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CALCULUS 1 PROBLEMS & SOLUTIONS

7.2
Areas Of Plane Regions


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Review


1. Areas Between A Curve And The x-Axis

A plane is a 2-dimensional space. A plane region is, well, a region on a plane, as opposed to, for
example, a region in a 3-dimensional space. We'll find the area A of a plane region bounded by the
graph of a function f continuous on [a, b] where a < b, the x-axis, the vertical line x = a, and the
vertical line x = b. See Figs. 1 and 2.

Fig. 1

Fig. 2 · Total area: A = A1 + A2 + A3. ·

Fig. 3

Integrating Absolute Values

Since there's no special “ technique” to integrate an absolute value, we must split an integral into a sum
of integrals on sub-intervals where f(x) ³ 0, and thus | f(x)| = f(x), and sub-intervals where f(x) < 0,
and thus | f(x)| = – f(x), as shown above and illustrated in Fig. 3. To find those sub-intervals, find the
points where the graph of f intersects the x-axis, ie, solve the equation f(x) = 0 for x. Also, a graph of
f should be sketched. The sign of f(x) can then be examined in the graph.

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2. Areas Between Two Curves

Let's find the area A of the region between the curves which are the graphs of the functions f and g
and between the vertical lines x = a and x = b, where a < b and f and g are continuous on [a, b]. See
Figs. 4, 5, and 6.

Suppose f(x) ³ 0, g(x) ³ 0, and f(x) ³ g(x) for all x in [a, b]. Refer to Fig. 4. Clearly the area A is
equal to the area under f over [a, b] minus the area under g over [a, b]. So:

Fig. 4

Fig. 5

Fig. 6

To integrate the absolute value | f(x) – g(x)| we split the integral into a sum of integrals on
sub-intervals where f(x) – g(x) ³ 0 or f(x) ³ g(x), so that | f(x) – g(x)| = f(x) – g(x), and on
sub-intervals where f(x) – g(x) < 0 or f(x) < g(x), so that | f(x) – g(x)| = g(x) – f(x).

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3. Area Elements

Let's examine the functions f and g in Fig. 4, reproduced in Fig. 7. At any point x between a and b,
the product ( f(x) – g(x)) dx is the area of an infinitely thin vertical rectangle whose width is the
infinitesimal dx and height is f(x) – g(x). Let dA be that infinitesimal area, so that dA =
( f(x) – g(x)) dx. As we've demonstrated above, the total area between the graphs of f and g is A =

Fig. 7 · Area element is dA = ( f(x) – g(x)) dx. It's gray shaded. · Total area is integral of area elements.

Fig. 8 · Area element is gray shaded and is dA = | f(x) – g(x)| dx. · Total area is integral of area elements.

Fig. 9 · h = | f – g|. · Aj = area of gray shaded rectangle, and Aj = dA.

Area As Integral Of Area Element

Consequently, the area between two curves on [a, b] is the definite integral of the area element
between x = a and x = b. Remark that the height of the thin vertical rectangle over any sub-interval is
equal to the height y of the upper curve minus that of the lower curve. Hence, in abbreviation, the area
element over any sub-interval is:

dA = ( y_upper – y_lower) dx.

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4. Finding Area Between Two Curves

Let's use the notion of area element to find the area between two curves.

Example 1

Find the area of the plane region bounded above by the graph of y = 2x and below by that of y = x² .

Solution

Fig. 10 Region of interest is over [0, 2] and is shaded. Area element is gray shaded. Note: area element isn't exactly a “ rectangle ”, but it's easiest to draw and more suggestive of the fact that it's a part of total area. So we can call it “ strip” instead of “ rectangle”.

The graphs of y = 2x and y = x² are sketched in Fig. 10. We have: 2x = x² Û x² – 2x = 0 Û
x(x – 2) = 0 Û x = 0 or x = 2. So the graphs intersect at x = 0 and x = 2. The region of interest is
over [0, 2]. The area element is dA = (2x – x²) dx. Thus, the area of interest is:


EOS


Example 2

Find the area of the plane region bounded by y = 2x and y = x³ from the smallest of the x-coordinates
of their points of intersection to the largest.

Solution

Fig. 11 The region in question is shaded.

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5. Areas Of Unbounded Regions

Like improper integrals, the area of an unbounded region between two curves is defined to be the limit
of area of bounded region, and thus is expressed as an improper integral. An example is shown in Fig.
12. Let f and g be functions continuous on [a, ¥). Suppose lim_x®¥ f(x) = lim_x®¥ g(x) = L. The region
between the graphs of f and g and to the right of the vertical line x = a is unbounded. Observe that the
graph of g is above that of f on the entire interval [a, ¥). The area A of the unbounded region is:

Fig. 12 Area of unbounded region is limit of area of bounded region.

