Equation of a Sphere
Key Questions

Answer:
#(x, y, z) = (rho cos theta sin phi, rho sin theta sin phi, rho cos phi)# Explanation:
One common form of parametric equation of a sphere is:
#(x, y, z) = (rho cos theta sin phi, rho sin theta sin phi, rho cos phi)# where
#rho# is the constant radius,#theta in [0, 2pi)# is the longitude and#phi in [0, pi]# is the colatitude.Since the surface of a sphere is two dimensional, parametric equations usually have two variables (in this case
#theta# and#phi# ).#color(white)()#
FootnoteIf you are determined to have a parametric equation with just one variable
#t# (say), then it is possible. For example, you can construct a surjection from the interval#[0, 1]# onto the rectangle#[0, 2pi] xx [0, pi]# and hence onto the surface of the sphere.Such a surjection can even be made continuous. I'll see if I can put together a simple short formulation.

An equation of the sphere with radius
#R# centered at the origin is#x^2+y^2+z^2=R^2# .Since
#x^2+y^2=r^2# in cylindrical coordinates, an equation of the same sphere in cylindrical coordinates can be written as#r^2+z^2=R^2# .
I hope that this was helpful.

The answer is:
#x^2+y^2+z^2+ax+by+cz+d=0# ,This is because the sphere is the locus of all
points#P(x,y,z)# in the space whose distance from#C(x_c,y_c,z_c)# is equal to r.So we can use the formula of distance from
#P# to#C# , that says:#sqrt((xx_c)^2+(yy_c)^2+(zz_c)^2)=r# and so:#(xx_c)^2+(yy_c)^2+(zz_c)^2=r^2# ,#x^2+2(x)(x_c) + x_c^2+y^2+2(y)(y_c)+y_c^2+z^2+2(z)(z_c)+z_c^2=r^2# ,#x^2+y^2+z^2+ax+by+cz+d=0# ,in which
#a=2x_c# ;
#b=2y_c# ;
#c=2z_c# ;
#d=x_c^2+y_c^2+z_c^2r^2# ;So:
#C(a/2,b/2,c/2)# and
#r# , if it exists, is:#r=sqrt(x_c^2+y_c^2+z_c^2d)# .If the center is in the Origin, than the equation is:
#x^2+y^2+z^2=r^2# , 
Answer:
#v = 4/3 pir^3# Explanation:
The surface area of a sphere of radius
#r# is#4pi r^2# Imagine dissecting a sphere into a large number of slender pyramids, with apices at the centre and (slightly rounded) bases tesselating the surface. As you use more pyramids, the bases get flatter.
The volume of each pyramid is
#1/3 xx "base" xx "height"# So the total volume of all the pyramids is:
#v = sum 1/3 xx "base" xx "height" = r/3 sum "base" = r/3 * 4pir^2 = 4/3 pir^3#