How can we derive expressions A. 6-31, A.6-32, A.6-33? Can you show me a step by step solution? Thanks We now tabulate for reference the same kind of information for spherical coordinates r,0, p. These coordinates are shown in Figure A.6-1b. They are related to the Cartesian co-ordinates byx = r sin 0 cos oy = r sin 0 sin ø(А.6-19)(A.6-20)(A.6-21)+Vx² + y? + z?0 = arctan(Vx² + y²/z)$ = arctan(y/x)(A.6-22)(A.6-23)(A.6-24)r =z = r cos 0For the spherical coordinates we have the following relations for the derivativeoperators:Cos 8 cos p+sin o(sin 0 cos 4)+(A.6-25)dxr sin 0) docos e sin oCos o(sin 0 sin ø)dydrr sin 0) do(A.6-26)sin 0(cos 0)++ (0)(A.6-27)dzarThe relations between the unit vectors ared, = ( sin 0 cos 4)ô, + (sin 0 sin ø)ô, + ( cos 0)8,de = ( cos 0 cos 4)d; + ( cos 0 sin )ô, + (- sin 0)d,ds = (– sin ø)ô, + ( cos 4)ô, + (0)8,(A.6-28)(A.6-29)(A.6-30)and&; = (sin 0 cos 4)ô, + ( cos 0 cos 4)d, + (- sin ø)dô, = ( sin 0 sin 4)ô, + ( cos 0 sin þ)ô, + (cos 4)dsd, = ( cos 0)ô, + (– sin 0)d, + (0)8s(A.6-31)(A.6-32)(А.6-33)And, finally, some sample operations in spherical coordinates are(0:7) = 0„T + OgTer + Or&T¢r++or' ro+(A.6-34)PP , ÞPu,Ug(u • [v X w]) =v,Vo(A.6-35)

We need to derive expressions A. 6-31, A.6-32, and A.6-33 by considering that A. 6-28, A.6-29, and…

The relations between the unit vectors are ô, = ( sin 0 cos 4)ô, + ( sin 0 sin ø)ô, + ( cos 0)8, do = ( cos 0 cos 4)ô̟ + ( cos 0 sin ø)ồ, + (- sin 0)d, | d4 = (- sin ø)ô,; + ( cos 4)ô, + (0)8, (A.6-28) (A.6-29) (A.6-30) %3D and 8x = ( sin 0 cos $)ô, + ( cos 0 cos 4)d, + (- sin 4)ôs 8, = ( sin 0 sin 4)ô, + (cos 0 sin 4)ô, + ( cos p)& 8, = ( cos 0)8, +(- sin 0)ô, + (0)8, (A.6-31) (A.6-32) (A.6-33) %3D