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6. Integration Along The y-Axis

To find the area A of the plane region lying between the curves x = f( y) and x = g( y), above the
horizontal line y = c, and below the horizontal line y = d, we can integrate along the y-axis over [c, d].
See Fig. 13. The area element dA is the area of a thin horizontal rectangle stretching from the left
curve to the right curve. The rectangle's length is | f( y) – g( y)| and its width is dy. So we have dA =
| f( y) – g( y)| dy. Thus, the area A is:



Integration along the y-axis is valid because it's simply the case of interchanging the roles of x and y
and integrating over an interval of the y-axis instead of the x-axis.

Fig. 13 Integrating along the y-axis.

As in the case of the integration along the x-axis, to integrate the absolute value we split the integral
into a sum of integrals on y-sub-intervals where f( y) ³ g( y), so that | f( y) – g( y)| = f( y) – g( y),
and on y-sub-intervals where f( y) < g( y), so that | f( y) – g( y)| = g( y) – f( y).

Remark that the length of the thin horizontal rectangle over any y-sub-interval is equal to the x-position
of the right curve minus that of the left curve. Hence, in abbreviation, the area element over any
y-sub-interval is:

dA = ( x_right – x_left) dy.


Example 3

Find the area of the region which lies to the right of the parabola x = y² – 12 and to the left of the line
y = x.

Solution

Fig. 14 Area of interest is shaded. Area element is area of thin horizontal gray shaded strip.

The graphs of x = y² – 12 and y = x are sketched in Fig. 14. We solve the equations x = y² – 12 and
x = y simultaneously to find the y-coordinates of the points of intersection: y² – 12 = y Û
y² – y – 12 = 0 Û ( y + 3)( y – 4) = 0 Û y = –3 or 4. Let's integrate along the y-axis from y = –3 to
y = 4. The area element is dA = ( y – ( y² – 12)) dy = (–y² + y + 12) dy. So the area A of interest is:


EOS


Remark 1

The area in the above example can also be found by integrating along the x-axis. Let's see what
happens if we attempt to do so:



If we evaluate these component integrals, we will get the same result as found in the example. Now,
what happens is that attempting to integrate along the x-axis leads to these more complicated
component integrals. For regions bounded by graphs of functions of y, ie, functions of the form x =
f( y), it's generally easier to integrate along the y-axis than along the x-axis.

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7. Finding The Area Of A Plane Region

In each of the following four steps, the group of symbols [ | ] is used as follows: [A|B] means that A
is the case for the integration along the x-axis and B is the case for the integration along the y-axis.

i. Sketch the curves. Find the [x|y]-coordinates of their points of intersection. Know where the region
is. Notice the [x|y]-sub-intervals where the curves change their relative positions if any.

ii. Draw a thin [vertical|horizontal] rectangle or strip of width [dx|dy]. It's enough to do this over
just one sub-interval.

iii. For each sub-interval, write down the area element (area of the rectangle), dA = [(height) dx|
(length) dy]. Make sure that [height|length] is in terms of [x| y] (in accordance with [dx|dy]).

iv. Integrate the expression for the area element along the [x|y]-axis. If necessary, split the integral
into a sum of integrals on the sub-intervals.


Remarks 2

i. Step iii above can be skipped. You may want to skip it when you're sure you've got a good grasp of
the topic.

ii. Integration along the x-axis is also called integration in the x-direction or using dx-increment.
Integration along the y-axis is also called integration in the y-direction or using dy-increment.

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Problems & Solutions


1. Find the area of the plane region bounded by the graph of y = cos x and the x-axis from x = 0 to
x = 3p/2.

Solution

On [0, 3p/2] we have cos x = 0 Û x = p/2 or 3p/2. So the area is:



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2. Compute the area of the plane region bounded by y = x² and y = – 2x + 3.

Solution



We have x² = – 2x + 3 Û x² + 2x – 3 = 0 Û (x – 1)(x + 3) = 0 Û x = 1 or –3. So the area is:



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3. Find the area of the region bounded by the curves with equations y = sin x and y = cos x,
respectively, from x = 0 to x = 2p.

Solution

On [0, 2p] we have sin x = cos x Û (tan x = (sin x)/(cos x) = 1 and cos x ¹ 0) Û ((x = p/4 or
5p/4) and (x ¹ p/2 and 3p/2)) Û x = p/4 or 5p/4. So the x-coordinates of the points of intersection
of the two curves are x = p/4 and x = 5p/4. From the sketch of the graphs the area A in question is:



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4. Calculate the area of the plane region bounded by y = (1 + e^x)/e^x and y = 1 and to the right of the
y-axis.

Solution



The equation (1 + e^x)/e^x = 1 or 1 + e^x = e^x has no solution. So the graphs have no points of intersection.
The area A of interest is:



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5. Find the area of the plane region bounded by the curve x = y² – 2 and the line y = – x using
dy-increments.

Solution



We have y² – 2 = – y Û y² + y – 2 = 0 Û ( y – 1)( y + 2) = 0 Û y = 1 or –2. Thus, the area A in
question is:



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