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In geometry, the napkin-ring problem involves finding the volume of a "band" of specified height around a sphere, i.e. the part that remains after a hole in the shape of a circular cylinder is drilled through the center of the sphere.
The theory was made rigorous a few decades later by Eudoxus of Cnidus, who used it to calculate areas and volumes. It was later reinvented in China by Liu Hui in the 3rd century AD in order to find the area of a circle. [2] The first use of the term was in 1647 by Gregory of Saint Vincent in Opus geometricum quadraturae circuli et sectionum.
The area of the base of a cylinder is the area of a circle (in this case we define that the circle has a radius with measure ): B = π r 2 {\displaystyle B=\pi r^{2}} . To calculate the total area of a right circular cylinder, you simply add the lateral area to the area of the two bases:
In computational geometry, the smallest enclosing box problem is that of finding the oriented minimum bounding box enclosing a set of points. It is a type of bounding volume. "Smallest" may refer to volume, area, perimeter, etc. of the box. It is sufficient to find the smallest enclosing box for the convex hull of the objects in question. It is ...
The ratio of the volume of a sphere to the volume of its circumscribed cylinder is 2:3, as was determined by Archimedes. The principal formulae derived in On the Sphere and Cylinder are those mentioned above: the surface area of the sphere, the volume of the contained ball, and surface area and volume of the cylinder.
Two common methods for finding the volume of a solid of revolution are the disc method and the shell method of integration.To apply these methods, it is easiest to draw the graph in question; identify the area that is to be revolved about the axis of revolution; determine the volume of either a disc-shaped slice of the solid, with thickness δx, or a cylindrical shell of width δx; and then ...
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Much more work is needed to find the volume if we use disc integration. First, we would need to solve y = 8 ( x − 1 ) 2 ( x − 2 ) 2 {\displaystyle y=8(x-1)^{2}(x-2)^{2}} for x . Next, because the volume is hollow in the middle, we would need two functions: one that defined an outer solid and one that defined the inner hollow.