m1120 class 5 - pi.math.cornell.edupi.math.cornell.edu/~web1120/slides/fall12/sep4.pdf · hints to...

M1120 Class 5

Dan Barbasch

September 4, 2011

Dan Barbasch () M1120 Class 5 September 4, 2011 1 / 16

Course Website

http://www.math.cornell.edu/˜web1120/index.html


Method of Slices

A special case of what is called Cavalieri’s Principle.(graphics courtesy of Allen Back)

V =

∫ b

aA(x) dx

Start with a region (e.g. a disk) of Area A


Method of Slices

Thicken it up vertically a distance ∆y .


Method of Slices

The volume is A∆y .


Method of Slices

This even works if you thicken up at a slant as long as the (vertical) heightis ∆y .


Method of Slices

This is the basis of both Cavalieri and the method of disks.


Method of Slices

Start with a line where x measures distance along the line.


Method of Slices

Start with a line where x measures distance along the line.

Keep track of the (perpendicular to the line) cross sectional area A(x).


Method of Slices

Volume(Body) =

∫ b

aA(x) dx .


Problems using Slices

Exercise 1: [(12) in the text] Find the volume of a pyramid with a squarebase of area 9 and height 5.

Exercise 2: [(6a) in the text] Find the volume of the solid which liesbetween planes perpendicular to the x-axis between x = ±π

3 , and crosssections (perpendicular to the x−axis) circular disks with diameter runningfrom the curve y = tan x to y = sec x .


Exercise 2

y = tan x


Exercise 2

y = tan x

y = sec x


Exercise 2

y = tan x

y = sec x

y = sec x and y = tan x together.


Exercise 2

The diameter at x is d(x) = sec x − tan x . The area of the cross section atx is

A(x) = πr(x)2 = π (d(x)/2)2 =π(sec x − tan x)2

4.

The volume is

V =

∫ π/3

−π/3

π

4(sec x − tan x)2 dx .

The value of the integral is

(6√

3− π)π

6.


Exercise 2

Some integrals ∫sec2 x dx = tan x + C ,∫sec x · tan x dx = sec x + C ,∫tan2 x dx =

∫ (sec2 x − 1

)dx .

Very useful formula:

tan2 x =sin2x

cos2 x=

1− cos2 x

cos2 x=

1

cos2 x− 1 = sec2 x − 1.


Disks and Washers

Rotate the area in the picture on the left about the x−axis. The volumeof the resulting body on the right is computed by the method of disks:


Disks and Washers

Rotate the area in the picture on the left about the x−axis. The volumeof the resulting body on the right is computed by the method of disks:

Volume =

∫ b

aπ(r(x)2

)dx =

∫ b

aπ (f (x))2 dx .


Washers

The method of washers is a (simple) variant of the method of disks.

V =

∫ b

aπ[r2(x)2 − r1(x)2

]dx .

Exercise 1: Compute the volume of the region bounded by x = 3y2 ,

x = 0, and y = 2 revolved about the y−axis.

Exercise 2: Write a definite integral which computes the volume of thesolid obtained by revolving the region in the previous exercise about theaxis x = 5.


Hints to ExercisesThe graph of the region is

The formula for the volume is V =

∫ b

aπf (x)2 dx =

∫ 2

0π[3y/2]2 dy .

The formula in blue is the general formula; but applied to this problem,integration is in y not x ; so we adjust accordingly.For the second problem the region is the same, and the general formula is“the same” too.Question: What changes?Answer: In this case we are dealing with washers, and the radius must beadjusted according to the axis of rotation.

V =

∫ b

aπ[r2(x)2 − r1(x)2

]dx =

∫ 2

0π[52 − (5− 3y/2)2

]dy .


Question: What if we want to integrate in x?


Method of Shells

This is the heart of the method of shells. Here is a picture

Question: What is the volume of the body between two concentriccylinders of height h, and respective radii r + ∆r and r?Answer: 2πr(x)h(x)∆x .


Method of ShellsThe formula is “obtained by opening up the shell:”

∆V ≈ length × height × width

A somewhat more rigorous argument is the following calculation:The volume of the region between two concentric cylinders of height h,and respective radii r + ∆r and r is

∆V = π(r + ∆r)2h − πr2h =π[(r2 + 2r∆rh + (∆r)2)− r2

]=

=2πr∆rh + π(∆r)2h ≈ 2πrh∆r .

The general formula is

V =

∫ b

a2πr(x)h(x) dx .

I use blue again to emphasize that this is the “general formula”. You haveto adjust according to the problem.


Method of Shells

For the body of revolution obtained by rotating the region in the pictureon the left about the y -axis, the method of shells gives:

Volume =

∫ b

a2πxf (x) dx .


