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Cones and Cylinders Help (page 2)

By — McGraw-Hill Professional
Updated on Oct 3, 2011

Volume Of Frustum Of Right Circular Cone

Imagine a frustum of a right circular cone as defined above and in Fig. 8-7. The volume, V , of the interior of the object is given by this formula:

Surface Area and Volume Cones and Cylinders Volume Of Frustum Of Right Circular Cone

The Slant Circular Cone

A slant circular cone has a base that is a circle, and an apex point such that a normal line from the apex point to the plane of the base does not pass through the center of the base (Fig. 8-8).

Surface Area and Volume Cones and Cylinders The Slant Circular Cone

 

Fig. 8-8 . A slant circular cone.

Volume Of Slant Circular Cone

Imagine a cone whose base is a circle. Let P be the apex of the cone, and let Q be a point in the plane X containing the base such that line segment PQ is perpendicular to X (Fig. 8-8). Let h be the height of the cone (the length of line segment PQ ). Let r be the radius of the base. Then the volume, V , of the corresponding cone is given by:

V = π r 2 h /3

This is the same as the formula for the volume of a right circular cone.

The Right Circular Cylinder

A right circular cylinder has a circular base and a circular top. The base and the top lie in parallel planes. The center of the base and the center of the top lie along a line that is normal to both the plane containing the base and the plane containing the top (Fig. 8-9).

Surface Area and Volume Cones and Cylinders The Right Circular Cylinder

 

Fig. 8-9 . A right circular cylinder.

Surface Area Of Right Circular Cylinder

Imagine a right circular cylinder where P is the center of the top and Q is the center of the base (Fig. 8-9). Let r be the radius of the cylinder, and let h be the height (the length of line segment PQ ). Then the surface area S 1 of the cylinder, including the base and the top, is given by:

S 1 = 2π rh + 2π r 2 = 2π r ( h + r )

The lateral surface area S 2 (not including the base or the top) is given by:

S 2 = 2π rh

Volume Of Right Circular Cylinder

Imagine a right circular cylinder as defined above and shown in Fig. 8-9. The volume, V , of the cylinder is given by:

V = π r 2 h

The Slant Circular Cylinder

A slant circular cylinder has a circular base and a circular top. The base and the top lie in parallel planes. The center of the base and the center of the top lie along a line that is not perpendicular to the planes that contain them (Fig. 8-10).

Surface Area and Volume Cones and Cylinders The Slant Circular Cylinder

 

Fig. 8-10 . A slant circular cylinder.

Volume Of Slant Circular Cylinder

Imagine a slant circular cylinder as defined above and in Fig. 8-10. The volume, V , of the corresponding solid is given by:

V = π r 2 h

Cones and Cylinders Practice Problems

PROBLEM 1

A cylindrical water tower is exactly 30 meters high and exactly 10 meters in radius. How many liters of water can it hold, assuming the entire interior can be filled with water? (One liter is equal to a cubic decimeter, or the volume of a cube measuring 0.1 meters on an edge.) Round the answer off to the nearest liter.

SOLUTION 1

Use the formula for the volume of a right circular cylinder to find the volume in cubic meters:

V = π r 2 h

Plugging in the numbers, let r = 10, h = 30, and π = 3.14159:

V = 3.14159 × 10 2 × 30

= 3.14159 × 100 × 30

= 9424.77

One liter is the volume of a cube measuring 10 centimeters, or 0.1 meter, on an edge (believe it or not!). Thus, there are 1000 liters in a cubic meter. This means that the amount of water the tower can hold, in liters, is equal to 9424.77 × 1000, or 9,424,770.

PROBLEM 2

A circus tent is shaped like a right circular cone. Its diameter is 50 meters and the height at the center is 20 meters. How much canvas is in the tent? Express the answer to the nearest square meter.

SOLUTION 2

Use the formula for the lateral surface area, S , of the right circular cone:

S = π r ( r 2 + h 2 ) 1/2

We know that the diameter is 50 meters, so the radius is 25 meters. Therefore, r = 25. We also know that h = 20. Let π = 3.14159. Then:

S = 3.14159 × 25 × (25 2 + 20 2 ) 1/2

= 3.14159 × 25 × (625 + 400) 1/2

= 3.14159 × 25 × 1025 1/2

= 3.14159 × 25 × 32.0156

= 2514.4972201

There are 2514 square meters of canvas, rounded off to the nearest square meter.

 Practice problems for these concepts can be found at:  Surface Area And Volume Practice Test.

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