areas solution - arizona state universityterri/courses/271resources/271ch6suex.pdf · 4x 2x = 3 x2...

1

Chapter 6 Set-up examples

The purpose of this document is to demonstrate the work that will be required if you areasked to set-up integrals on an exam and/or quiz.

1. Areas

(1) Set up, do not evaluate, any integrals needed to find the area of the region enclosedby the curves

y = cosx, y = sin 2x, 0 ≤ x ≤ π

2Solution:

0

dx

dx

For the intersections:

cosx = sin 2x = 2 sin x cosx (double angle identity)

cosx = 0 or sinx =1

2

x =π

2,π

6(since 0 ≤ x ≤ π/2)

there are two regions, the first (A1) for 0 ≤ x ≤ π/6:top curve: y = cosxbottom curve: y = sin 2xwidth of slice: dxarea of slice: [cosx− sin 2x] dx

in the second region (A2), for π/6 ≤ x ≤ π/2, they switchtop curve: y = sin 2xbottom curve: y = cosx.width of slice: dxarea of slice: [sin 2x− cosx] dx

A = A1 + A2 =

∫ π/6

0

[cosx− sin 2x] dx+

∫ π/2

π/6

[sin 2x− cosx] dx

c©2018Arizona State University School of Mathematical & Statistical Sciences

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(2) Set up, do not evaluate, an integral to find the area of the region enclosed by thecurves

y = ex, y = x2 − 1, x = −1, x = 1

Solution:

0

dx

The graph shows that when x = −1 the curve y = ex is at y = 1/e, which ispositive, and the curve y = x2 − 1 is at the x-axis.

When x = 1 the curve y = ex is at y = e and the curve y = x2 − 1 is again at thex-axis.top curve: y = ex

bottom curve: y = x2 − 1width of slice: dxarea of slice: [ex − (x2 − 1)] dx

A =

∫ 1

−1

[ex − (x2 − 1)] dx =

∫ 1

−1

[ex − x2 + 1] dx


3

(3) Set up, do not evaluate, a single integral to find the area of the region enclosed bythe curves

y = (x− 2)2, y = x.

Solution:

0

dx


(x− 2)2 = x

x2 − 4x+ 4 = x

x2 − 5x+ 4 = 0

(x− 4)(x− 1) = 0

x = 1, 4

When x = 1 the curve y = ex is at y = e and the curve y = x2 − 1 is again at thex-axis.top curve: y = ex

bottom curve: y = x2 − 1width of slice: dxarea of slice: [x− (x− 2)2] dx

A =

∫ 4

1

[x− (x− 2)2] dx =

∫ 4

1

[x− (x2 − 4x+ 4)] dx =

∫ 4

1

[−x2 + 5x− 4)] dx


4

(4) Set up, do not evaluate, a single integral to find the area of the region enclosed bythe curves

x = 2y2, x = 4 + y2.

0 (note: this graph would be provided)

Solution:

0

dy


2y2 = 4 + y2

y2 = 4

y = ±2

right curve: x = 4 + y2

left curve: x = 2y2

width of slice: dyarea of slice: [4 + y2 − 2y2] dy

A =

∫ 2

−2

[4 + y2 − 2y2] dy =

∫ 2

−2

[4− y2] dy


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2. Volumes of Revolution

(1) Use the method of washers to set up, do not evaluate, an integral to find the volumeof the solid obtained by rotating the region bounded by

y2 = x, x = 2y

about the y-axis.

