7 2 Volume 01

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Transcript 7 2 Volume 01

7-2: Volume
Objectives:
Assignment:
1. To find the volume of a
solid of revolution using
disks and washers
β€’ P. 463-465: 1-11 odd, 12,
23, 27, 33, 45, 49, 53, 55
2. To find the volume of a
solid with known cross
sections
β€’ P. 465-466: 61-63
β€’ Homework Supplement
Warm Up 1
Find the volume of
the given figure in
cubic units.
Cavalieri’s Principle
If two solids have the same height and the
same cross-sectional area at every level,
then they have the same volume.
All 3 of these shapes have the same volume.
𝑉 = π΅β„Ž
Warm Up 2
Draw and name the 3-D solid of revolution
formed by revolving the given 2-D shape
around the y-axis.
Solids of Revolution
Draw and name the 3-D solid of revolution
formed by revolving the given 2-D shape
around the y-axis.
Sphere
Hemisphere
Torus
Warm Up 3
Draw and name the
3-D solid of
revolution formed
by revolving the
given 2-D shape
around the
indicated axis.
What is its volume?
π‘Ÿ
β„Ž
Solids of Revolution
Draw and name the
3-D solid of
revolution formed
by revolving the
given 2-D shape
around the
indicated axis.
What is its volume?
Cylinder
𝑉 = πœ‹π‘Ÿ 2 β„Ž
Objective 1
You will be able to find the
volume of a solid of revolution
using disks and washers
Disk Method
Much like finding the
area between two
curves, to find the
volume of a solid of
revolution, we first
consider a representative
rectangle.
Disk Method
Revolving this
rectangle around an
axis generates a
disk.
Volume of Disk:
𝑉 = πœ‹π‘Ÿ 2 β„Ž
𝑉 = πœ‹π‘…2 βˆ†π‘₯
𝑅 is a function of π‘₯
Disk Method
We can approximate
the volume of the
solid by summing 𝑛
such disks in a
Riemann sum.
𝑛
π‘‰β‰ˆ
πœ‹ 𝑅 π‘₯𝑖
𝑖=1
2 βˆ†π‘₯
Disk Method
Taking the limit as
𝑛 β†’ ∞ of this sum
yields the exact
volume of the solid.
𝑛
𝑉 = lim
π‘›β†’βˆž
πœ‹ 𝑅 π‘₯𝑖
𝑖=1
𝑏
𝑉=
πœ‹π‘… π‘₯
π‘Ž
2
𝑑π‘₯
2 βˆ†π‘₯
Disk Method
Taking the limit as
𝑛 β†’ ∞ of this sum
yields the exact
volume of the solid.
𝑛
𝑉 = lim
π‘›β†’βˆž
πœ‹ 𝑅 π‘₯𝑖
𝑖=1
𝑏
𝑉=πœ‹
𝑅 π‘₯
π‘Ž
2
𝑑π‘₯
2 βˆ†π‘₯
Vertical vs. Horizontal Disks
Depending on the axis of rotation, you may
have vertical or horizontal disks.
Exercise 1
Find the volume of the
solid formed by
revolving the region
bounded by 𝑦 = sin π‘₯
and the π‘₯-axis on the
interval 0, πœ‹ about the
π‘₯-axis.
Exercise 1
Find the volume of the
solid formed by
revolving the region
bounded by 𝑦 = sin π‘₯
and the π‘₯-axis on the
interval 0, πœ‹ about the
π‘₯-axis.
Exercise 2
Find the volume of the solid formed by
revolving the region bounded by 𝑦 = π‘₯ 2 and
the π‘₯-axis on the interval 0,2 about the
π‘₯-axis.
Exercise 3
Find the volume of the solid formed by
1
revolving the region bounded by 𝑦 = π‘₯ βˆ’ 1,
2
𝑦 = 0, 𝑦 = 2, and π‘₯ = 0 about the 𝑦-axis.
Exercise 4
Find the volume of the
solid formed by revolving
the region bounded by
𝑓 π‘₯ = 2 βˆ’ π‘₯ 2 and
𝑔 π‘₯ = 1 about the line
𝑦 = 1.
