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 24 Page 653

Use the formula for volume of a sphere to find the volume of the hemisphere.

288 π ≈ 904.78 ft^3

Practice makes perfect

We want to find the volume of a composite solid in the shape of a cone with a hemisphere attached to it.

Cone with a hemisphere attached to it.

We will first find the volume of the cone and then the volume of the hemisphere. Then we can add the results to find the volume of the solid. Let's do it!

Volume of the Cone

Let's begin by finding the volume of the cone.

Cone with a hemisphere attached to it. The cone is highlighted.
The volume of a cone is one third the product of the area of the base and the height. V=1/3 Bh Our cone has a circular base, B, and its area is the product of π times the radius squared. V=1/3 Bh ⇒ V = 1/3 π r^2 h Let's substitute the corresponding values into the formula and find its volume.
V = 1/3 π r^2 h
V = 1/3 π ( 6)^2 ( 12)
Simplify right-hand side
V = 1/3 π (36) (12)
V = 1/3 (36) (12)π
V = 36 (12)/3 π
V = 144π
The cone has a volume of 144π ft^3.

Volume of the Hemisphere

The top section of the solid is in the shape of a hemisphere.

Cone with a hemisphere attached to it. The hemisphere is highlighted.
The volume of a hemisphere is half the volume of a sphere. V=43π r^3/2 ⇒ V = 2/3π r^3 We can now find the volume our hemisphere, which has a radius of 6 feet.
V = 2/3π r^3
V = 2/3π ( 6)^3
Simplify right-hand side
V = 2/3π (216)
V = 2/3(216)π
V = 2(216)/3π
V= 144π
The hemisphere has a volume of 144πft^3.

Volume of the Solid

To find the volume of the composite solid we will use the Volume Addition Postulate.

The volume of a solid is the sum of the volumes of all its non-overlapping parts.

The solid has two non-overlapping parts — one cone and one hemisphere. Let's add their volumes. V_(solid)= V_(cone)+ V_(hemisphere) ⇓ V_(solid)= 144π+ 144π = 288 π The solid has the volume 288 π ≈ 904.78 ft^3.