Summary

For the region between x = a, x = b, bounded above by y = f (x) ≥ 0 andbelow by the x−axis,

the volume obtained by revolving about the x-axis is given by the methodof disks ∫ b

0π(f (x))2 dx

The volume of the body obtained by revolving the region before witha ≥ 0, b ≥ 0 rotated about the y−axis is given by the method of shells:∫ b

02πxf (x) dx .


Solution to Exercises 1,2 Method of Shells

The general formula is∫ ba 2πh(x)r(x) dx .

For the first problem, h(x) = 5− y = 5− 2x/3, r(x) = x :

V =

∫ 3

02πx

(5− 2x

3

)dx = 6π.

For the second problem, h(x) = 5− y = 5− 2x/3 as before, but the radiusis r(x) = 5− x . The volume is

V =

∫ 3

02π(5− x)(5− 2x/3) dx = 24π.


Problems from section 6.1

Exercise 1: [(22) in the text] Find the volume of the body obtained byrotating the region bounded by y = 2

√x , x = 0, y = 2, about the x−axis.

Exercise 2: [(40) in the text] Find the volume of the body obtained byrotating the region bounded by y = 2

√x , x = 0, y = 2, is revolved about

the x-axis.

Exercise 3: [(56) in the text] Find the volume of the bowl which has a

shape generated by revolving the graph of y = x2

2 between y = 0 andy = 5 about the y -axis.Water is running in into the bowl at 3 cubic units per second.How fast will the water be rising when it is 4 units deep?


Solution to Exercise 3: The bowl is obtained by rotating the region in thefirst quadrant between the y−axis and y = x2/2.

Method of Disks: We integrate in y . The volume element is

∆V = πr2 ∆y . So V =

∫ 5

0πx2 dy = π

∫ 5

02y dy = 25π .

Method of Shells: We integrate in x . The general formula is∆V = 2πrh∆x , and the volume is

V =

∫ √100

2πx

(5− x2

2

)dx = 2π

(5x2

2− x4

8

)∣∣∣∣√10

0

= 25π.

Check the calculations, I skipped some arithmetic!


For the second part of the problem, we need the volume V as a functionof the height h. We know that dV

dt = 3. We can use h = y ; FTC implies

V (h) =∫ h0 π(2y) dy = πh2. So

dV

dt= 2πh

dh

dt,

dh

dt=

1

2πh

dV

dt,

and we can evaluatedh

dt=

3

8πin/sec .


Problems from Section 6.2

Use the method of shells in the following exercises.

Exercise 1: [(6) in the text] Compute the volume of the body generated by

rotating the region bounded by y =9x√x3 + 9

, x = 0, x = 3 about the

y -axis.

Exercise 2: [(10) in the text] Compute the volume of the body generatedby rotating the region bounded by y = 2− x2, x = 0, y = x2 about they -axis.

Exercise 3: [(22) in the text] Compute the volume of the body generatedby rotating the region bounded by y =

√x , y = 0, y = 2− x about the

x-axis.


Problems from Section 6.2

Use the method of shells in the following exercises.

Exercise 4: [(24) in the text] Consider the region bounded by y = x3,y = 8, x = 0. Compute the volume of the body generated by this regionwhen rotated about the

1 y -axis.

2 line x = 3.

3 line x = −2.

4 x-axis.

5 line y = 8.

6 line y = −1.


Volume of a Right Circular Cone

Exercise: Compute the volume of a right circular cone of radius r andheight h.Solution:Step 1: Find a region which when rotated about and axis gives thedesired cone.Answer: Rotate the region in the first quadrant bounded above by y = h,below by x = r

hy . The picture is “the same” as for exercise 1.

Step 2:

Disks: V =

∫ h

0π( rhy)2

dy = πr2

h2y3

3

∣∣∣∣h0

=πr2h

3.

Shells: V =

∫ r

02πx

(h − h

rx

)dx = 2π

(hx2

2− h

r

x3

3

)∣∣∣∣r0

=πr2h

3.


Volume of a Right Circular Cone

Slices: The circular cone of radius r and height h can be written asequation z2 = h2

r2

(x2 + y2

)with 0 ≤ z ≤ h. Slice it perpendicular to the

x−axis. The cross sections are hyperbolas; at x = a, you getz2 = h2

r2y2 + h2a2

r2. The volume is then

V =

∫ r

0A(x) dx .

The area bounded by such a hyperbola is challenging to compute.


m1120 class 5 - pi.math.cornell.edupi.math.cornell.edu/~web1120/slides/fall12/sep4.pdf · hints to...

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