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.
Cavalieriβs 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.
π = π΅β
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
β
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
π = ππ 2 β
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:
π = ππ 2 β
π = ππ
2 βπ₯
π
is a function of π₯
Disk Method
We can approximate
the volume of the
solid by summing π
such disks in a
Riemann sum.
π
πβ
π π
π₯π
π=1
2 βπ₯
Disk Method
Taking the limit as
π β β of this sum
yields the exact
volume of the solid.
π
π = lim
πββ
π π
π₯π
π=1
π
π=
ππ
π₯
π
2
ππ₯
2 βπ₯
Disk Method
Taking the limit as
π β β of this sum
yields the exact
volume of the solid.
π
π = lim
πββ
π π
π₯π
π=1
π
π=π
π
π₯
π
2
ππ₯
2 βπ₯
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 π¦ = sin π₯
and the π₯-axis on the
interval 0, π about the
π₯-axis.
Exercise 1
Find the volume of the
solid formed by
revolving the region
bounded by π¦ = sin π₯
and the π₯-axis on the
interval 0, π about the
π₯-axis.
Exercise 2
Find the volume of the solid formed by
revolving the region bounded by π¦ = 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 π¦ = π₯ β 1,
2
π¦ = 0, π¦ = 2, and π₯ = 0 about the π¦-axis.
Exercise 4
Find the volume of the
solid formed by revolving
the region bounded by
π π₯ = 2 β π₯ 2 and
π π₯ = 1 about the line
π¦ = 1.
Exercise 4
Find the volume of the
solid formed by revolving
the region bounded by
π π₯ = 2 β π₯ 2 and
π π₯ = 1 about the line
π¦ = 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:
π = ππ
2 π€ β ππ 2 π€
π = π π
2 β π 2 π€
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.
π = π π
2 β π 2 βπ₯
Washer Method
The volume can be
approximated by
summing π such
washers.
π
πβ
π π
π₯π
π=1
2
β π π₯π
2
βπ₯
Washer Method
Taking the limit as
π β β yields the
volume of the solid.
π
π = lim
πββ
π π
π₯π
π=1
2
β π π₯π
2
βπ₯
Washer Method
Taking the limit as
π β β yields the
volume of the solid.
Outer Radius
Inner radius
π
π=
π π
π₯
π
2
β π π₯
2
ππ₯
Exercise 6
Find the volume of
the solid formed by
revolving the region
bounded by the
graphs of π¦ = π₯
and π¦ = π₯ 2 about
the π₯-axis.
Exercise 6
Find the volume of
the solid formed by
revolving the region
bounded by the
graphs of π¦ = π₯
and π¦ = π₯ 2 about
the π₯-axis.
Exercise 7
Find the volume of
the solid formed by
revolving the region
bounded by the
graphs of π¦ = π₯ 2 + 1,
π¦ = 0, π₯ = 0, and
π₯ = 1 about the π¦axis.
Exercise 7
Find the volume of
the solid formed by
revolving the region
bounded by the
graphs of π¦ = π₯ 2 + 1,
π¦ = 0, π₯ = 0, and
π₯ = 1 about the π¦axis.
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 π΄ π₯ ,
the area of one cross section. Multiplying
this by its width βπ₯ gives the volume of one
cross section.
Other Cross Sections
The volume can
then be
approximated by
summing π of
these cross
sections in a
Riemann sum.
π
πβ
π΄ π₯π βπ₯
π=1
Other Cross Sections
Finally, taking the
limit as π β β
yields the exact
volume of the
solid.
π
π = lim
πββ
π΄ π₯π βπ₯
π=1
π
π=
π΄ π₯ ππ₯
π
Exercise 8
Consider a region π
bounded below by the
curve π π₯ = π₯ 2 β 1
and above by the
curve π π₯ = π₯ + 1.
Find the area of π
.
Consider a region π
bounded below by the
curve π π₯ = π₯ 2 β 1 and
above by the curve π π₯ =
π₯ + 1. Region π
is the
base of a solid. Find the
volume of the solid formed
by square cross sections
perpendicular to the π₯axis.
π π₯ βπ π₯
Exercise 9
π π₯ βπ π₯
Exercise 10
Consider a region π
bounded below by the
curve π π₯ = π₯ 2 β 1 and
above by the curve π π₯ =
π₯ + 1. Region π
is the base
of a solid. Find the volume
of the solid formed by
semicircular cross sections
perpendicular to the π₯-axis.
1
π π₯ βπ π₯
2
Consider a region π
bounded below by the
curve π π₯ = π₯ 2 β 1 and
above by the curve π π₯ =
π₯ + 1. Region π
is the
base of a solid. Find the
volume of the solid formed
by equilateral triangular
cross sections
perpendicular to the π₯-axis.
3
π π₯ βπ π₯
2
Exercise 11
π π₯ βπ π₯
Consider a region π
bounded by π¦ = π π₯
and π¦ = π₯ + 2.
Region π is the base
of a solid. Find the
volume of the solid
formed by squares
cross sections
perpendicular to the
π¦-axis.
ln π¦ β π¦ β 2
Exercise 12
ln π¦ β π¦ β 2
Exercise 13: AP FRQ
Let π and π be functions
defined by
2 β2π₯
π₯
π π₯ =1+π₯+π
and
π π₯ = π₯ 4 β 6.5π₯ 2 + 6π₯ + 2.
Let π
and π be the two
regions enclosed by the
graphs of π and π shown in
the figure.
Region π 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 π and π be functions
defined by
2 β2π₯
π₯
π π₯ =1+π₯+π
and
π π₯ = π₯ 4 β 6.5π₯ 2 + 6π₯ + 2.
Let π
and π be the two
regions enclosed by the
graphs of π and π shown in
the figure.
Let β be the vertical
distance between
the graphs of π and
π in the region π.
Find the rate at
which β changes
with respect to π₯
when π₯ = 1.8.
Exercise 15: AP FRQ
Let π and π be functions
defined by
2 β2π₯
π₯
π π₯ =1+π₯+π
and
π π₯ = π₯ 4 β 6.5π₯ 2 + 6π₯ + 2.
Let π
and π be the two
regions enclosed by the
graphs of π and π shown in
the figure.
Find the volume of
the solid formed by
rotating region π
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