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Sphere and Ellipsoid Practice Problems

— McGraw-Hill Professional
Updated on Jun 17, 2014

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Sphere and Ellipsoid Practice Problems

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.

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