completing these steps, we find the area, resulting in \( A = \frac{1}{12} \pi^4 \), which represents the area of the region under consideration bounded by the spiral and the x-axis.