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.

21.4

Fig. 21-3 Fig. 21-4

163 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

21.7

Fig. 21-5

21.6 The region bounded by the curves y = Vx, y = l, and x = 4.

Fig. 21-6

164 _{CHAPTER 21}

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 = or x = l. So, the points of intersection are(1,1) and (1,4). Hence, the area is il,2[(6x-2)~

See Fig. 21-6. The region is bounded above by and below by y = 1. Hence, the area is given by

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

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, y2 + 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.

U, x=k

4*2]<ic = (3*2-2;c-tx3)]|/2 = ( 3 - 2 - i ) - ( ! - l - i ) = i .

(cos x — sin x) dx = (sin x + cos x) ], -(0+1) = - 1 y]

AREA AND ARC LENGTH

Fig. 21-9 Fig. 21-10

21.11 The bounded region between the curve y =

See Fig. 21-11. First we find the points of intersection:

jc = l. Thus, the points of intersection are (0,0) and (1,1). The upper curve is y —

• y = x³.

is Hence, the area is /„' (Vx - x3) dx = (I*3'2 - U") M = i - I = n •

Fig. 21-11 Fig. 21-12

21.12 The bounded region in the first quadrant between the curves 4y + 3x = 7 and y = x 2. See Fig. 21-12. First we find the points of intersection:

x-l is an obvious root. Dividing 3x³ - 7x² + 4 by x — l, we obtain 3x² — 4x -4 = (3x + 2)(x - 2).

Hence, the other roots are x = 2 and x = — f . So, the intersection points in the first quadrant are (1,1) and (2, |). The upper boundary is the line and the lower boundary is y = x~². The area is given by 165

and the lower curve^or and y = x .

21.13 The region bounded by the parabolas y = x2 and y — —x2 + 6x.

See Fig. 2113. First let us find the intersection points: x2 = x2 + 6x, x2 = 3x, x23x = 0, x(x -3) = 0, AC = 0 or jt = 3. Hence, the points of intersection are (0,0) and (3,9). The second parabola y= -x2 + 6x = -(x2 -6x)= -[(*-3)2 -9] = -(jtr-3)2 + 9 has its vertex at (3,9), x = 3 is its axis of symmetry, and it opens downward. That parabola is the upper boundary of our region, and y = x² is the lower boundary. The area is given by J³ [(-x² + 6x) - x²] dx = J⁰³ (6* - 2*2) dx = (3x2 - §x3) ]3 = 27 - 18 = 9.

21.14 The region bounded by the parabola x = y2 + 2 and the line y = x — 8.

See Fig. 21-14. Let us find the points of intersection: y + 8 = y2+2, y²-y-6 = Q, (y -3)(y + 2) = 0, y = 3 or y=-2. So, the points of intersection are (11,3) and (6,-2). It is more convenient to integrate with respect toy. The area is J!2 [(y + 8) - (y2 + 2)] dy = J!2 (y + 6 - y2) dy = ($y* + 6y- \y3) ]3_2 = (f + 18-9)-(2-12-f 1)=^.

x = x⁶, *(jt⁵-l) = 0, A - = 0

+ 3x = 7, 4 + 3*3 = lx\ 3*3 - lx- + 4 = 0.

CHAPTER 21

21.15 The region bounded by the parabolas y = x2 — x and y = x — x2.

See Fig. 21-15. Let us find the points of intersection: x2 - x = x - x2, 2x2 - 2x = 0, x(x - 1) = 0, x = 0 or x = 1. Thus, the intersection points are (0,0) and (1,0). The parabola y = x2 - x = (x - \ )2 - \ has its vertex at ( \, - \ ) and opens upward, while y = x - x2 has its vertex at (|, J) and opens downward. The latter parabola is the upper boundary, and the first parabola is the lower boundary. Hence, the area is

Fig. 21-15 Fig. 21-16

21.16 The region in the first quadrant bounded by the curves y = x² and y — x*.

See Fig. 21-16. Let us find the points of intersection: x* = x1, x4 - x2 = 0, x2(x2 - 1) = 0, * = 0 or x = ±1. So, the intersection points in the first quadrant are (0,0) and (1,1). y = x2 is the upper curve.

Hence, the area is /„' (x2 - x4) dx = (Jjt3 - |*5) ] J = } - \ = *.

In Problems 21.17-21.21, find the arc length of the given curve.

21.17 y =

Recall that the arc length formula is

Hence, Thus,

to

dx. In this case, from x = 1

Fig. 21-13 Fig. 21-14

166

x = 2.