Exercise 4
Find the volume of the
solid formed by revolving
the region bounded by
𝑓 π‘₯ = 2 βˆ’ π‘₯ 2 and
𝑔 π‘₯ = 1 about the line
𝑦 = 1.
Exercise 5
Draw and name the
3-D solid of
revolution formed
by revolving the
given 2-D shape
about the indicated
axis. What is its
volume?
Exercise 5
Draw and name the
3-D solid of
revolution formed
by revolving the
given 2-D shape
about the indicated
axis. What is its
volume?
Volume of Washer:
𝑉 = πœ‹π‘…2 𝑀 βˆ’ πœ‹π‘Ÿ 2 𝑀
𝑉 = πœ‹ 𝑅2 βˆ’ π‘Ÿ 2 𝑀
Outer Radius
Inner radius
Washer Method
Sometimes
revolving a 2-D
shape around an
axis produces a
solid with a hole.
As before, we still
need to consider a
representative
rectangle.
Washer Method
However, rotating
this rectangle about
an axis produces a
washer.
𝑉 = πœ‹ 𝑅2 βˆ’ π‘Ÿ 2 βˆ†π‘₯
Washer Method
The volume can be
approximated by
summing 𝑛 such
washers.
𝑛
π‘‰β‰ˆ
πœ‹ 𝑅 π‘₯𝑖
𝑖=1
2
βˆ’ π‘Ÿ π‘₯𝑖
2
βˆ†π‘₯
Washer Method
Taking the limit as
𝑛 β†’ ∞ yields the
volume of the solid.
𝑛
𝑉 = lim
π‘›β†’βˆž
πœ‹ 𝑅 π‘₯𝑖
𝑖=1
2
βˆ’ π‘Ÿ π‘₯𝑖
2
βˆ†π‘₯
Washer Method
Taking the limit as
𝑛 β†’ ∞ yields the
volume of the solid.
Outer Radius
Inner radius
𝑏
𝑉=
πœ‹ 𝑅 π‘₯
π‘Ž
2
βˆ’ π‘Ÿ π‘₯
2
𝑑π‘₯
Exercise 6
Find the volume of
the solid formed by
revolving the region
bounded by the
graphs of 𝑦 = π‘₯
and 𝑦 = π‘₯ 2 about
the π‘₯-axis.
Exercise 6
Find the volume of
the solid formed by
revolving the region
bounded by the
graphs of 𝑦 = π‘₯
and 𝑦 = π‘₯ 2 about
the π‘₯-axis.
Exercise 7
Find the volume of
the solid formed by
revolving the region
bounded by the
graphs of 𝑦 = π‘₯ 2 + 1,
𝑦 = 0, π‘₯ = 0, and
π‘₯ = 1 about the 𝑦axis.
Exercise 7
Find the volume of
the solid formed by
revolving the region
bounded by the
graphs of 𝑦 = π‘₯ 2 + 1,
𝑦 = 0, π‘₯ = 0, and
π‘₯ = 1 about the 𝑦axis.
Objective 2
You will be able to find the volume
of a solid with known cross
sections
Other Cross Sections
Finding the volume of a solid of revolution
can be extended to include solids of any
similar cross sectional shape, assuming you
can find its area.
Common Cross
Sections:
Squares
Rectangles
Triangles
Trapezoids
Semicircles
Other Cross Sections
To find the volume of these solids, find 𝐴 π‘₯ ,
the area of one cross section. Multiplying
this by its width βˆ†π‘₯ gives the volume of one
cross section.
Other Cross Sections
The volume can
then be
approximated by
summing 𝑛 of
these cross
sections in a
Riemann sum.
𝑛
π‘‰β‰ˆ
𝐴 π‘₯𝑖 βˆ†π‘₯
𝑖=1
Other Cross Sections
Finally, taking the
limit as 𝑛 β†’ ∞
yields the exact
volume of the
solid.
𝑛
𝑉 = lim
π‘›β†’βˆž
𝐴 π‘₯𝑖 βˆ†π‘₯
𝑖=1
𝑏
𝑉=
𝐴 π‘₯ 𝑑π‘₯
π‘Ž
Exercise 8
Consider a region 𝑅
bounded below by the
curve 𝑓 π‘₯ = π‘₯ 2 βˆ’ 1
and above by the
curve 𝑔 π‘₯ = π‘₯ + 1.
