M1120 Class 5 Dan Barbasch September 4, 2011 Dan Barbasch () - - PowerPoint PPT Presentation

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

Dan Barbasch () M1120 Class 5 September 4, 2011 2 / 16

Method of Slices

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

a

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

Dan Barbasch () M1120 Class 5 September 4, 2011 3 / 16

Method of Slices

Thicken it up vertically a distance ∆y.

Dan Barbasch () M1120 Class 5 September 4, 2011 3 / 16

Method of Slices

The volume is A∆y.

Dan Barbasch () M1120 Class 5 September 4, 2011 3 / 16

Method of Slices

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

Dan Barbasch () M1120 Class 5 September 4, 2011 3 / 16

Method of Slices

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

Dan Barbasch () M1120 Class 5 September 4, 2011 3 / 16

Method of Slices

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

Dan Barbasch () M1120 Class 5 September 4, 2011 3 / 16

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).

Dan Barbasch () M1120 Class 5 September 4, 2011 3 / 16

Method of Slices

Volume(Body) = b

a

A(x) dx.

Dan Barbasch () M1120 Class 5 September 4, 2011 3 / 16

Problems using Slices

Exercise 1: [(12) in the text] Find the volume of a pyramid with a square base of area 9 and height 5. Exercise 2: [(6a) in the text] Find the volume of the solid which lies between planes perpendicular to the x-axis between x = ± π

3 , and cross

sections (perpendicular to the x−axis) circular disks with diameter running from the curve y = tan x to y = sec x.

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

Exercise 2

y = tan x

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

Exercise 2

y = tan x y = sec x

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

Exercise 2

y = tan x y = sec x y = sec x and y = tan x together.

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

Exercise 2

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

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

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.

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

Disks and Washers

Rotate the area in the picture on the left about the x−axis. The volume

• f the resulting body on the right is computed by the method of disks:

Dan Barbasch () M1120 Class 5 September 4, 2011 6 / 16

Disks and Washers

Rotate the area in the picture on the left about the x−axis. The volume

• f the resulting body on the right is computed by the method of disks:

Volume = b

a

π

• r(x)2

dx = b

a

π (f (x))2 dx.

Dan Barbasch () M1120 Class 5 September 4, 2011 6 / 16

Washers

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

a

π

• r2(x)2 − r1(x)2

dx. Exercise 1: Compute the volume of the region bounded by x = 3y

2 ,

x = 0, and y = 2 revolved about the y−axis. Exercise 2: Write a definite integral which computes the volume of the solid obtained by revolving the region in the previous exercise about the axis x = 5.

Dan Barbasch () M1120 Class 5 September 4, 2011 7 / 16

Hints to Exercises

The graph of the region is The formula for the volume is V = b

a

πf (x)2 dx = 2 π[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 be adjusted according to the axis of rotation. V = b

a

π

• r2(x)2 − r1(x)2

dx = 2 π

• 52 − (5 − 3y/2)2

dy.

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Question: What if we want to integrate in x?

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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 concentric cylinders of height h, and respective radii r + ∆r and r? Answer: 2πr(x)h(x)∆x.

Dan Barbasch () M1120 Class 5 September 4, 2011 10 / 16

Method of Shells

The 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

a

2πr(x)h(x) dx. I use blue again to emphasize that this is the “general formula”. You have to adjust according to the problem.

Dan Barbasch () M1120 Class 5 September 4, 2011 10 / 16

Method of Shells

For the body of revolution obtained by rotating the region in the picture

• n the left about the y-axis, the method of shells gives:

Volume = b

a

2πxf (x) dx.

Dan Barbasch () M1120 Class 5 September 4, 2011 10 / 16

Summary

For the region between x = a, x = b, bounded above by y = f (x) ≥ 0 and below by the x−axis, the volume obtained by revolving about the x-axis is given by the method

• f disks

b π(f (x))2 dx The volume of the body obtained by revolving the region before with a ≥ 0, b ≥ 0 rotated about the y−axis is given by the method of shells: b 2πxf (x) dx.

Dan Barbasch () M1120 Class 5 September 4, 2011 11 / 16

Solution to Exercises 1,2 Method of Shells

The general formula is b

a 2πh(x)r(x) dx.

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

• 5 − 2x

3

• dx = 6π.

For the second problem, h(x) = 5 − y = 5 − 2x/3 as before, but the radius is r(x) = 5 − x. The volume is V = 3 2π(5 − x)(5 − 2x/3) dx = 24π.

Dan Barbasch () M1120 Class 5 September 4, 2011 12 / 16

Problems from section 6.1

Exercise 1: [(22) in the text] Find the volume of the body obtained by rotating 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 by rotating 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 and

y = 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?

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Solution to Exercise 3: The bowl is obtained by rotating the region in the first 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 πx2 dy = π 5 2y dy = 25π . Method of Shells: We integrate in x. The general formula is ∆V = 2πrh∆x, and the volume is V = √

10

2πx

• 5 − x2

2

• dx = 2π
• 5x2

2 − x4 8

• √

10

= 25π. Check the calculations, I skipped some arithmetic!

Dan Barbasch () M1120 Class 5 September 4, 2011 14 / 16

For the second part of the problem, we need the volume V as a function

• f the height h. We know that dV

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

V (h) = h

0 π(2y) dy = πh2. So

dV dt = 2πhdh dt , dh dt = 1 2πh dV dt , and we can evaluate dh dt = 3 8πin/sec.

Dan Barbasch () M1120 Class 5 September 4, 2011 14 / 16

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 generated by rotating the region bounded by y = 2 − x2, x = 0, y = x2 about the y-axis. Exercise 3: [(22) in the text] Compute the volume of the body generated by rotating the region bounded by y = √x, y = 0, y = 2 − x about the x-axis.

Dan Barbasch () M1120 Class 5 September 4, 2011 15 / 16

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 region when 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. Dan Barbasch () M1120 Class 5 September 4, 2011 15 / 16

Volume of a Right Circular Cone

Exercise: Compute the volume of a right circular cone of radius r and height h. Solution: Step 1: Find a region which when rotated about and axis gives the desired 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 π r hy 2 dy = π r2 h2 y3 3

• h

= πr2h 3 . Shells: V = r 2πx

• h − h

r x

• dx = 2π
• hx2

2 − h r x3 3

• r

= πr2h 3 .

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

Volume of a Right Circular Cone

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

r2 y2 + h2a2 r2 . The volume is then

V = r A(x) dx. The area bounded by such a hyperbola is challenging to compute.

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