Problem 15 - Thomas Calculus

Question

Find the areas of the regions Inside the circle \(r=6\) above the line \(r=3 \csc \theta\)

Step-by-Step Solution

Answer

The area is \( 12\pi - 6\sqrt{3} \).

Step 1: Understand the graphical representation

To solve this problem, visualize the circle and the line. We have the circle represented by the polar equation \( r = 6 \), which is a circle of radius 6 centered at the origin. The line is represented by \( r = 3 \csc \theta \), or \( r \sin \theta = 3 \), which is a horizontal line at \( y = 3 \) in Cartesian coordinates.

Step 2: Convert polar equation to Cartesian (for line)

The polar equation \( r = 3 \csc \theta \) translates to \( r \sin \theta = 3 \) in Cartesian terms. So, \( y = 3 \), which is a straight horizontal line cutting through the circle.

Step 3: Set limits of integration

Determine the points of intersection of these two curves to set the bounds for integration. At \( y = 3 \) (\( r \sin \theta = 3 \)), we need to find the corresponding \( \theta \) values where the line intersects the circle. \( y = 3 \) intersects the circle where \( x^2 + 3^2 = 6^2 \), solving gives \( x = \pm \sqrt{27} = \pm 3\sqrt{3} \). In terms of \( \theta \), this means the \( \theta \) range is from \( \theta_1 = \arcsin{\left(\frac{1}{2}\right)} \) to \( \theta_2 = \pi - \arcsin{\left(\frac{1}{2}\right)} \). So, \( \theta_1 = \frac{\pi}{6}\) and \(\theta_2 = \frac{5\pi}{6}\).

Step 4: Set up the integral

The area of interest is the region where the circle is above the line. This requires calculating the integral of the radial function \( (circle - line) = (6) - (3 \csc \theta) \) with respect to \( \theta \) from \( \theta_1 \) to \( \theta_2 \). The integral forms to: \[ A = \int_{\theta_1}^{\theta_2} \frac{1}{2} \left[ (6)^2 - (3 \csc \theta)^2 \right] \, d\theta \]

Step 5: Evaluate the integral

Evaluate the integral based on the given limits:\[ A = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \frac{1}{2} \left[ 36 - \frac{9}{\sin^2\theta} \right] \, d\theta \]This simplifies to:\[ A = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \frac{1}{2} \left(36 - 9\csc^2\theta \right) \, d\theta \]Now, compute each:- \( \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} 36 \, d\theta = 36\left[\theta\right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} = 36\left(\frac{5\pi}{6} - \frac{\pi}{6}\right) = 24\pi \).- \( \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} 9\csc^2\theta \, d\theta = 9 \left[ -\cot \theta \right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \).Compute these integrals to find the area.

Step 6: Combine results to find total area

Combine the results from the integral computations:- The contribution from \( 36 \) is \( 24\pi \).- The contribution from \( 9\csc^2\theta \) is \( 9\left( -\cot \left(\frac{5\pi}{6}\right) + \cot \left(\frac{\pi}{6}\right) \right) \).Solving:- \( -\cot \left( \frac{5\pi}{6} \right) = 1/\sqrt{3} \) - \( \cot \left( \frac{\pi}{6} \right) = \sqrt{3} \)Thus, the result from the \( \cot \) terms is \( 9 \left( \sqrt{3} - 1/\sqrt{3} \right) = 9 \cdot \frac{2\sqrt{3}}{3} = 6\sqrt{3} \).Finally, total area \( A = 12\pi - 6\sqrt{3} \).

Key Concepts

Area of Region

Understanding how to find the area of a region in polar coordinates involves a different approach compared to using Cartesian coordinates. In polar coordinates, the key is to determine the bounds for integration, represented in terms of the angle \( \theta \). We first visualize the region of interest.In this exercise, our task is to find the area of the region within the circle \( r=6 \) that lies above the line \( r=3 \csc \theta \). This requires identifying the points of intersection between these two curves. By converting the polar equation for the line to Cartesian form, we find it corresponds to a horizontal line at \( y=3 \). This intersection helps in setting up our limits of integration based on the angles formed. Finally, calculating the area in polar coordinates involves integrating with respect to \( \theta \) over the specified range, which gives us the exact area of the region where the circle is above the line.

Polar Integration

Polar integration is a method to calculate the area of curved regions defined in polar coordinates. The integral is set up by considering segments of the region that form a wedge-shaped area. The general formula for the area \( A \) in polar coordinates is:

• \[ A = \int_{\theta_1}^{\theta_2} \frac{1}{2} [f(\theta)]^2 \, d\theta \]

This formula directly relates to the area of a sector of a circle, where \( f(\theta) \) is the radial distance.In this problem, we are integrating the difference between the circle and the line to find the area where the circle lies above the line. Therefore, the integral becomes:

• \[ A = \int_{\theta_1}^{\theta_2} \frac{1}{2} [(6)^2 - (3 \csc \theta)^2] \, d\theta \]

Evaluating this integral involves dealing with constants and the trigonometric \( \csc \theta \) function, which affects the integral's complexity. However, the steps need careful trigonometric manipulation and integration techniques to find the final area.

Geometry of Curves

The geometry of curves in polar coordinates often provides an intuitive way to explore circular and spiral shaped regions. In this exercise, the curves are described by the equations \( r=6 \) (a perfect circle) and \( r=3 \csc \theta \) (a curve that transforms into a line in Cartesian coordinates). By understanding these forms, we can see how different curves interact over regions of the plane.To interpret the geometry, it's crucial to convert between polar and Cartesian forms to grasp how these shapes overlap. The circle centered at the origin with radius 6 is straightforward in polar form. The second equation, which translates to the horizontal line \( y=3 \), leads to intersecting points that help determine boundaries for integration.By analyzing the shapes and positions of these curves, we can understand their geometry, helping to better visualize the exact region in which the two curves define an area.