Sphere, Ellipsoid, and Torus Help (page 2)
Introduction to the Sphere, Ellipsoid, and Torus (Donut)
There exists an incredible variety of geometric solids that have curved surfaces throughout. Here, we’ll look at three of the most common: the sphere , the ellipsoid , and the torus .
Surface Area and Volume of a Sphere
Consider a specific point P in 3D space. The surface of a sphere S consists of the set of all points at a specific distance or radius r from point P . The interior of sphere S , including the surface, consists of the set of all points whose distance from point P is less than or equal to r . The interior of sphere S , not including the surface, consists of the set of all points whose distance from P is less than r .
Surface Area Of Sphere
Imagine a sphere S having radius r as shown in Fig. 8-11. The surface area, A , of the sphere is given by:
A = 4π r 2
Volume Of Sphere
Imagine a sphere S as defined above and in Fig. 8-11. The volume, V , of the solid enclosed by the sphere is given by:
V = 4π r 3 /3
This volume applies to the interior of sphere S , either including the surface or not including it, because the surface has zero volume.
Surface Area and Volume of an Ellipsoid
Let E be a set of points that forms a closed surface. Then E is an ellipsoid if and only if, for any plane X that intersects E, the intersection between E and X is either a single point, a circle, or an ellipse.
Figure 8-12 shows an ellipsoid E with center point P and radii r 1 , r 2 , and r 3 as specified in a 3D rectangular coordinate system with P at the origin. If r 1 , r 2 , and r 3 are all equal, then E is a sphere, which is a special case of the ellipsoid.
Volume Of Ellipsoid
Imagine an ellipsoid whose semi-axes are r 1 , r 2 , and r 3 (Fig. 8-12). The volume, V , of the enclosed solid is given by:
V = 4π r 1 r 2 r 3 /3
Surface Area and Volume of a Torus (Donut)
Imagine a ray PQ , and a small circle C centered on point Q whose radius is less than half of the distance between points P and Q . Suppose ray PQ , along with the small circle C centered at point Q , is rotated around its end point, P , so that point Q describes a circle that lies in a plane perpendicular to the small circle C . The resulting set of points in 3D space, “traced out” by circle C , is a torus.
Figure 8-13 shows a torus T thus constructed, with center point P . The inside radius is r 1 and the outside radius is r 2 . The torus is sometimes informally called a “donut.”
Surface Area Of Torus
Imagine a torus with an inner radius of r 1 and an outer radius of r 2 as shown in Fig. 8-13. The surface area, A , of the torus is given by:
A = π 2 ( r 2 + r 1 )( r 2 − r 1 )
Volume Of Torus
Let T be a torus as defined above and in Fig. 8-13. The volume, V , of the enclosed solid is given by:
V = π 2 ( r 2 + r 1 )( r 2 − r 1 ) 2 /4
Sphere, Ellipsoid, and Torus Practice Problems
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.
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.
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?
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
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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