K[(x-x2)-(x2-x)]dx = 2ti(x-x2)dx = 2(±)]1X0 = 2tt-l)=:li.

21.19 y = x2'3

AREA AND ARC LENGTH 0 167

21.21

and

Fig. 21-17 Fig. 21-18

21.18 y = 3x-2 from * = 0 to jc = l.

So,

from x = 1 to x = 8.

Hence, 1 + (y')2 = 1 + (4/9x2'3) = (9x213 + 4)/9x2'3. Thus, Then,

Let

21.20 jc2'3 + y2'3 = 4 from x = 1 By implicit differentiation,

to x = 8.

So,

Hence, Therefore,

from x = 1 to x = 2.

So,

Hence,

21.22 Let &t consist of all points in the plane that are above the x-axis and below the curve whose equation is

y ~ -x2 + 2* + 8. Find the area of $.

(see Fig. 21-17). To find where it cuts the*-axis, let -x2+2* + 8 = 0, x - 2 x - 8 = 0, (x - 4)(* + 2) = 0, or x = —2. Hence, the area of &i is J_2

21.23 Find the area bounded by the curves y = 2x2 - 2 and y = ::2 + x.

See Fig. 21-18. y = 2x2 - 2 is a parabola with vertex at (0, -2). On the other hand, y = x2 + x = y' = 3.

L = M = 9^x2/3 + 4, du = 6x~ⁱⁿ dx.

y = -(X-8) = -[(X-l)2-9]=-(X-lY + 9.The parabola's vertex is (1,9) and it opens downward2

x = 4

32)-(§+4-16) = 36. (-x* + 2x + 8)dx = (- ;U + x* + 8x) f_² = (- f + 16 +

(x + j)2 - I is a parabola with vertex (-5, - j). To find the points of intersection, set 2x2 - 2 = x2 + x, x2-x-2 = 0, (x - 2)(x + 1) = 0, x = 2 or *=-!. Thus, the points are (2,6) and (-1,0). Hence, the area is J2, [(x2 + x) - (2x2 -2)] dx = /!, (2 + x - x2) dx = (2x + \x2 - ^3) ]2_, = (4 + 2- f) - (-2 + J + J) = -§.

168 CHAPTER 21

21.24 Find the area of the region between the jc-axis and y = (x — I)3 from x = 0 to x = 2.

Fig. 21-19 Fig. 21-20

21.25 Find the area bounded by the curves y = 3x2 — 2x and y = 1 — 4x.

21.26 Find the area of the region bounded by the curves y = x2 — 4x and x + y = 0.

Fig. 21-21 Fig. 21-22

21.27 Find the area of the bounded region between the curve y = x3 — 6x2 + 8x and the x-axis.

y = x³ - 6x² + 8x = x(x² - 6x + 8) = x(x - 2)(x - 4). So, the curve cuts the x-axis at x = 0, x = 2, and x = 4. Since Km f(x) = +°° and lim f(x) = —<*, the graph can be roughly sketched as in Fig. 21-22.

Hence the required area is A, + A2 = Jo (x3 6x2 + 8x) dx + J2 (x3 6x2 + 8x)dx = ( \x4 2x3 + 4x2) ]„ -( \x4 - 2x3 + 4x2) ]42 = -(4 - 16 + 16) - [-(64 - 128 + 64) - -(4 - 16 + 16)] = 4 - -(-4) = 8.

See Fig. 21-20. y = 3x2 - 2x = 3(x2 - \x) = 3[(* - \)2 - 5] = 3(x - |)2 - \. Thus, that curve is a parabola with vertex (i,-j)- To find the intersection, let 3x2 - 2x = 1 - 4x, 3x2 + 2x - 1 = 0, (3x - l)(x + 1) = 0, x= j or x=—\. Hence, the intersection points are (5, —3) and (—1,5). Thus, the area is Jlj3 [(1 — 4X)-(3XX)]dx = S1-?(l-2x-3x2)dX = (X-X)]^ = (li-%-lf)-(-l-l + l)=%.

See Fig. 21-21. The parabola y = x² - 4x = (x - 2)² - 4 has vertex (2, -4). Let us find the intersection of the curves: x2 — 4x = —x, x2 — 3x = 0, x(x — 3) = 0, x = 0 or x = 3. So, the points of intersection are (0,0) and (3, -3). Hence, the area is J⁰³ [-x - (x2 - 4x)] dx = J⁰³ (3* - x2) dx = (|x² - ³x³) ]³ = ¥ -9=|.

