CHAPTER 21
Area and Arc Length
21.1 Sketch and find the area of the region to the left of the parabola x = 2y2, to the right of the y-axis, and between
y — 1 and y — 3.
See Fig. 21-1. The base of the region is the y-axis. The area is given by the integral
Fig. 21-1 Fig. 21-2
21.2 Sketch and find the area of the region above the line y = 3x - 2, in the first quadrant, and below the line
y = 4.
See Fig. 21-2, The region has a base on the y-axis. We must solve y = 3x - 2 for x:
Then the area is
21.3 Sketch and find the area of the region between the curve y = x3 and the lines y = — x and y = 1.
See Fig. 21-3. The lower boundary of the region is y=—x and the upper boundary is y = x3. Hence,
the area is given by the integral
In Problems 21.4-21.16, sketch the indicated region and find its area.
Fig. 21-3 Fig. 21-4
21.4 The bounded region between the curves y = x2 and y = x3.
See Fig. 21-4. The curves intersect at (0,0) and(l, 1). Between x = 0 and x = 1, y = x2 lies above
y = x3. The area of the region between them is
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21.5 The bounded region between the parabola y = 4x2 and the line y - 6x - 2.
See Fig. 21-5. First we find the points of intersection: 4x2 = 6x-2, 2x2 - 3x + I = 0, (2x - l)(x - 11 =
U, x=k or x = l. So, the points of intersection are (1,1) and (1,4). Hence, the area is il,2[(6x-2)~
4*2]<ic = (3* 2 -2;c-tx 3 )]| / 2 = ( 3 - 2 - i ) - ( ! - l - i ) = i .
Fig. 21-5 Fig. 21-6
21.6 The region bounded by the curves y = Vx, y = l, and x = 4.
See Fig. 21-6. The region is bounded above by y] and below by y = 1. Hence, the area is given
by
21.7 The region under the curve and in the first quadrant.
See Fig. 21-7. The region has its base on the x-axis. The area is given by
Fig. 21-7 Fig. 21-8
21.8 The region bounded by the curves y = sin x, y = cos x, x = 0, and x = 7T/4,
See Fie. 21-8. The upper boundary is y = cos x, the lower boundary is y = sin x, and the left side is
the y-axis. The area is given by (cos x — sin x) dx = (sin x + cos x) ], -(0+1) = -1
21.9 The bounded region between the parabola x = -y2 and the line y = x + 6.
See Fig. 21-9. First we find the points of intersection: y = -y2 + 6, y 2 + y - 6 = 0, (y -2)(y + 3) = 0,
y = 2 or y = - 3 . Thus, the points of intersection are (-4,2) and (-9,-3). It is more convenient to
integrate with respect to y, with the parabola as the upper boundary and the line as the lower boundary. The
area is given by the integral f* [-y2 - (y - 6)1 dy = (- iy3 - ^y2 + 6y) ]2_, = (- f - 2 + 12) - (9 - 1 - 18) =
21.10 The bounded region between the parabola y = x2 - x - 6 and the line y = -4.
See Fig. 21-10. First we find the points of intersection: -4 = x2 - x - 6, x2 - x - 2 = 0, (x - 2)(x +
1) = 0, x = 2 or x = -I. Thus, the intersection points are (2, -4) and (-1, -4). The upper boundary of
the region is y = —4, and the lower boundary is the parabola. The area is given by J^j [-4 — (x2 — x -
6)]dx = $2_l(2-x2 + x)dx = (2x-lx3+kx2)t1 = (4-l+2)-(-2+l + i2)=92.
Area and Arc Length
21.1 Sketch and find the area of the region to the left of the parabola x = 2y2, to the right of the y-axis, and between
y — 1 and y — 3.
See Fig. 21-1. The base of the region is the y-axis. The area is given by the integral
Fig. 21-1 Fig. 21-2
21.2 Sketch and find the area of the region above the line y = 3x - 2, in the first quadrant, and below the line
y = 4.
See Fig. 21-2, The region has a base on the y-axis. We must solve y = 3x - 2 for x:
Then the area is
21.3 Sketch and find the area of the region between the curve y = x3 and the lines y = — x and y = 1.
See Fig. 21-3. The lower boundary of the region is y=—x and the upper boundary is y = x3. Hence,
the area is given by the integral
In Problems 21.4-21.16, sketch the indicated region and find its area.
Fig. 21-3 Fig. 21-4
21.4 The bounded region between the curves y = x2 and y = x3.
See Fig. 21-4. The curves intersect at (0,0) and(l, 1). Between x = 0 and x = 1, y = x2 lies above
y = x3. The area of the region between them is
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21.5 The bounded region between the parabola y = 4x2 and the line y - 6x - 2.
See Fig. 21-5. First we find the points of intersection: 4x2 = 6x-2, 2x2 - 3x + I = 0, (2x - l)(x - 11 =
U, x=k or x = l. So, the points of intersection are (1,1) and (1,4). Hence, the area is il,2[(6x-2)~
4*2]<ic = (3* 2 -2;c-tx 3 )]| / 2 = ( 3 - 2 - i ) - ( ! - l - i ) = i .
Fig. 21-5 Fig. 21-6
21.6 The region bounded by the curves y = Vx, y = l, and x = 4.
See Fig. 21-6. The region is bounded above by y] and below by y = 1. Hence, the area is given
by
21.7 The region under the curve and in the first quadrant.
See Fig. 21-7. The region has its base on the x-axis. The area is given by
Fig. 21-7 Fig. 21-8
21.8 The region bounded by the curves y = sin x, y = cos x, x = 0, and x = 7T/4,
See Fie. 21-8. The upper boundary is y = cos x, the lower boundary is y = sin x, and the left side is
the y-axis. The area is given by (cos x — sin x) dx = (sin x + cos x) ], -(0+1) = -1
21.9 The bounded region between the parabola x = -y2 and the line y = x + 6.
See Fig. 21-9. First we find the points of intersection: y = -y2 + 6, y 2 + y - 6 = 0, (y -2)(y + 3) = 0,
y = 2 or y = - 3 . Thus, the points of intersection are (-4,2) and (-9,-3). It is more convenient to
integrate with respect to y, with the parabola as the upper boundary and the line as the lower boundary. The
area is given by the integral f* [-y2 - (y - 6)1 dy = (- iy3 - ^y2 + 6y) ]2_, = (- f - 2 + 12) - (9 - 1 - 18) =
21.10 The bounded region between the parabola y = x2 - x - 6 and the line y = -4.
See Fig. 21-10. First we find the points of intersection: -4 = x2 - x - 6, x2 - x - 2 = 0, (x - 2)(x +
1) = 0, x = 2 or x = -I. Thus, the intersection points are (2, -4) and (-1, -4). The upper boundary of
the region is y = —4, and the lower boundary is the parabola. The area is given by J^j [-4 — (x2 — x -
6)]dx = $2_l(2-x2 + x)dx = (2x-lx3+kx2)t1 = (4-l+2)-(-2+l + i2)=92.
