6 049 Volume

6049 Volume

AP Calculus

Volume

Volume = the sum of the quantities in each layer

𝑙∗𝑤∗h

where h is # layers

x

x

y

xx

x-axis

Volume by Cross Sections

n

( )x thickness h

foundation

BE7250

Axial (cross sectional) magnetic resonance image of a brain with a large region of acute infarction, formation of dying or dead tissue, with bleeding. This infarct involves the middle and posterior cerebral artery territories.Credit: Neil Borden / Photo Researchers, Inc.

Volume by Slicing(Finding the volume of a solid built on the base in the x – y plane)

METHOD:

1.) Graph the “BASE”

 2.) Sketch the line segment across the base.

That is the representative slice “n”

Use “n” to find: a.) x or y (Perpendicular to axis)

b.) the length of “n”

 3.) Sketch the “Cross Sectional Region” - the shape of the slice (in 3-D )

from Geometry V = B*h h = or is the thickness of the slice

B = the Area of the cross section

 4.) Find the area of the region

 5) Write a Riemann’s Sum for the total Volume of all the Regions

∆ 𝑥 𝑎𝑛𝑑𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑖

𝑒𝑠

Example 1: The base of a solid is the region in the x-y plane bounded by the graph

and the y – axis. Find the volume of the solid if every cross section by a plane perpendicular to the x-axis is a square.

2 3x y

Base

Cross -Section

x

y

𝑦 2=3 −𝑥𝑦=±√3 −𝑥

𝑇𝑉= lim𝑛→ ∞

∑0

3

(2√3 −𝑥 )2 ∆ 𝑥

∫0

3

(2√3−𝑥 )2𝑑𝑥

𝑦=√3 −𝑥

𝑦=−√3− 𝑥

h

𝑛=(√3 −𝑥 ) − (−√3−𝑥 )

𝑛=2√3 −𝑥𝑉=𝐵∗h𝑉=𝑛2 ∆𝑥𝑉= (2√3−𝑥 )2∆ 𝑥

Example 2: The base of a solid is the region in the x-y plane bounded by the graph

and the y – axis. Find the volume of the solid if every cross section by a plane perpendicular to the x-axis is an Isosceles Rt. Triangle (leg on the base).

2 3x y

x

y

nn 𝑉=𝐵∗h

𝑉=12

(𝑛2 ) ∆ 𝑥

𝑦=√3 −𝑥

𝑦=−√3− 𝑥

h

𝑛=(√3 −𝑥 ) − (−√3 −𝑥 )

𝑛=2√3 −𝑥

𝑇𝑉= lim𝑛→ ∞

∑0

3

( 12

(( 2√3− 𝑥 )2 ))∆𝑥

∫0

3

( 12

( (2√3 −𝑥 )2 ))𝑑𝑥

Some Important Area FormulasSquare-

side on base

Square-

diagonal on base

Isosceles rt Δ

leg on base

Isosceles rt Δ

hypotenuse on base

Equilateral Δ

Semi - Circle Circle

diameter on the base

𝐵=𝑛2

n

nnn

n

n

n

𝐵=𝑛2

2

𝐵=12 𝑛

2𝐵=

𝑛2

4

𝑛2 √3

𝐵=12𝑛(𝑛2 )√3

𝐵=𝑛2

4 √3

𝐵=𝜋 𝑟2

4𝐵=

12 (𝑛2 )

2

𝜋

𝐵=𝜋 𝑟 2

8

EXAMPLE #4/406The solid lies between planes perpendicular to the x - axis at x = -1 and x = 1 . The cross sections perpendicular to the x – axis are circular disks whose diameters run from the parabola y = x2 to the parabola y = 2 – x2.

𝑥2=2 −𝑥2

2 𝑥2=2𝑥2=1𝑥±√1

𝑏=∆ 𝑥 [− 1,1 ]𝑛=(2 −𝑥2 ) − (𝑥2 )𝑛=2 −2 𝑥2n

𝑉=𝐵∗h𝑉=

𝜋𝑛2

4∆ 𝑥

𝑉=𝜋4

(2 −2 𝑥2 )2 ∆𝑥

𝑇𝑉= lim𝑛→ ∞

∑ 𝜋4

(2 −2𝑥2 )2∆ 𝑥  

𝑇𝑉=∫ 𝜋4

(2 −2 𝑥2)2𝑑𝑥  

𝑇𝑉=𝜋4 ∫ ( 4 − 8𝑥2+4 𝑥4 )𝑑𝑥

The base of a solid is the region bounded by The cross sections, perpendicular to the x-axis, are rectangles whose height is 3x

