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Draw the polar region \(\\{(r, \theta): 1 \leq r \leq 2,0 \leq \theta \leq \pi / 2\\} .\) Why is it called a polar rectangle?

Short Answer

Expert verified
Short Answer: The polar region defined by the inequalities \(1 \leq r \leq 2, \, 0 \leq \theta \leq \pi / 2\) is called a polar rectangle because it can be parameterized as a Cartesian product of two intervals: \(R = [1, 2] \times [0, \pi / 2]\). This parallelism between rectangles in Cartesian coordinates and such regions in polar coordinates emphasizes their similar structure. In the polar coordinate system, the region appears as a sector within the first quadrant and between two circles of radius 1 and 2.

Step by step solution

01

Understand the inequalities

The given inequalities are: $$1 \leq r \leq 2, \, 0 \leq \theta \leq \pi / 2.$$ These inequalities tell us the bounds of the polar region in terms of the radial coordinate \(r\) and the angular coordinate \(\theta\). Specifically, the radial coordinate is bounded between 1 and 2, and the angular coordinate is bounded between 0 and \(\pi / 2\).
02

Draw the radial bounds

To draw the radial bounds, we will use two circles of radius 1 and 2, both centered at the origin. The region described by \(1 \leq r \leq 2\) will be the area between these two circles.
03

Draw the angular bounds

To draw the angular bounds, we will use the angles defined by the inequality in \(\theta\). The bounds \(0 \leq \theta \leq \pi / 2\) tell us that the region should be in the first quadrant of the coordinate system, since the angles are between 0 and \(\pi / 2\).
04

Combine the radial and angular bounds

Now that we have drawn both the radial and angular bounds, we will find the intersection of the two regions to determine the polar region defined by the given inequalities. The polar region will be the area within the first quadrant and between the two circles of radius 1 and 2, which forms a sector.
05

Explain why it is called a polar rectangle

The region is called a polar rectangle because it can be parameterized in the following way: $$R = [1, 2] \times [0, \pi / 2].$$ This means that the region is the Cartesian product of two intervals, which is similar to how we describe a rectangle in the Cartesian coordinate system. The polar region forms a sector in the \((r, \theta)\) coordinate system, but its rectangular nature comes from the structure of the inequalities that define it. The term "polar rectangle" highlights the parallelism between rectangles in Cartesian coordinates and such regions in polar coordinates.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polar Region
When we approach geometry in the polar coordinate system, the term polar region refers to a specific area on a plane defined by both radial and angular constraints. A polar region can assume various shapes depending on these constraints, ranging from sectors, to the more complex forms shaped by curves like spirals.

Consider the example \(\{(r, \theta): 1 \leq r \leq 2,0 \leq \theta \leq \pi / 2\} \). Here, the region described is bound between two circles (radial bounds) and within a slice of a pie (angular bounds). It's a sector, or a 'slice' of the area between the circles from the origin to \(\pi / 2\), resembling a piece of pie positioned in the first quadrant of the polar plane. This visualization is crucial for students to understand how radial and angular bounds shape a region.
Inequalities in Polar Coordinates
In polar coordinates, inequalities serve to define regions just like they do in Cartesian coordinates. However, instead of defining boundaries in terms of x and y, polar inequalities use the radius \(r\) and the angle \(\theta\) to set the limits.

For instance, the inequality \(1 \leq r \leq 2\) specifies that the region in question includes all points whose distance from the origin is at least 1 but no more than 2. Similarly, \(0 \leq \theta \leq \pi / 2\) defines that the region lies within the first quadrant, as the angle \(\theta\) is restricted to angles from the positive x-axis to \(90^{\circ}\) or \(\pi / 2\) radians. These inequalities are essential for plotting the region accurately and understanding the scope of the area they encompass.
Polar Rectangle
The polar rectangle might sound like a contradiction at first because we usually think of rectangles as strictly Cartesian shapes. However, in polar terms, when a region has simple, straight-line radial and angular bounds, it is analogous to a rectangle because each point within the region can be described by a range of radial and angular coordinates—much like the x,y-coordinates describe a rectangle in Cartesian terms.

The example given previously, \(\{(r, \theta): 1 \leq r \leq 2,0 \leq \theta \leq \pi / 2\} \) forms a polar rectangle because it corresponds to the Cartesian product of two intervals, creating a set of points that form a sector. This sector has ‘straight sides’ when viewed in terms of \(r\) and \(\theta\), making it the polar equivalent of a rectangle.
Radial and Angular Bounds
Clearly delineating a region requires setting radial and angular bounds. The radial bound is the set of all points that lie a certain distance from the origin. In our exercise, this distance is between 1 and 2 units inclusive. The shape traced by these bounds is two concentric circles centered at the origin.

On the other hand, the angular bound is determined by the spread of angles included in the region. In our example, we consider angles starting from 0 (the positive x-axis) to \(\pi / 2\) radians. Hence, only the first quadrant is included in this angular range. By combining these radial and angular constraints, we define a precise region in the polar coordinate system that adheres to these limits.

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Most popular questions from this chapter

Find the volume of the solid bounded by the surface \(z=f(x, y)\) and the \(x y\)-plane. (Check your book to see figure) $$f(x, y)=e^{-\left(x^{2}+y^{2}\right) / 8}-e^{-2}$$

A cylindrical soda can has a radius of \(4 \mathrm{cm}\) and a height of \(12 \mathrm{cm} .\) When the can is full of soda, the center of mass of the contents of the can is \(6 \mathrm{cm}\) above the base on the axis of the can (halfway along the axis of the can). As the can is drained, the center of mass descends for a while. However, when the can is empty (filled only with air), the center of mass is once again \(6 \mathrm{cm}\) above the base on the axis of the can. Find the depth of soda in the can for which the center of mass is at its lowest point. Neglect the mass of the can, and assume the density of the soda is \(1 \mathrm{g} / \mathrm{cm}^{3}\) and the density of air is \(0.001 \mathrm{g} / \mathrm{cm}^{3} .\)

Sketch the following regions \(R\). Then express \(\iint_{R} g(r, \theta) d A\) as an iterated integral over \(R\) in polar coordinates. The region inside both the cardioid \(r=1+\sin \theta\) and the cardioid \(r=1+\cos \theta\)

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Suppose the density of a thin plate represented by the polar region \(R\) is \(\rho(r, \theta)\) (in units of mass per area). The mass of the plate is \(\iint_{R} \rho(r, \theta) d A .\) Find the mass of the thin half annulus \(R=\\{(r, \theta): 1 \leq r \leq 4,0 \leq \theta \leq \pi\\}\) with a density \(\rho(r, \theta)=4+r \sin \theta\).

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