§ 14.4 Surface Areas
Surface Area
Definition
The surface area of a solid figure is the sum of all of the areas of the
faces, whether polygons or curved figures.
Surface Area
Definition
The surface area of a solid figure is the sum of all of the areas of the
faces, whether polygons or curved figures.
Definition
The lateral surface area of a solid figure is the sum of the areas of the
faces, not including the bases.
Prisms
What is the formula for the surface area of a right prism? Can we
generalize the formula so that it applies to all right prisms?
Prisms
What is the formula for the surface area of a right prism? Can we
generalize the formula so that it applies to all right prisms?
Formula
Right Prism
The surface area of a right prism is 2AB + PB h.
Pyramids
Can we do similar for right pyramids?
Pyramids
Can we do similar for right pyramids?
Formula
Right Pyramid
The surface area of a right pyramid is AB + 12 PB l.
Cylinders
We said cylinders are not polyhedra. Why?
Cylinders
We said cylinders are not polyhedra. Why?
Can we generalize the surface area formula of the right cylinder?
Cylinders
We said cylinders are not polyhedra. Why?
Can we generalize the surface area formula of the right cylinder?
Formula
Right Circular Cylinder
The surface area of a right circular cylinder is 2πr2 + 2πrh.
Cones
And finally, our cones. Can we do the same for these?
Cones
And finally, our cones. Can we do the same for these?
Formula
Right Circular Cone
The surface area of a right circular cone is πr2 + πrl where l is the
slant height.
Justification for the Cone Formula
There are two parts to the formula
Base
πr2
Justification for the Cone Formula
There are two parts to the formula
Base
πr2
This is the area of a circle.
Lateral Face
πrl
Justification for the Cone Formula
There are two parts to the formula
Base
πr2
This is the area of a circle.
Lateral Face
πrl
If the lateral face was a whole circle, we would have an area of πl2 .
But ... we don’t. So what part do we have?
Justification for the Cone Formula
There are two parts to the formula
Base
πr2
This is the area of a circle.
Lateral Face
πrl
If the lateral face was a whole circle, we would have an area of πl2 .
But ... we don’t. So what part do we have?
πl2 ×
2πr
r
= πl2 × = πrl
2πl
l
Sphere
Formula
Sphere
The surface area of a sphere is 4πr2 .
Sphere
Formula
Sphere
The surface area of a sphere is 4πr2 .
Lateral SAcylinder = 2πrh = 2πr × 2r = 4πr2
Surface Area of a Cone
Example
Find the surface area of the cone if the net corresponding to the lateral
surface area of the cone has a radius of 3 inches and this portion of the
net has a central angle of 180◦ .
Surface Area of a Cone
Example
Find the surface area of the cone if the net corresponding to the lateral
surface area of the cone has a radius of 3 inches and this portion of the
net has a central angle of 180◦ .
This is
1
2
of a circle with radius 3 inches, so the lateral surface area is
1
9
π(3)2 = π in2
2
2
Base of the Cone
Now the dimensions of the base isn’t given. We know it is a circle,
and we know it’s circumference must be as long as the rolled up
lateral face. So, the circumference must be
Base of the Cone
Now the dimensions of the base isn’t given. We know it is a circle,
and we know it’s circumference must be as long as the rolled up
lateral face. So, the circumference must be
1
2π(3) = 3π in
2
Base of the Cone
Now the dimensions of the base isn’t given. We know it is a circle,
and we know it’s circumference must be as long as the rolled up
lateral face. So, the circumference must be
1
2π(3) = 3π in
2
Now, we need the radius of the base so that we can find the area.
Base of the Cone
Now the dimensions of the base isn’t given. We know it is a circle,
and we know it’s circumference must be as long as the rolled up
lateral face. So, the circumference must be
1
2π(3) = 3π in
2
Now, we need the radius of the base so that we can find the area.
3π = 2πr ⇒ r =
3
in
2
Base of the Cone
Now the dimensions of the base isn’t given. We know it is a circle,
and we know it’s circumference must be as long as the rolled up
lateral face. So, the circumference must be
1
2π(3) = 3π in
2
Now, we need the radius of the base so that we can find the area.
3π = 2πr ⇒ r =
Therefore, the base has an area of
3
in
2
Base of the Cone
Now the dimensions of the base isn’t given. We know it is a circle,
and we know it’s circumference must be as long as the rolled up
lateral face. So, the circumference must be
1
2π(3) = 3π in
2
Now, we need the radius of the base so that we can find the area.
