四维球的表面积、体积求解.pdf

The Volume and Surface Area of a 4D Sphere Feixiang XU Aug 30 2016 Abstract In this paper I m trying to calculate the volume of spheres from one dimension to four dimension using basic integral calculation step by step Then I derive the surface area of 4D sphere from the volume 1The Volume Calculation 1 11D case Here we consider a line segment with length 2R Take its midpoint as center 1 22D case In 2 dimensional case a sphere is a circle Now I try to expand the segment into a circle First we construct the per pendicular line of the line segment through its midpoint Then we move the segment along the perpendicular line upward At the same time we shrink the segment to keep the distance between the segment s endpoint and the center un changed which is R Obviously when moving a distance R the segment shrinks into a point The area which the shrinking segment has been scanning becomes a half of a 2 dimensional sphere namely a semi circle Connect the moving segment s endpoint and the center into a segment assume the angle between the new segment and the initial segment is varies from 0 to 2 Then we can calculate the circle s area which called the volume of the 2D sphere 1 Figure 1 In the process of moving the length of segment is 2Rcos consider that it moves a little distance d Rsin so the area it scans is 2Rcos d Rsin Thus the semi circle s area is Z 2 0 2Rcos d Rsin R2 2 So the circle s area is R2 which is the volume of the 2D sphere 1 33D case In a similar way we can expand the circle into a sphere First construct the perpendicular line of the circle through its center Then move the circle along the perpendicular line upward and shrink the circle to keep the distance between the circle s edge and the center unchanged which is R as well The space which the shrinking circle has been scanning becomes a half of a 3 dimensional sphere Similarly the radius of the moving circle is 2Rcos consider that it moves a little distance d Rsin so the space it scans is Rcos 2 d Rsin Thus the hemisphere s volume is Z 2 0 Rcos 2d Rsin 2 3 R 3 So the sphere s volume is 4 3 R 3 2 1 44D case From the derivation above we can easily derive that the volume of the 4D hemisphere is Z 2 0 4 3 Rcos 3d Rsin 1 4 2R4 So the 4D sphere s volume is 1 2 2R4 2The Surface Area Calculation 2 1Low dimensions case In the two dimension the 2D sphere s volume the circle s area is R2 the surface area the circle s perimeter is 2 R We can easily fi nd out that the surface area is the derivative of the volume relative to R d R2 dR 2 R In the three dimension the sphere s volume is 4 3 R 3 the surface area is 4 R2 We can easily fi nd out that the surface area is the derivative of the volume relative to R as well d 4 3 R 3 dR 4 R2 By such analogy in the four dimension the sphere s volume is 1 2 2R4 then the surface area should be d 1 2 2R4 dR 2 2R3 Here is the proof below 2 2Proof 2 2 12D Case Now we have a circle with radius R increase the radius by R then we get an annulus The area of annulus is R R 2 R2 when R is very small we can consider the annulus as a rectangle with width R Then the length of the rectangle is R R 2 R2 R When R 0 the length is the perimeter of the circle with radius R which is lim R 0 R R 2 R2 R 2 R So the perimeter of the circle with radius R is 2 R 3 2 2 23D Case Now we have a sphere with radius R increase the radius by R then we get a spherical shell The volume of spherical shell is 4 3 R R 3 4 3 R 3 when R is very small we can consider the annulus as a cuboid with height R Then the base area of the cuboid is 4 3 R R 3 4 3 R 3 R When R 0 the base area is the surface area of the sphere with radius R which is lim R 0 4 3 R R 3 4 3 R 3 R 4 R2 So the surface area of the sphere with radius R is 4 R2 2 2 34D Case Similarly we can prove that the surface area of a 4D sphere with radius R is 2 2R3 3Conclusion Sphere with radius RVolumeSurface Area 1D 2D R22 R 3D 4 3 R 3 4 R2