As shown in Fig. 21-19, the region consists of two pieces, one below the *-axis from x = Q to x = 1, and the other above the jc-axis from x — 1 to x = 2. Hence, the total area is Jo — (x — I)³ dx +

;

¹²

(^-i)

³

dx = -Ux-i)

⁴

]o+U^-i)

⁴

]i = [-Uo-i)] + [i(i-o)] = i.

AREA AND ARC LENGTH 169 21.28 Find the area enclosed by the curve y2 = x2 - x⁴.

Since y = x (1 —x)(l +x), the curve intersects the Jt-axis at x = Q, x = \, and x = —\. Since the graph is symmetric with respect to the coordinate axes, it is as indicated in Fig. 21-23. The total area is four times the area in the first quadrant, which is JJ *Vl -x2 dx = -\ JJ (1 - *2)I/2D,(1 - x2) dx = \ • |(1 -x²)³'² ]¹⁰ = -HO- 1) = 3. Hence, the total area is f .

Fig. 21-23 Fig. 21-24

21.29 Find the area of the loop of the curve y2 = x4(4 + x) between x = -4 and x = 0. (See Fig. 21-24.) By symmetry with respect to the x-axis, the required area is 2 J!4 y dx = 2 J°4 *2V4 + x dx. Let u = V4 + x, u2 = 4 + x, x = u2-4, dx = 2udu. Hence, we have 2 ft (u2 - 4)2w • 2u du = 4 J"02 (u6 - 8w4 + 16^M2)ciH = 4 ( ^⁷- | M⁵+ f^{M 3}) ] ^ = 4(^-²? + ^ ) = ^ .

21.30 Find the length of the arc of the curve x = 3y3'² - 1 from y = 0 to y = 4.

The arc length L = J04 \/l + (dx/dy)2 dy, d x / d y = % y1'2, 1 + (dx/dy)2 = 1+ 81y/4 = (4 + 8ly)/4. So, L=|J^{0 4}V4~+Wrf>'. Let w = 4 + 81v, du = 81 <fy. Then

2^ (328 • 2V82 - 8) = 2S (82V82 - 1).

21.31 Find the length of the arc of 24xy = x4 + 48 from x = 2 to x = 4.

Then Hence,

So,

21.32 Find the length of the arc of y3 = 8x2 from x = 1 to x = 8.

So, Hence,

21.33 Find the length of the arc of 6xy = x4 + 3 from x = 1 to x = 2.

Then and Then

21.34 Find the length of the arc of 27 y2 = 4(x - 2)3 from (2,0) to (11,6V5).

21.35 Find the area bounded by the curve y = l-x 2 and the lines y = l, x = l, and x=4. (See Fig.

21-25.)

The upper boundary is the line y = l. So, the area is J\⁴ [1 - (1 - x~²)] dx = f,⁴ x~² dx = -x~^l I⁴ = - a - l ) = f .

Fig. 21-25

21.36 Find the area in the first quadrant lying under the arc from the _y-axis to the first point where the curve x + y + y2 = 2 cuts the positive A:-axis.

It hits the x-axis when y = 0, that is, when x = 2. Hence, the arc extends in the first quadrant from (0,1) to (2,0), and the required area is ft (2 - y - y2) dy = (2y - \y2 - iy3) ]J = 2 - \ - \ = J.

21.37 Find the area under the arch of y = sin;t between x = Q and x = IT.

The area is J0" sin xdx= -cos x ]„ = -(-1 - 1) = 2.

21.38 Find the area of the bounded region between y = x and y = 2* (see Fig. 21-26).

Setting x2 = 2x, we find x = 0 or * = 2. Hence, the curves intersect at (0,0) and (2,4). For 0 < j r < 2 , x2<2, and, therefore, y = 2x is the upper curve. The area is J02 (2* - x2) dx = (x2

-1 ,,3\ ^2 — A 8 _ 4 3 * ) J o ~4~ 3 3

-Fig. 21-26 Fig. 21-27

21.39 Find the area of the region bounded by the parabolas y = x2 and x = y2.

I See Fig. 21-27. Solving simultaneously, * = y2 = ;t4, x = Q or j c = l . Hence, the curves intersect at (0,0) and (1,1). Since Vx > x2 for 0<A:<1, x = y2 is the upper curve. The area is f,!(*1/2

-^ r f r - O -^ - J -^ J i - i - l - l .

