How do you find the volume of the sphere using integration?

Answer 1

Required:

1. -Knowlege of how Integration works.
2. -Knowlege of derivatives
3. -Knowlege of Triganometry.
4. -Knowlege of Basic Geometry.

Given:

1. The volume of a sqhere and be represented in infinitely small cross sections [dx].
2. Each cross section looks like a circle [pi(r)2]
3. If you made a graph of the radius of the cross sections from one side of the sphere to the other it would look like a half cirlce [Sin(u/a) from 0 to Pi where a is the inverse of the radius of the sqhere] (radius of the sphere=1=u/a=x=hypotenuse, radius of the crossection equals=opposite=sin(x), because sin(x)=opp/hyp=opp/1=opp. the center of the sphere is located at x=pi/2)
4. Combine the above equations and integrate to find the sum of the cross sections.[integral from 0 to pi of (pi(sin(x))2dx)]=[integral from 0 to pi of (pi (1-cos(2x))/2 dx)]=[integral of (pi/2 dx)]-[integral of (pi/2)*(cos(2x))dx)]= (pi/2(x)-pi/4(sin(2x)) evlauated from 0 to pi.
5. volume of a sphere V=4/3pi(r)3=[(pi/2)(u/a)-(pi/4)(sin2(u/a)) evaluated from 0 to (a)Pi