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Verified Answer :

Edufirst

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:

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 \(0\le t\le \pi\).

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