Find the area of 𝑅.
Consider a region 𝑅
bounded below by the
curve 𝑓 π‘₯ = π‘₯ 2 βˆ’ 1 and
above by the curve 𝑔 π‘₯ =
π‘₯ + 1. Region 𝑅 is the
base of a solid. Find the
volume of the solid formed
by square cross sections
perpendicular to the π‘₯axis.
𝑔 π‘₯ βˆ’π‘“ π‘₯
Exercise 9
𝑔 π‘₯ βˆ’π‘“ π‘₯
Exercise 10
Consider a region 𝑅
bounded below by the
curve 𝑓 π‘₯ = π‘₯ 2 βˆ’ 1 and
above by the curve 𝑔 π‘₯ =
π‘₯ + 1. Region 𝑅 is the base
of a solid. Find the volume
of the solid formed by
semicircular cross sections
perpendicular to the π‘₯-axis.
1
𝑔 π‘₯ βˆ’π‘“ π‘₯
2
Consider a region 𝑅
bounded below by the
curve 𝑓 π‘₯ = π‘₯ 2 βˆ’ 1 and
above by the curve 𝑔 π‘₯ =
π‘₯ + 1. Region 𝑅 is the
base of a solid. Find the
volume of the solid formed
by equilateral triangular
cross sections
perpendicular to the π‘₯-axis.
3
𝑔 π‘₯ βˆ’π‘“ π‘₯
2
Exercise 11
𝑔 π‘₯ βˆ’π‘“ π‘₯
Consider a region 𝑆
bounded by 𝑦 = 𝑒 π‘₯
and 𝑦 = π‘₯ + 2.
Region 𝑆 is the base
of a solid. Find the
volume of the solid
formed by squares
cross sections
perpendicular to the
𝑦-axis.
ln 𝑦 βˆ’ 𝑦 βˆ’ 2
Exercise 12
ln 𝑦 βˆ’ 𝑦 βˆ’ 2
Exercise 13: AP FRQ
Let 𝑓 and 𝑔 be functions
defined by
2 βˆ’2π‘₯
π‘₯
𝑓 π‘₯ =1+π‘₯+𝑒
and
𝑔 π‘₯ = π‘₯ 4 βˆ’ 6.5π‘₯ 2 + 6π‘₯ + 2.
Let 𝑅 and 𝑆 be the two
regions enclosed by the
graphs of 𝑓 and 𝑔 shown in
the figure.
Region 𝑆 is the base
of a solid whose
cross sections
perpendicular to the
π‘₯-axis are squares.
Find the volume of
the solid.
Exercise 14: AP FRQ
Let 𝑓 and 𝑔 be functions
defined by
2 βˆ’2π‘₯
π‘₯
𝑓 π‘₯ =1+π‘₯+𝑒
and
𝑔 π‘₯ = π‘₯ 4 βˆ’ 6.5π‘₯ 2 + 6π‘₯ + 2.
Let 𝑅 and 𝑆 be the two
regions enclosed by the
graphs of 𝑓 and 𝑔 shown in
the figure.
Let β„Ž be the vertical
distance between
the graphs of 𝑓 and
𝑔 in the region 𝑆.
Find the rate at
which β„Ž changes
with respect to π‘₯
when π‘₯ = 1.8.
Exercise 15: AP FRQ
Let 𝑓 and 𝑔 be functions
defined by
2 βˆ’2π‘₯
π‘₯
𝑓 π‘₯ =1+π‘₯+𝑒
and
𝑔 π‘₯ = π‘₯ 4 βˆ’ 6.5π‘₯ 2 + 6π‘₯ + 2.
Let 𝑅 and 𝑆 be the two
regions enclosed by the
graphs of 𝑓 and 𝑔 shown in
the figure.
Find the volume of
the solid formed by
rotating region 𝑅
about the π‘₯-axis.
7-2: Volume
Objectives:
Assignment:
1. To find the volume of a
solid of revolution using
disks and washers
β€’ P. 463-465: 1-11 odd, 12,
23, 27, 33, 45, 49, 53, 55
2. To find the volume of a
solid with known cross
sections
β€’ P. 465-466: 61-63
β€’ Homework Supplement