Solution:

0

R

r


y2 = 2y

y2 − 2y = 0

y(y − 2) = 0

y = 0, 2

We want washers, so we need our slice to be perpendicular to the axis of rotation; dythickness of the washer: dybig radius is determined by the x value for the line: 2yarea of the big disc: π(2y)2

volume of the big disc: π(2y)2 dysmall radius is determined by the x value for the parabola: y2

area of the hole to be removed: π(y2)2

volume of the hole to be removed: π(y2)2 dyvolume of the washer: π(2y)2 dy − π(y2)2 dy = π[(2y)2 − (y2)2] dy

V =

∫ 2

0

π[(2y)2 − (y2)2] dy = π

∫ 2

0

[4y2 − y4] dy


6

(2) Use the method of washers to set up, do not evaluate, an integral to find the volumeof the solid obtained by rotating the region bounded by

x− y = 1, y = x2 − 4x+ 3

about the line y = 4.

0

(note: this graph would be provided)

Solution:

0

dx

R

r


x2 − 4x+ 3 = x− 1 (since x− y = 1 =⇒ y = x− 1)

x2 − 5x+ 4 = 0

(x− 4)(x− 1) = 0

x = 1, 4

We want washers, so we need our slice to be perpendicular to the axis of rotation; dxthickness of the washer: dxbig radius is determined by the line y = 4 and the x value for the parabola:

4− (x2 − 4x+ 3) = 4− x2 + 4x− 3 = −x2 + 4x+ 1area of the big disc: π(−x2 + 4x+ 1)2

volume of the big disc: π(−x2 + 4x+ 1)2 dxsmall radius is determined by the line y = 4 and the x value for the line y = x − 1:4− (x− 1) = 5− xarea of the hole to be removed: π(5− x)2

volume of the hole to be removed: π(5− x)2 dxvolume of the washer: π(−x2 + 4x+ 1)2 dx− π(5− x)2 dx

V = π

∫ 4

1

[(−x2 + 4x+ 1)2 − (5− x)2] dx = π

∫ 4

1

[(−x2 + 4x+ 1)2 − (5− x)2] dx


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(3) Use the method shells to set up, do not evaluate, an integral to find the volume ofthe solid obtained by rotating the region bounded by

y = x3, y = 8, x = 0

about the x-axis.

Solution:

0

dy

r

h

We want shells, hence we need our slice to be parallel to the axis of rotation; dyradius of the shell: yheight of the shell: x for x = 3

√y

thickness of shell: dyvolume of shell: 2πy( 3

√y)

V = 2π

∫ 8

0

y4/3dy


8

(4) Use the method shells to set up, do not evaluate, an integral to find the volume ofthe solid obtained by rotating the region bounded by

y = 4x− x2, y = 3

about the x-axis.

0

(note: this graph would be provided)

Solution:

0

dx

h

rx


4x− x2 = 3

x2 − 4x+ 3 = 0

(x− 3)(x− 1) = 0

x = 1, 3

We want shells, hence we need our slice to be parallel to the axis of rotation; dxradius of the shell: x− 1height of the shell = top curve - bottom curve: 4x− x2 − 3thickness of shell: dxvolume of shell: 2π(x− 1)(4x− x2 − 3)dx

V = 2π

∫ 3

1

(x− 1)(4x− x2 − 3)dx


9

3. Arc Length

As there is not a lot of work to be shown here, I will just do one example. You can let meknow if you want another.

(1) Find the exact length of the curve

y = 1 + 6x3/2, 0 ≤ x ≤ 1

Solution:

L =

∫ 1

0

√1 +

(dy

dx

)2

dx

=

∫ 1

0

√1 + (9x1/2)2 dx

=

∫ 1

0

√1 + 81x dx

=1

81

∫ 82

1

√u du (u = 1 + 81x, du = 81 dx)

=1

81

∫ 82

1

u1/2 du

=1

81

u3/2

3/2

]82

1

=2

81 · 3

((82)3/2 − 1

)

4. Work

4.1. Springs.

(1) A spring has a natural length of 20 cm. If a 25-N force is required to keep it stretchedto a length of 30 cm, how much work is required to stretch it from 20 cm to 25 cm?

Solution: We must first convert centimeters to meters: the natural length of thespring is .2 m, etc.

