6049 Volume
AP Calculus

Volume
Volume = the sum of the quantities in each layer 𝑙 ∗ 𝑤 ∗ ℎ where h is # layers

 x y x x
x

Volume by Cross Sections
 x
 n 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

Example 1:
The base of a solid is the region in the x-y plane bounded by the graph x
 y
2 
3 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.
Base y 𝑦 2 = 3 − 𝑥
 𝑦 = 3 − 𝑥 𝑦 = ± 3 − 𝑥
 x

3
𝑇𝑉 = lim 𝑛→∞
0
2 3 − 𝑥
2
∆𝑥
3
2 3 − 𝑥
2 𝑑𝑥
0
 𝑦 = − 3 − 𝑥 h = ∆𝑥 0,3
Cross -Section 𝑛 = 3 − 𝑥 − − 3 − 𝑥
𝑉 = 𝐵 ∗ ℎ
𝑉 = 𝑛 2 ∆𝑥
𝑉 = 2 3 − 𝑥
2
∆𝑥 𝑛 = 2 3 − 𝑥

Example 2: The base of a solid is the region in the x-y plane bounded by the graph x
 y
2 
3 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).
y 𝑦 = 3 − 𝑥

3
𝑇𝑉 = lim 𝑛→∞
0
3
1
2
0
1
2
2 3 − 𝑥
2 3 − 𝑥
2
2 𝑑𝑥
∆𝑥
 x

 𝑦 = − 3 − 𝑥 n h = ∆𝑥 0,3 𝑛 = 3 − 𝑥 − − 3 − 𝑥 𝑛 = 2 3 − 𝑥 n
𝑉 = 𝐵 ∗ ℎ
𝑉 =
1
2 𝑛 2 ∆𝑥

Some Important Area Formulas
Square-
Squaren side on base
𝐵 = 𝑛 2
Isosceles rt Δ n
𝐵 =
1
2 𝑛 2 leg on base
Semi - Circle
𝐵 =
1
2 𝑛
2
2 𝜋 n
𝐵 = 𝜋𝑟
2
8
Isosceles rt Δ 𝑛 2 n
𝐵 = hypotenuse on base
4 n n 𝐵 = 𝑛 2
2 diagonal on base
Equilateral Δ 𝑛
3
2 1
𝐵 =
2 n 𝑛 𝑛 2
𝐵 = 𝑛
2
4
3
3
Circle
𝐵 = 𝜋𝑟 2
4 diameter on the base

EXAMPLE #4/406
The 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 = x 2 to the parabola y = 2 – x 2 .
𝑇𝑉 = lim 𝑛→∞ 𝜋
4
𝑇𝑉 = 𝜋
4
2 − 2𝑥
2 2
∆𝑥
2 − 2𝑥 2 2 𝑑𝑥 𝑥 2 = 2 − 𝑥 2
2𝑥 2 𝑥 2
= 2
= 1 𝑥 ± 1
Type equation here.
𝑏 = ∆𝑥 −1,1
𝑇𝑉 = 𝜋
4
4 − 8𝑥
2
+ 4𝑥
4 𝑑𝑥 n 𝑛 = 2 − 𝑥 2 − 𝑥 2 𝑛 = 2 − 2𝑥 2
𝑉 = 𝐵 ∗ ℎ
𝑉 = 𝜋𝑛 2
∆𝑥
4
𝑉 = 𝜋
4
2 − 2𝑥 2 2 ∆𝑥

The base of a solid is the region bounded by 𝑦 = 𝑥 2 𝑎𝑛𝑑 𝑦 = 4.
The cross sections, perpendicular to the xaxis, are rectangles whose height is 3x lim 𝑛→∞
12𝑥 − 3𝑥 3 ∆𝑥
4 = 𝑥 2
±2 = 𝑥
12𝑥 − 3𝑥 3 𝑑𝑥
𝑉 = 𝐵 ∗ ℎ
𝑉 = 4 − 𝑥 2 3𝑥 ∆𝑥 ℎ = ∆𝑥 −2,2 𝑛 = 4 − 𝑥 2

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

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 >>
R o
= outer radius r i
= inner radius

Rotate the region bounded by f(x) = 4 – x 2 i n the first quadrant about the y - axis
The Formula: The formula is based on the
𝑉 = 𝜋𝑟 2 ℎ ℎ = ∆𝑦 0,4 𝑟 = 4 − 𝑦 − 0
Right-left 𝑦 = 4 − 𝑥 2
4 − 𝑦 = 𝑥 2
± 4 − 𝑦 = 𝑥 𝜋 4 − 𝑦
2 𝑑𝑦

Rotate the region bounded by f(x) = x 2 , x = 2 , and y = 0 about the y - axis
Separate from
The region is _______________ __________ the axis of rotation.
The Formula: The formula is based on ℎ = ∆𝑦 0,4
𝑅 = 2 − 0 𝑟 = 𝑦 − 0
𝑉 = 𝜋𝑅
2 ℎ − 𝜋𝑟
2 ℎ 𝜋 𝑅 2 − 𝑟 2 ℎ lim 𝑛→∞
𝜋 2
4
2 − 𝑦 2 ∆𝑦 𝜋 4 − 𝑦 𝑑𝑦
0
𝑉 = 8𝜋

Rotate the region bounded by f(x) = 2x – 2 , x = 4 , and y = 0 about the line x = 4
Adjacent to
The region is _______________ _______ the axis of rotation.
𝑦 = 2𝑥 − 2
𝐷𝑖𝑠𝑘 𝜋𝑟 2 ℎ
∆𝑦 0,6 𝑥 = 𝑦 + 2
2 𝑟 = 4 − 𝑦 + 2
2
𝑉 =
1
3 𝜋𝑟 2 ℎ 𝑟 = 4 −
1
2 𝑦 − 1
𝑉 =
1
3 𝜋 3 2 6 𝑟 = 3 −
1
2 𝑦
𝑉 = 18𝜋
6 𝜋
0
3 −
1
2 𝑦
2 𝑑𝑦

Rotate the region bounded by f(x) = -2x + 10 , x = 2 , and y = 0 about the y - axis
Separate ∴ 𝑤𝑎𝑠ℎ𝑒𝑟 𝑦 = −2𝑥 + 10 𝑥 = 5 −
1
2 𝑦
𝑅 = 5 −
1
2 𝑦 ℎ = ∆𝑦 0,6 𝜋 𝑅 2 − 𝑟 2 ℎ lim 𝑛→∞
5 −
1
2 𝑦
2
− 2 2 𝑟 = 2
V= π
0
6
∆𝑦
5 −
1
2 𝑦
2
− 2 2 𝑑𝑦

Example 1:
The region is bounded by y the x-axis , and the y-axis x
2
Rotated about: a) The x-axis b) The y-axis c) x = 3 d) y = 4

Example 2:
The region is bounded by: f(x) = x and g(x) = x 2 in the first quadrant
Rotated about: a) the x-axis b) the y-axis c) x = 2 d) y = 2

Example 2: The base of a solid is the region in the x-y plane bounded by the graph x
 y
2 
3 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).