4=𝑥2

± 2=𝑥

h=∆𝑥 [− 2,2 ]𝑛=4 −𝑥2

𝑉=𝐵∗h𝑉= [ ( 4 −𝑥2 ) 3𝑥 ] ∆ 𝑥

lim𝑛→ ∞

∑ [ (12 𝑥− 3𝑥3 ) ] ∆𝑥  

∫ [ (12𝑥−3 𝑥3 ) ]𝑑𝑥

Assignment:

P. 406 # 1 - 6 all If the problem has multiple parts work “ a “ then set up only ( to the definite integral) the other parts.

Volumes of Revolution: Disk and Washer Method

AP Calculus

Volume of Revolution: Method Lengths of Segments:In revolving solids about a line, the lengths of several segments are needed for the radii of disks, washers, and for the heights of cylinders.A). DISKS AND WASHERS 1) Shade the region in the first quadrant (to be rotated) 2) Indicate the line the region is to be revolved about.

3) Sketch the solid when the region is rotated about the indicated line.4) Draw the representative radii, its disk or washer and give their lengths.<<REM: Length must be positive! Top – Bottom or Right – Left >>

Ro = outer radius

ri = inner radius

Disk MethodRotate the region bounded by f(x) = 4 – x2 in the first quadrant about the y - axis

The region is _______________ _______ the axis of rotation.

The Formula: The formula is based on the

_____________________________________________

adjacent to

Volume of a cylinder

𝑉=𝜋 𝑟2 h

h=∆ 𝑦 [ 0,4 ]𝑟=√4 −𝑦− 0

Right-left

𝑦=4 −𝑥2

4 − 𝑦=𝑥2

±√4 −𝑦=𝑥

𝜋∫ (√4 −𝑦 )2𝑑𝑦

Washer MethodRotate the region bounded by f(x) = x2, x = 2 , and y = 0 about the y - axis

The region is _______________ __________ the axis of rotation.

The Formula: The formula is based on

_____________________________________________

Separate from

Big cylinder-small cylinder𝑉=𝜋 𝑅❑

2 h −𝜋𝑟 2h𝜋 (𝑅2 −𝑟2 ) h

lim𝑛→ ∞

∑ 𝜋 ( (2 )2− (√𝑦 )2 )∆ 𝑦h=∆ 𝑦 [ 0,4 ]𝑅=2 −0𝑟=√𝑦− 0

𝜋∫0

4

( 4 − 𝑦 )𝑑𝑦

𝑉=8𝜋

Disk MethodRotate the region bounded by f(x) = 2x – 2 , x = 4 , and y = 0 about the line x = 4

The region is _______________ _______ the axis of rotation.Adjacent to

𝐷𝑖𝑠𝑘𝜋 𝑟2 h 𝑦=2 𝑥−2

𝑥=𝑦+2

2∆ 𝑦 [ 0,6 ]

𝑟=4 −( 𝑦+22 )

𝑟=4 − 12 𝑦− 1

𝑟=(3 − 12𝑦 )

𝑉=13 𝜋𝑟2 h

𝑉=13 𝜋 (3 )2 (6 )

𝑉=18𝜋

𝜋∫0

6

(3− 12𝑦 )

2

𝑑𝑦

Washer MethodRotate the region bounded by f(x) = -2x + 10 , x = 2 , and y = 0 about the y - axis

The region is _______________ __________ the axis of rotation.Separate from

𝑦=− 2𝑥+10

𝑥=5− 12 𝑦

𝑅=5 − 12 𝑦

𝑟=2

h=∆ 𝑦 [ 0,6 ]𝜋 (𝑅2 −𝑟 2 ) h

lim𝑛→ ∞

∑ ((5 − 12𝑦 )

2

− (2 )2)∆ 𝑦

V=

Separate

Example 1: The region is bounded by Rotated about:

the x-axis, and the y-axis a) The x-axis

b) The y-axis

c) x = 3

d) y = 4

24y x

Example 2:The region is bounded by: Rotated about:

f(x) = x and g(x) = x2 a) the x-axis in the first quadrant b) the y-axis

c) x = 2d) y = 2

Example 2: The base of a solid is the region in the x-y plane bounded by the graph

and the x – axis. Find the volume of the solid if every cross section by

a plane perpendicular to the x-axis is an Isosceles Rt. Triangle (leg on the base).

2 3x y