6.1. USING DEFINITE INTEGRALS TO FIND AREA AND LENGTH
337
Activity 6.1.
In each of the following problems, our goal is to determine the area of the region
described. For each region, (i) determine the intersection points of the curves, (ii) sketch
the region whose area is being found, (iii) draw and label a representative slice, and (iv)
state the area of the representative slice. Then, state a definite integral whose value is
the exact area of the region, and evaluate the integral to find the numeric value of the
region’s area.
(a) The finite region bounded by y =
√
x and y =
1
4 x.
(b) The finite region bounded by y = 12 − 2x 2 and y = x 2 − 8.
(c) The area bounded by the y-axis, f (x) = cos(x), and g(x) = sin(x), where we
consider the region formed by the first positive value of x for which f and g
intersect.
(d) The finite regions between the curves y = x 3 − x and y = x 2 .
⊳
Finding Area with Horizontal Slices
At times, the shape of a geometric region may dictate that we need to use horizontal
rectangular slices, rather than vertical ones. For instance, consider the region bounded by
the parabola x = y 2 − 1 and the line y = x − 1, pictured in Figure 6.4. First, we observe
that by solving the second equation for x and writing x = y + 1, we can eliminate a variable
through substitution and find that y + 1 = y 2 − 1, and hence the curves intersect where
y 2 − y − 2 = 0. Thus, we find y = −1 or y = 2, so the intersection points of the two curves
are (0, −1) and (3, 2).
We see that if we attempt to use vertical rectangles to slice up the area, at certain values
of x (specifically from x = −1 to x = 0, as seen in the center graph of Figure 6.4), the curves
that govern the top and bottom of the rectangle are one and the same. This suggests, as
shown in the rightmost graph in the figure, that we try using horizontal rectangles as a way
to think about the area of the region. For such a horizontal rectangle, note that its width
depends on y, the height at which the rectangle is constructed. In particular, at a height
y between y = −1 and y = 2, the right end of a representative rectangle is determined
by the line, x = y + 1, while the left end of the rectangle is determined by the parabola,
x = y 2 − 1, and the thickness of the rectangle is △y.
Therefore, the area of the rectangle is
A rect = [(y + 1) − (y
2 − 1)]△y,
from which it follows that the area between the two curves on the y-interval [−1, 2] is
337
Activity 6.1.
In each of the following problems, our goal is to determine the area of the region
described. For each region, (i) determine the intersection points of the curves, (ii) sketch
the region whose area is being found, (iii) draw and label a representative slice, and (iv)
state the area of the representative slice. Then, state a definite integral whose value is
the exact area of the region, and evaluate the integral to find the numeric value of the
region’s area.
(a) The finite region bounded by y =
√
x and y =
1
4 x.
(b) The finite region bounded by y = 12 − 2x 2 and y = x 2 − 8.
(c) The area bounded by the y-axis, f (x) = cos(x), and g(x) = sin(x), where we
consider the region formed by the first positive value of x for which f and g
intersect.
(d) The finite regions between the curves y = x 3 − x and y = x 2 .
⊳
Finding Area with Horizontal Slices
At times, the shape of a geometric region may dictate that we need to use horizontal
rectangular slices, rather than vertical ones. For instance, consider the region bounded by
the parabola x = y 2 − 1 and the line y = x − 1, pictured in Figure 6.4. First, we observe
that by solving the second equation for x and writing x = y + 1, we can eliminate a variable
through substitution and find that y + 1 = y 2 − 1, and hence the curves intersect where
y 2 − y − 2 = 0. Thus, we find y = −1 or y = 2, so the intersection points of the two curves
are (0, −1) and (3, 2).
We see that if we attempt to use vertical rectangles to slice up the area, at certain values
of x (specifically from x = −1 to x = 0, as seen in the center graph of Figure 6.4), the curves
that govern the top and bottom of the rectangle are one and the same. This suggests, as
shown in the rightmost graph in the figure, that we try using horizontal rectangles as a way
to think about the area of the region. For such a horizontal rectangle, note that its width
depends on y, the height at which the rectangle is constructed. In particular, at a height
y between y = −1 and y = 2, the right end of a representative rectangle is determined
by the line, x = y + 1, while the left end of the rectangle is determined by the parabola,
x = y 2 − 1, and the thickness of the rectangle is △y.
Therefore, the area of the rectangle is
A rect = [(y + 1) − (y
2 − 1)]△y,
from which it follows that the area between the two curves on the y-interval [−1, 2] is