91Ó°ÊÓ

Iterated integrals are a powerful tool for calculating areas and volumes in polar coordinates. An iterated integral is essentially an integral within another integral. It allows us to determine the area of a region by breaking the calculation into simpler parts. In polar coordinates, these integrals often take the form \( \text{Area} = \int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} r \, dr \, d\theta \). This setup indicates two integrations: \( r \) is integrated first, representing radial distances, followed by \( \theta \), representing angles around the origin.

For example, when finding the area right of \( r = \frac{1}{2 \cos \theta} \) and inside \( r = 1 \), we use specific \( r \) and \( \theta \) limits. The limits \( r_1 = \frac{1}{2} \) to \( r_2 = 1 \) reflect the radial distance from the line to the circle. The angle limits, derived from the line's orientation, are \( \theta_1 = \arccos\left(\frac{1}{2}\right) \) to \( \theta_2 = \frac{\pi}{2} \). This iterated integral approach simplifies the process of area calculation in complex polar regions.

Calculating the area of a section in polar coordinates involves both visualizing the region and setting up the correct limits for integration. To determine the area inside a circle and to the right of a line, we first translate these conditions into polar coordinates. The problem specifies the circle \( r = 1 \) and the line \( x = \frac{1}{2} \).

Firstly, note that the circle is centered at the origin with a constant radius. The line given in Cartesian coordinates transforms into \( r = \frac{1}{2\cos\theta} \) in polar terms. This line is vertical at \( x = \frac{1}{2} \). The area of interest lies where \( \frac{1}{2} \leq r \leq 1 \) and between the angles \( \theta = \arccos\left(\frac{1}{2}\right) \) to \( \theta = \frac{\pi}{2} \).

By setting up the integral with these limits, we are effectively measuring the pie-slice segment of the circle that sits between the circle and the line. By evaluating the inner integral over \( r \) and then the outer integral over \( \theta \), we calculate the desired area.

Graphing polar equations can initially seem complex but is a skillful method for understanding where different curves lie in relation to one another. For the provided exercise, the graphical focus is on a line and a circle.

The line described by \( r = \frac{1}{2 \cos \theta} \) is a line parallel to the y-axis, crossing the x-axis at \( x = \frac{1}{2} \). This occurs because, when transforming this polar equation into Cartesian coordinates, it aligns with a vertical line.

Meanwhile, the graph of \( r = 1 \) is a perfect circle centered at the origin. The graph shows all points that are exactly one unit away from the origin, forming a circle.

To graph these:

• Plot the circle by marking all points one unit from the origin.
• Draw the vertical line at \( x = \frac{1}{2} \), noting that this is applicable only within \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \).

Seeing these graphs superimposed reveals which regions you need for calculating integrals, such as the segment that lies inside the circle and to the right of the line.