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Complex Areas

Complex regions

Some regions in the plane are more complicated and cannot be evaluated with a single integral. This happens when the area element is not bounded by the same curves throughout the region. For instance, consider the region bounded by a parabola and two lines:

In this case, the only way to find the area of the region is to divide it into regions which can be integrated separately:

Horizontal strips

Other regions are difficult to integrate using vertical strips as the area element, but work well with horizontal strips as the area element. For example, consider the following region bounded on the left by and on the right by :

In this case, the area of a horizontal strip is a function of , namely , where is the curve on the right and is the curve on the left.

Example

Find the area between the curves

Expressing these curves as functions of , we find

Graphing these curves, one finds the bounded region:

To find the intersections, set the curves equal to one another. This gives

A rearranging and factoring gives

and so we find that the intersection points are at and (the x-coordinates are the same, since they are on the line ). Note that using a vertical rectangle as the area element here would not be so easy, because the area element depends on the value of . Sometimes the strip goes from the parabola below to the line above, as shown in blue, and sometimes the strip goes from parabola to parabola, shown in red:

In particular, the area element for a vertical strip is

But using a horizontal strip as the area element works much better because throughout the region the strip is always going from the line on the left to the parabola on the right. So using a horizontal strip gives the area element

Integrating this over the range of gives the area:

Example

Find the area of the region bounded by and .

The region looks roughly as in the following:

By setting , collecting like terms, and factoring, one finds the intersection points at and , as indicated in the figure. The area element is a horizontal rectangle, which has area .

Thus, the area between the curves is

Example

Find the area of the region bounded by and the lines , , and .

The region looks roughly like the following:

Note that using vertical rectangles would not be ideal because this would require two integrals (for from 0 to 1 and from 1 to ). Instead, one can express the curve as . Now, using horizontal rectangles gives an area element of . Thus

Polar shapes

A polar shape is the graph of a polar function . Here, the input to the function is , which is the angle formed with the positive -axis (known as the pole). The output is the distance from the origin (or radial distance). For example, the following shows the graph of the polar function , which is known as a cardioid:

The area of such a region is not usually easy to compute by integrating with respect to or (for one thing, the polar equation would need to be expressed in terms of and first!). Instead, the way to integrate over such regions is to use a polar area element, which is a wedge shaped region. Here are several examples of the polar area element for various values of :

To compute what the polar area element is in terms of , , and , note that the region is roughly triangular (the curved portion at the base of the triangle can be ignored since it is a higher order term). The angle at the tip of the triangle is , the height of the triangle is , and the base of the triangle is :

Thus, the polar area element is

since . Thus, the area of a polar region defined by , where , is

Example

Compute the area of the cardioid .

In this case, , and so the area element is

Because ranges from 0 to to trace out the entire cardioid, it follows that the area is

Example

Find the area of a single petal of the polar curve :

Hint: To find the bounds on , compute when .

The area element is . To find the bounds on , set , which gives . The smallest values of for which this occurs is and :

Thus, the area of a single petal is

Example

Find the area inside the circle and outside the circle :

First, we find the intersections by setting the curves equal, which gives

and so the intersections are at and . The area element of the region is the polar area element of the circle minus the polar area element of the circle :

So we have that

Thus, the area is

From the second to the third line above, we used the power reduction formula for sine:


EXERCISES

  • Find the area enclosed by the curves , , and .
  • Find the area of the bounded region enclosed by the -axis, the lines and and the hyperbola .
  • Compute the area in the bounded (i.e., finite) regions between and the -axis.
  • Find the area of the sector of a circular disc of radius (centered at the origin) given by (as usual, is in radians).
  • Use polar coordinates to find an area within and outside .
  • Find the area of the overlap between two circles of radius 2 that pass through each others' centers. You can do so with either cartesian or polar coordinates (though one might be easier than the other!).
  • Find the area bound by the curves and .
  • Kepler's First Law states that the orbit of every planet is an ellipse with the Sun at one of its two foci. If we think of the Sun as being situated at the origin, we can describe the orbit with the equation:
The point at which the planet is closest to the Sun (the perihelion) corresponds to , while the planet is furthest away from the Sun at (the aphelion). Knowing the distance between the Sun and the planet at these two points would allow you to fix the values of the constants and . Notice that describes a perfect circle, so that the "eccentricity" measures how far the orbit is from being a circle.
Kepler's Second Law states that the line joining a planet and the Sun sweeps out equal areas during equal intervals of time. Another way of expressing this fact is by saying that the "areal velocity"
of that line is constant in time.
Express the area element in terms of the angle element and use Kepler's Second Law to deduce the differential equation governing the time evolution of .
  • Let be the circle given by . Let be the circle given by . Find the area of region in that is not in .