14.4 surface areasbtravers.weebly.com/uploads/6/7/2/9/6729909/surface_area_slides.pdfsurface area of...
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§ 14.4 Surface Areas
Surface Area
DefinitionThe surface area of a solid figure is the sum of all of the areas of thefaces, whether polygons or curved figures.
DefinitionThe lateral surface area of a solid figure is the sum of the areas of thefaces, not including the bases.
Surface Area
DefinitionThe surface area of a solid figure is the sum of all of the areas of thefaces, whether polygons or curved figures.
DefinitionThe lateral surface area of a solid figure is the sum of the areas of thefaces, not including the bases.
Prisms
What is the formula for the surface area of a right prism? Can wegeneralize the formula so that it applies to all right prisms?
FormulaRight PrismThe surface area of a right prism is 2AB + PBh.
Prisms
What is the formula for the surface area of a right prism? Can wegeneralize the formula so that it applies to all right prisms?
FormulaRight PrismThe surface area of a right prism is 2AB + PBh.
Pyramids
Can we do similar for right pyramids?
FormulaRight PyramidThe surface area of a right pyramid is AB + 1
2 PBl.
Pyramids
Can we do similar for right pyramids?
FormulaRight PyramidThe surface area of a right pyramid is AB + 1
2 PBl.
Cylinders
We said cylinders are not polyhedra. Why?
Can we generalize the surface area formula of the right cylinder?
FormulaRight Circular CylinderThe surface area of a right circular cylinder is 2πr2 + 2πrh.
Cylinders
We said cylinders are not polyhedra. Why?
Can we generalize the surface area formula of the right cylinder?
FormulaRight Circular CylinderThe surface area of a right circular cylinder is 2πr2 + 2πrh.
Cylinders
We said cylinders are not polyhedra. Why?
Can we generalize the surface area formula of the right cylinder?
FormulaRight Circular CylinderThe surface area of a right circular cylinder is 2πr2 + 2πrh.
Cones
And finally, our cones. Can we do the same for these?
FormulaRight Circular ConeThe surface area of a right circular cone is πr2 + πrl where l is theslant height.
Cones
And finally, our cones. Can we do the same for these?
FormulaRight Circular ConeThe surface area of a right circular cone is πr2 + πrl where l is theslant height.
Justification for the Cone Formula
There are two parts to the formula
Base
πr2
This is the area of a circle.
Lateral Faceπrl
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 × 2πr2πl
= πl2 × rl= πrl
Justification for the Cone Formula
There are two parts to the formula
Base
πr2
This is the area of a circle.
Lateral Faceπrl
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 × 2πr2πl
= πl2 × rl= πrl
Justification for the Cone Formula
There are two parts to the formula
Base
πr2
This is the area of a circle.
Lateral Faceπrl
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 × 2πr2πl
= πl2 × rl= πrl
Justification for the Cone Formula
There are two parts to the formula
Base
πr2
This is the area of a circle.
Lateral Faceπrl
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 × 2πr2πl
= πl2 × rl= πrl
Sphere
FormulaSphereThe surface area of a sphere is 4πr2.
Lateral SAcylinder = 2πrh = 2πr × 2r = 4πr2
Sphere
FormulaSphereThe surface area of a sphere is 4πr2.
Lateral SAcylinder = 2πrh = 2πr × 2r = 4πr2
Surface Area of a Cone
ExampleFind the surface area of the cone if the net corresponding to the lateralsurface area of the cone has a radius of 3 inches and this portion of thenet has a central angle of 180◦.
This is 12 of a circle with radius 3 inches, so the lateral surface area is
12π(3)2 =
92π in2
Surface Area of a Cone
ExampleFind the surface area of the cone if the net corresponding to the lateralsurface area of the cone has a radius of 3 inches and this portion of thenet has a central angle of 180◦.
This is 12 of a circle with radius 3 inches, so the lateral surface area is
12π(3)2 =
92π in2
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 uplateral face. So, the circumference must be
12
2π(3) = 3π in
Now, we need the radius of the base so that we can find the area.
3π = 2πr ⇒ r =32
in
Therefore, the base has an area of
π
(32
)2
=94π in2
So, the total surface area is92π in2 +
94π in2 =
274π in2
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 uplateral face. So, the circumference must be
12
2π(3) = 3π in
Now, we need the radius of the base so that we can find the area.
