Given point&nbsp;P(−2,6,3)&nbsp;and vector&nbsp;B=yax+(x+z)ay, Describe P and B in spherical and cylindrical coordinates. In the Cartesian, cylindrical, and spherical systems, evaluate A at P.

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Given point&amp;nbsp;P(−2,6,3)&amp;nbsp;and vector&amp;nbsp;B=yax+(x+z)ay, Describe P and B in spherical and cylindrical coordinates. In the Cartesian, cylindrical, and spherical systems, evaluate A at P.

mhalmantus · Accepted Answer

Step 1  The objective is to express the point P−2,6,3 and vector B=yax+x+zay in cylindrical and spherical coordinates.  Step 2  To convert Cartesian coordinates x,y,z to cylindrical coordinates r,θ,z  r=x2+y2θ=tan−1yxz=z  The point P-2,6,3 in cylindrical coordinates is,  r=4+36=40=210  And θ=tan−16−2=−1.25  Cylindrical coordinates is 210,−1.25,3  To convert Cartesian coordinates  x,y,z to spherical coordinates ρθϕ  ρ=x2+y2+z2θ=tan−1yxϕ=tan−1x2+y2z  The point P−2,6,3 in spherical coordinates is,  ρ=x2+y2+z2=4+36+9=49=7  And θ=tan−16−2=−1.25  And ϕ=tan−14+363=tan−1403=1.3034  Spherical coordinates is 7,−1.25,1.3034  Step 3  The vector B=yax+x+zay in cylindrical and spherical coordinates.  In the cartesian system B at P is  B=6ax+ay  For vector B,  Bx=yBy=x+zBz=0  In cylindrical system  Ar=ycos⁡θ+x+zsin⁡θAθ=−ysin⁡θ+x+zcos⁡θAz=0  From step 2 we have substituting values of θ, we get  A=−0.9487ar−6.008aθ  Similarly in spherical system  Aρ=ysin⁡θcos⁡ϕ+x+zsin⁡θsin⁡ϕAθ=ycos⁡θcos⁡ϕ+x+zcos⁡θsin⁡ϕAϕ=−ysin⁡ϕ+(x+z)cos⁡ϕ  Since,  x=ρsin⁡θcos⁡ϕy=ρsin⁡θsin⁡ϕz=ρcos⁡θ  From step 2, substitute ρ,θ,ϕ  we get,  A=−0.8571aρ−0.4066aθ−6.008aϕ

Given point (-2,6,3) and vector B=ya_{x}+(x+z)a_{y}, Express P and B in cylindrical and spherical coordinates.

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