When the spring is stretched to a length of .3 m, this is .1 m beyond the naturallength, it takes 25=N of force, so

25 = f(.1) = .1k =⇒ k =25

.1= 250


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We want to stretch the spring from .2 m to .25 m, that is, from x = 0 to x = .05,and the work done is

W =

∫ .05

0

f(x) dx

=

∫ .05

0

250x dx

= 250x2

2

].05

0

= 125(.05)2 (ok if “do not simplify”)

= 125(.025) J

(2) Suppose that 2 J of work is needed to stretch a spring from its natural length of 30cm to a length of 42 cm. How much work is needed to stretch the spring from 35 cmto 40 cm?

Solution: We must change units from cm to m, so that the natural length of thespring is .3 m. We are given that the work done in stretching the spring from x = 0to x = .12 is 2, so

2 = W =

∫ .12

0

kx dx =k

2(.12)2

and we can solve for k:

k =4

(.12)2.

Thus the work done in stretching the spring from x = .05 to x = .1 is

W =

∫ .1

.05

4

.122x dx

=4

(.12)2

∫ .1

.05

x dx

=4

(.12)2

x2

2

].1.05

=2

(.12)2

((.1)2 − (.05)2

)J

4.2. Cables.

(1) A heavy rope, 50 ft long, weighs .5 lb/ft and hangs over the edge of a building 120ft high. How much work is done in pulling the rope to the top of the building?

Solution: Note: the information that the building is 120 ft hight is not reallyrelevant — the only important thing is that the building is at least tall enough sothat the rope hangs vertically.

Introduce an x-axis pointing down, with x = 0 at the top of the rope and x = 50at the bottom of the rope. So, we have 0 ≤ x ≤ 50.


11

Consider a short segment of the rope x ft below the top:length: dx ftdensity: .5 lb/ftweight: .5 dx lbraised: x ftwork on segment: .5x dx ft-lb

Therefore the total work is

W =

∫ 50

0

.5x dx

=1

2

∫ 50

0

x dx

=1

2

(502

2

)= 625 ft-lb

4.3. Pumping.

(1) Consider a hemispherical tank of radius 3 m, with a pipe of length 1 m extendingvertically from the top of the tank, and with a spout at the top of this pipe. If thetank is full of water, find the work required to pump the water out of the spout.Acceleration due to gravity is 9.8 m/sec2; the density of water is 1000 kg/m3.

Solution: ALWAYS draw your picture, put in axes and label them (if you donot do so, the problem cannot be graded.

0

x

y

dx

h

Note that the tank is 3 m deep (the radius of the sphere). The equation of the circlethat forms the sides of the tank (determines the relationship between x and y) is then

(x−3)2+y2 = 32 which can be solved for y to get y = −√

9− (x− 3)2 = −√

6x− x2.

Thus a horizontal cross section of the tank is a circle of radius =√

6x− x2


12

area of the cross section:π(√

6x− x2)2 = π(6x− x2) m2

thickness of the cross section: dx mvolume of a horizontal slice is: π(6x− x2)dx m3

its mass is: 1000π(6x− x2)dx kgthe force on the slice is: 9.8(1000)π(6x− x2)dx kg-m/sec2=Nthis slice is lifted: 4− x mso the work done for the slice is: 9800π(4 − x)(6x − x2)dx N-m=J Thus the totalwork is

W =

∫ 3

0

9800π(4− x)(6x− x2) dx

= 9800π

∫ 3

0

(24x− 10x2 + x3

)dx

= (9800)π

[12x2 − 10

3x3 +

1

4x4

]3

0

= (9800)π

(108− 90 +

81

4

)= (9800)π

(153

4

)= 2450(153)π

Make sure you look at the lecture notes for moreexamples of pumping problems.

the first person to report a typo will get $1 for each typo


areas solution - arizona state universityterri/courses/271resources/271ch6suex.pdf · 4x 2x = 3 x2...

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