3π = 2πr ⇒ r =32
in
Therefore, the base has an area of
π
(32
)2
=94π in2
So, the total surface area is92π in2 +
94π in2 =
274π in2
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 uplateral face. So, the circumference must be
12
2π(3) = 3π in
Now, we need the radius of the base so that we can find the area.
3π = 2πr ⇒ r =32
in
Therefore, the base has an area of
π
(32
)2
=94π in2
So, the total surface area is92π in2 +
94π in2 =
274π in2
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 uplateral face. So, the circumference must be
12
2π(3) = 3π in
Now, we need the radius of the base so that we can find the area.
3π = 2πr ⇒ r =32
in
Therefore, the base has an area of
π
(32
)2
=94π in2
So, the total surface area is92π in2 +
94π in2 =
274π in2
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 uplateral face. So, the circumference must be
12
2π(3) = 3π in
Now, we need the radius of the base so that we can find the area.
3π = 2πr ⇒ r =32
in
Therefore, the base has an area of
π
(32
)2
=94π in2
So, the total surface area is92π in2 +
94π in2 =
274π in2
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 uplateral face. So, the circumference must be
12
2π(3) = 3π in
Now, we need the radius of the base so that we can find the area.
3π = 2πr ⇒ r =32
in
Therefore, the base has an area of
π
(32
)2
=94π in2
So, the total surface area is92π in2 +
94π in2 =
274π in2
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 uplateral face. So, the circumference must be
12
2π(3) = 3π in
Now, we need the radius of the base so that we can find the area.
3π = 2πr ⇒ r =32
in
Therefore, the base has an area of
π
(32
)2
=94π in2
So, the total surface area is92π in2 +
94π in2 =
274π in2
The Cone and the Cylinder
ExampleFind the surface area of the cylinder inscribed in a cone with height 20inches and base with radius 10 inches if the cylinder is 10 inches tall.
1020
=r
10⇒ r = 5 inches
So, a cylinder with radius 5 inches and height 10 inches has surfacearea
2πr2 + 2πrh = 2π(5)2 + 2π(5)(10) = 150π in2
The Cone and the Cylinder
ExampleFind the surface area of the cylinder inscribed in a cone with height 20inches and base with radius 10 inches if the cylinder is 10 inches tall.
1020
=r
10⇒ r = 5 inches
So, a cylinder with radius 5 inches and height 10 inches has surfacearea
2πr2 + 2πrh = 2π(5)2 + 2π(5)(10) = 150π in2
The Cone and the Cylinder
ExampleFind the surface area of the cylinder inscribed in a cone with height 20inches and base with radius 10 inches if the cylinder is 10 inches tall.
1020
=r
10⇒ r = 5 inches
So, a cylinder with radius 5 inches and height 10 inches has surfacearea
2πr2 + 2πrh = 2π(5)2 + 2π(5)(10) = 150π in2
The Cone and the Cylinder
ExampleFind the surface area of the cylinder inscribed in a cone with height 20inches and base with radius 10 inches if the cylinder is 10 inches tall.
1020
=r
10⇒ r = 5 inches
So, a cylinder with radius 5 inches and height 10 inches has surfacearea
2πr2 + 2πrh = 2π(5)2 + 2π(5)(10) = 150π in2
Surface Area of a Ring
ExampleSuppose you have a ring that you want to get replated. The jewelerneeds to know the surface area of the ring in order to determine howmuch to charge you. If the ring has an inner radius of 25 mm and thering 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’ssurface 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 ...
Surface Area of a Ring
ExampleSuppose you have a ring that you want to get replated. The jewelerneeds to know the surface area of the ring in order to determine howmuch to charge you. If the ring has an inner radius of 25 mm and thering 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’ssurface 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 ...
Surface Area of a Ring
ExampleSuppose you have a ring that you want to get replated. The jewelerneeds to know the surface area of the ring in order to determine howmuch to charge you. If the ring has an inner radius of 25 mm and thering 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’ssurface 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 ...
Surface Area of a Ring
ExampleSuppose you have a ring that you want to get replated. The jewelerneeds to know the surface area of the ring in order to determine howmuch to charge you. If the ring has an inner radius of 25 mm and thering 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’ssurface 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 ...
Surface Area of a Ring
ExampleSuppose you have a ring that you want to get replated. The jewelerneeds to know the surface area of the ring in order to determine howmuch to charge you. If the ring has an inner radius of 25 mm and thering 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’ssurface 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
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
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
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
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