Big Ideas Math Geometry, 2014
BI
Big Ideas Math Geometry, 2014 View details
8. Surface Areas and Volumes of Spheres
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Exercise 46 Page 654

Practice makes perfect
a
Let's look into the formula for the volume of a spherical cap. V=π h/6(3 a^2 + h^2 ) We want to find the volume of a spherical cap that we know has a base with a radius a of 4 feet. We also know that the sphere has a radius of 5 feet. To find the volume, we also need to know the spherical cap's height, h. Let's consider a diagram of a sphere and a spherical cap.
A sphere with a spherical cap and several measures marked

Notice that the segments x and h together make a radius of the sphere. h+ x= r ⇔ x= r- h Let's write this into the diagram.

A sphere with a spherical cap and several measures marked
We have a right triangle in the diagram. Let's write the measures of its sides into into the Pythagorean Theorem. a^2+b^2=c^2 ⇒ a^2+( r- h)^2= r^2 We can find h by using this equation.
a^2+(r-h)^2=r^2
4^2+( 5- h)^2= 5^2
Solve for h
16+(5-h)^2=25
(5-h)^2=9
5-h=3
- h=- 2
h= 2
Since 5-h is the side of a triangle it must be non-negative, which is why we only kept the principal root when solving the equation. Now that we know that h= 2 we can find the volume of the spherical cap. Let's do it!
V=π h/6(3a^2 +h^2 )
V=π ( 2)/6(3( 4)^2 + 2^2 )
Simplify right-hand side
V=π/3(3(4)^2 +2^2 )
V=π/3(3(16) +4 )
V=π/3(48 +4 )
V=π/3(52 )
V=π (52 )/3
V=52π/3
The volume of the spherical cap is 52π3≈ 54.45 ft^3.
b
We have a sphere with a radius of 34 cm. This sphere has a spherical cap for which we know that the radius of the base a is 30 cm. To find the volume of the spherical cap we also need to find the height of the cap, h. Let's use the formula we found in Part A to find the cap's height.
a^2+(r-h)^2=r^2
30^2+( 34- h)^2= 34^2
Solve for h
900+(34-h)^2=1156
(34-h)^2=256
34-h=16
- h= - 16
h= 16
Recall the formula for the volume of a spherical cap. V=π h/6(3 a^2 + h^2 ) Let's substitute the corresponding values and calculate the volume.
V=π h/6(3a^2 +h^2 )
V=π ( 16)/6(3( 30)^2 + 16^2 )
Simplify right-hand side
V=π (8)/3(3(30)^2 +16^2 )
V=π (8)/3(3(900) +256 )
V=π (8)/3(2700 +256 )
V=π (8)/3(2956 )
V=π (8)(2956)/3
V=π (23 648)/3
V=23 648π/3
The spherical cap has a volume of 23 648π3≈ 24 764 cm^3.
c
A sphere with a radius of 13 meters has a spherical cap with a height of 8 meters. To find the volume we need to find the radius of the base, a. We can use the expression we found in Part A. a^2+( r- h)^2= r^2Let's substitute the corresponding values and solve for a.
a^2+(r-h)^2=r^2
a^2+( 13- 8)^2= 13^2
Solve for a
a^2+5^2=13^2
a^2+25=169
a^2=144
a=sqrt(144)
a= 12
Since a is a distance it must be non-negative, which is why we only kept the principal root when solving the equation. Now we have enough information about the spherical cap to find its volume.
V=π h/6(3a^2 +h^2 )
V=π ( 8)/6(3( 12)^2 + 8^2 )
Simplify right-hand side
V=π (4)/3(3(12)^2 +8^2 )
V=π (4)/3(3(144) +64 )
V=π (4)/3(432 +64 )
V=π (4)/3(496 )
V=π (4)(496)/3
V=4(496)π/3
V=1984π/3
The spherical cap's volume is 1984π 3≈ 2077.64 m^3.
d
For a sphere with a radius of 75 inches we want to know the volume of a spherical cap with a height of 54 inches. Let's begin by finding a, the radius of the base.
a^2+(r-h)^2=r^2
a^2+( 75- 54)^2= 75^2
Solve for a
a^2+21^2=75^2
a^2+441=5625
a^2=5184
a=sqrt(5184)
a= 72
Since a must be non-negative, we only kept the principal root when solving the equation. Let's find the volume of the spherical cap.
V=π h/6(3a^2 +h^2 )
V=π ( 54)/6(3( 72)^2 + 54^2 )
Simplify right-hand side
V=π (9)(3(72)^2 +54^2 )
V=π (9)(3(5184) +2916 )
V=π (9)(15 552 +2916 )
V=π (9)(18 468 )
V=9(18 468 )π
V=166 212π
We have now found the volume of the spherical cap and it is 166 212π ≈ 522 170 inches^3.