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An approximation for the volume of a thin spherical shell is the surface area of the inner sphere multiplied by the thickness t of the shell: [2] V ≈ 4 π r 2 t , {\displaystyle V\approx 4\pi r^{2}t,}
Propositions 70 and 71 consider the force acting on a particle from a hollow sphere with an infinitesimally thin surface, whose mass density is constant over the surface. The force on the particle from a small area of the surface of the sphere is proportional to the mass of the area and inversely as the square of its distance from the particle.
A sphere (from Greek σφαῖρα, sphaîra) [1] is a geometrical object that is a three-dimensional analogue to a two-dimensional circle. Formally, a sphere is the set of points that are all at the same distance r from a given point in three-dimensional space. [2] That given point is the center of the sphere, and r is the sphere's radius.
For example, assuming the Earth is a sphere of radius 6371 km, the surface area of the arctic (north of the Arctic Circle, at latitude 66.56° as of August 2016 [7]) is 2π ⋅ 6371 2 | sin 90° − sin 66.56° | = 21.04 million km 2 (8.12 million sq mi), or 0.5 ⋅ | sin 90° − sin 66.56° | = 4.125% of the total surface area of the Earth.
Intersection of a sphere and cone emanating from its center. A spherical sector (blue) A spherical sector. In geometry, a spherical sector, [1] also known as a spherical cone, [2] is a portion of a sphere or of a ball defined by a conical boundary with apex at the center of the sphere. It can be described as the union of a spherical cap and the ...
As can be seen, the area of the circle defined by the intersection with the sphere of a horizontal plane located at any height equals the area of the intersection of that plane with the part of the cylinder that is "outside" of the cone; thus, applying Cavalieri's principle, it could be said that the volume of the half sphere equals the volume ...
Much more work is needed to find the volume if we use disc integration. First, we would need to solve y = ( x − 1 ) 2 ( x − 2 ) 2 {\displaystyle y=(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.
For a larger sphere, the band will be thinner but longer. 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. It is a counterintuitive fact that this volume does ...