21.40 Find the area of the bounded region between the curves y = 2 and y = 4jc3 + 3x2 + 2.

For y = 4x³ + 3x² + 2, y'= l2x² + 6x = 6x(2x + 1), and y" = 24x + 6. Hence, the critical number j: = 0 yields a relative minimum, and the critical number -1 yields a relative maximum. To find intersection points, 4*3 + 3x2 = 0, x = 0 or x = -1. Thus, the region is as indicated in Fig. 21-28, and the area is J!3/4 [(4*3 + 3*2 + 2) - 2] dx = J!3/4 (4*3 + 3*2) dx = (*4 + *3) ]°_3M = -(& - g) = &.

170 CHAPTER 21

The curve hits they-axis when x = 0, that is, y2 + y-2 = 0, (y + 2)(y - 1) = 0, y = -2 or y = \.

AREA AND ARC LENGTH 171

Fig. 21-28 Fig. 21-29

21.41 Find the area of the region bounded by y — x — 3* and y = x.

For y = x3 — 3x, y' = 3x — 3 = 3(x - l)(x + 1), and y" = 6x. Hence, the critical number x = \ yields a relative minimum, and the critical number x=—\ yields a relative maximum. Setting x3 — 3>x = x, x3 = 4x, x = 0 or x = ±2. Hence, the intersection points are (0,0), (2, 2), and (-2, -2). Thus, the region consists of two equal pieces, as shown in Fig. 21-29. The piece between x = -2 and x = 0 has area J!2 [(x3 - 3x) -x)]dx = J!2 (*3 - 4x) dx = (i*4 - 2x2) ]°_2 = -(4 - 8) = 4. So the total area is 8.

21.42 The area bounded by y = x2 and y = 4 (two even functions) is divided into two equal parts by a line y = c. Determine c.

See Fig. 21-30. The upper part is, by symmetry, 2 Jc4 y112 dy = 2- iy3'2]* = 5(8- c3'2). The lower part is 2tiyll2dy = 2-%y3'2 ]C0= $c3'2. Hence, 8-c3 / 2 = c3'2, 8 = 2c3'2, 4=c3'2, c = 42/3=^16.

Fig. 21-30

21.43 Let What happens to the area above the *-axis bounded by v = x p, x = \, and x = b, as The area is

As the limit is

21.44 Find the arc length of for

So, and

Hence,

172 21.45

21.46 Find the area bounded by y = x3 and its tangent line at x = 1 (Fig. 21-31).

At x = l, y'=3x2=3. Hence, the tangent line is (y - !)/(* - 1) = 3, or y = 3x-2. To find out where this line intersects y = x3, we solve x3 = 3x - 2, or x3 - 3x + 2 = 0. One root is x = 1 (the point of tangency). Dividing x3-3x + 2 by x = l, we obtain x2 + x -2 = (x + 2)(x -1). Hence, x = — 2 is a root, and, therefore, the intersection points are (1,1) and (—2, —8). Hence, the required area is J12[*3 -(3* -2)] dx = ^2(x3 -3*+ 2) dx = (^-lx2 + 2x)t2 = ( l - § + 2 ) - ( 4 - 6 - 4 ) = ? .

Fig. 21-31 Fig. 21-32

21.47 Find the area of the region above the curve y = *2-6, below v = x, and above y=-x (Fig. 21-32).

We must find the area of region OPQ. To find P, solve y = -x and y = *2-6: x2-6=-x, x2 + x-6 = 0, (x + 3)(x -2) = 0, * = -3 or x = 2. Hence, P is (2,-2). To find Q, solve y = x and y = x2-6: x2-6 = x, x2-x-6 = Q, (x-3)(x + 2) = 0, * = 3 or x = -2. Hence, Q is (3,3). The area of the region is J2 [x - (-x)] dx + J3 [x - (x2 - 6)] dx = J02 2x dx + /23 (x - x2 + 6) dx = x2 ]2 + ({x2

-^x3 + 6x) }\.= 4 + (§ - 9 + 18) - (2 - f + 12) = f . Notice that the region had to be broken into two pieces before we could integrate.

Find the arc length of y=l(l + x2)3'2 for Os *=£ 3.

CHAPTER 21

y' = (l + x2)l'2-2x, and (y')2 = 4x2(l +x2). So, 1 + (y')2 = 1 + 4x2 + 4xA = (1 + 2x2)2. Hence, L = J03 (1 +2*2) <fc = (* + I*3) I'= 3 + 18 = 21.

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CHAPTER 22

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