This point is in cylindrical form #(r, theta, z)#. So let's first convert it to rectangular form #(x,y,z)# by using the formulas #x=rcos theta, y=r sin theta, z=z#
That is,
#x=1*cos(pi/4)=1/sqrt2, y=1*sin(pi/4)=1/sqrt2, z=2#
hence, the point is #(1/sqrt2, 1/sqrt2, 2)#.
Now let's use the formulas
#rho^2 = x^2 + y^2 +z^2, x=rho sin phi cos theta, y=rho sin phi sin theta, z=rho cos phi# to change the point to spherical coordinates.
#rho = sqrt ((1/sqrt2)^2 +(1/sqrt2)^2+(2)^2) = sqrt (1/2 + 1/2 + 4) = sqrt5 #
To find #phi# let's use the formula #z=rho cos phi#
Therefore,
#z=rho cos phi#
#2=sqrt 5 cos phi#
#cos^-1(2/sqrt 5)=phi#
#phi~~0.46#
#:. (rho, theta, phi)~~(sqrt5, pi/4,0.46) ~~(2.24, 0.79,0.46)#
