Question

Find parametric equations for the sphere centered at the origin and with radius 3. Use the parameters s and t in your answer.

Answer 1

radius, r = 3

The equation of a sphere entered at the origin in cartesian coordinates is

x^2 + y^2 + z^2 = r^2 

That in spherical coordinates is:

x = rcos(theta)*sin(phi)
y= r sin(theta)*sin(phi)
z = rcos(phi)

where you can make u = r cos(phi) to obtain the parametrical equations

x = √[r^2 - u^2] cos(theta)
y = √[r^2 - u^2] sin (theta)
z = u

where theta goes from 0 to 2π and u goes from -r to r.

In our case r = 3, so the parametrical equations are:

Answer:
x = √[9 - u^2] cos(theta)
y = √[9 - u^2] sin (theta)
z = u

Answer 2

Answer:

The parametric equation are

$x=3\cos (s)\sin (t)$

$y=3\sin (s)\cos (t)$

$z=3\cos (t)$

where, $0\le s\le 2\pi$ and $0\le t\le \pi$.

Step-by-step explanation:

The general equation of a sphere is

$(x-h)^2+(y-k)^2+(z-l)^2=r^2$

where, (h,k,l) is center of the sphere and r is radius.

It is given that the sphere centered at the origin and with radius 3. It means h=0, k=0, l=0, r=3.

$(x-0)^2+(y-0)^2+(z-0)^2=3^2$

$x^2+y^2+z^2=9$

The parametric equation of a sphere are

$x=h+r\cos (s)\sin (t)$

$y=k+r\sin (s)\cos (t)$

$z=l+r\cos (t)$

where, $0\le s\le 2\pi$ and $0\le t\le \pi$.

Substitute h=0, k=0, l=0, r=3 in the above equations. So, the parametric equation are

$x=0+3\cos (s)\sin (t)=3\cos (s)\sin (t)$

$y=0+3\sin (s)\cos (t)=3\sin (s)\cos (t)$

$z=0+3\cos (t)=3\cos (t)$

where, $0\le s\le 2\pi$ and $0\le t\le \pi$.