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# Sphere, Ellipsoid, and Torus Help (page 2)

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By McGraw-Hill Professional
Updated on Oct 3, 2011

#### PROBLEM 1

Suppose a football field is to be covered by an inflatable dome that takes the shape of a half-sphere. If the radius of the dome is 100 meters, what is the volume of air enclosed by the dome in cubic meters? Find the result to the nearest 1000 cubic meters.

#### SOLUTION 1

First, find the volume V of a sphere whose radius is 100 meters, and then divide the result by 2. Let π = 3.14159. Using the formula with r = 100 gives this result:

V = (4 × 3.14159 × 100 3 )/3

= (4 × 3.14159 × 1,000,000)/3

= 4,188,786.667 . . .

Thus V /2 = 4,188,786.667/2 = 2,094,393.333. Rounding off to the nearest 1000 cubic meters, we get 2,094,000 cubic meters as the volume of air enclosed by the dome.

#### PROBLEM 2

Suppose the dome in the previous example is not a half-sphere, but instead is a half-ellipsoid. Imagine that the height of the ellipsoid is 70 meters above its center point, which lies in the middle of the 50-yard line at field level. Suppose that the distance from the center of the 50-yard line to either end of the dome, as measured parallel to the sidelines, is 120 meters, and the distance from the center of the 50-yard line, as measured along the line containing the 50-yard line itself, is 90 meters. What is the volume of air, to the nearest 1000 cubic meters, enclosed by this dome?

#### SOLUTION 2

First, consider the radii r 1 , r 2 , and r 3 in meters, with respect to the center point, as follows:

r 1 = 120

r 2 = 90

r 3 = 70

Then use the formula for the volume V of an ellipsoid with these radii:

V = (4 × 3.14159 × 120 × 90 × 70)/3

= (4 × 3.14159 × 756,000)/3

= 3,166,722.72

Thus V /2 = 3,166,722.72/2 = 1,583,361.36. Rounding off to the nearest 1000 cubic meters, we get 1,583,000 cubic meters as the volume of air enclosed by the half-ellipsoidal dome.

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

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