3π = 2πr ⇒ r =
3
in
2
Therefore, the base has an area of
2
3
9
π
= π in2
2
4
Base of the Cone
Now the dimensions of the base isn’t given. We know it is a circle,
and we know it’s circumference must be as long as the rolled up
lateral face. So, the circumference must be
1
2π(3) = 3π in
2
Now, we need the radius of the base so that we can find the area.
3π = 2πr ⇒ r =
3
in
2
Therefore, the base has an area of
2
3
9
π
= π in2
2
4
So, the total surface area is
9
9
27
π in2 + π in2 = π in2
2
4
4
The Cone and the Cylinder
Example
Find the surface area of the cylinder inscribed in a cone with height 20
inches and base with radius 10 inches if the cylinder is 10 inches tall.
The Cone and the Cylinder
Example
Find the surface area of the cylinder inscribed in a cone with height 20
inches and base with radius 10 inches if the cylinder is 10 inches tall.
10
r
=
⇒ r = 5 inches
20
10
The Cone and the Cylinder
Example
Find the surface area of the cylinder inscribed in a cone with height 20
inches and base with radius 10 inches if the cylinder is 10 inches tall.
10
r
=
⇒ r = 5 inches
20
10
So, a cylinder with radius 5 inches and height 10 inches has surface
area
The Cone and the Cylinder
Example
Find the surface area of the cylinder inscribed in a cone with height 20
inches and base with radius 10 inches if the cylinder is 10 inches tall.
10
r
=
⇒ r = 5 inches
20
10
So, a cylinder with radius 5 inches and height 10 inches has surface
area
2πr2 + 2πrh = 2π(5)2 + 2π(5)(10) = 150π in2
Surface Area of a Ring
Example
Suppose you have a ring that you want to get replated. The jeweler
needs to know the surface area of the ring in order to determine how
much to charge you. If the ring has an inner radius of 25 mm and the
ring is 10 mm thick and is 10 mm high, find the surface area.
Surface Area of a Ring
Example
Suppose you have a ring that you want to get replated. The jeweler
needs to know the surface area of the ring in order to determine how
much to charge you. If the ring has an inner radius of 25 mm and the
ring is 10 mm thick and is 10 mm high, find the surface area.
Inner ring: We pretend we have a full cylinder and are finding it’s
surface area.
Surface Area of a Ring
Example
Suppose you have a ring that you want to get replated. The jeweler
needs to know the surface area of the ring in order to determine how
much to charge you. If the ring has an inner radius of 25 mm and the
ring is 10 mm thick and is 10 mm high, find the surface area.
Inner ring: We pretend we have a full cylinder and are finding it’s
surface area.
SAlat = 2πrh = 2π(25)(10) = 500π mm2
Surface Area of a Ring
Example
Suppose you have a ring that you want to get replated. The jeweler
needs to know the surface area of the ring in order to determine how
much to charge you. If the ring has an inner radius of 25 mm and the
ring is 10 mm thick and is 10 mm high, find the surface area.
Inner ring: We pretend we have a full cylinder and are finding it’s
surface area.
SAlat = 2πrh = 2π(25)(10) = 500π mm2
Outer ring: Again, we pretend we have a full cylinder.
SAlat = 2πrh = 2π(35)(10) = 700π mm2
Surface Area of a Ring
Example
Suppose you have a ring that you want to get replated. The jeweler
needs to know the surface area of the ring in order to determine how
much to charge you. If the ring has an inner radius of 25 mm and the
ring is 10 mm thick and is 10 mm high, find the surface area.
Inner ring: We pretend we have a full cylinder and are finding it’s
surface area.
SAlat = 2πrh = 2π(25)(10) = 500π mm2
Outer ring: Again, we pretend we have a full cylinder.
SAlat = 2πrh = 2π(35)(10) = 700π mm2
Now the circles on top and bottom ...
Solution (cont.)
Outer circle: The outer radius is 35 mm, so the area is
Solution (cont.)
Outer circle: The outer radius is 35 mm, so the area is
π(35)2 = 1225π mm2
The inner radius is 25 mm, so the area is
Solution (cont.)
Outer circle: The outer radius is 35 mm, so the area is
π(35)2 = 1225π mm2
The inner radius is 25 mm, so the area is
π(25)2 = 625π mm2
So, the area of the disc is
Solution (cont.)
Outer circle: The outer radius is 35 mm, so the area is
π(35)2 = 1225π mm2
The inner radius is 25 mm, so the area is
π(25)2 = 625π mm2
So, the area of the disc is
1225π mm2 − 625π mm2 = 600 mm2
Putting these together we see
SA = 2 × 600 mm2 + 700 mm2 + 500 mm2 = 2400 mm2